Free math worksheets

- What does “right triangle” tell you about \(\gamma\)? - What is the sum of the interior angles of a triangle? - Which angle measures are already known?

1. Since the triangle is a right triangle, \(\gamma = 90^\circ\). 2. The interior angles of a triangle total \(180^\circ\). 3. Therefore, \(\beta = 180^\circ - 90^\circ - 37.5^\circ = 52.5^\circ\).

\(\beta = 52.5^\circ\) and \(\gamma = 90^\circ\).

- Add the two known interior angles. - Subtract their sum from \(180^\circ\).

1. The interior angles of a triangle total \(180^\circ\). 2. Therefore, \(\beta=180^\circ-48^\circ-74^\circ=58^\circ\).

\(\beta=58^\circ\)

- What do the two acute angles of a right triangle add to? - Subtract the given acute angle from that total.

1. The angle at \(C\) is \(90^\circ\). 2. Therefore, \(\alpha=180^\circ-90^\circ-62^\circ=28^\circ\).

\(\alpha=28^\circ\)

- What total measure do the two angles in a linear pair have? - Use that total with the given angle to find the unknown.

1. A linear pair is supplementary, so \(\beta+\gamma=180^\circ\). 2. Substitute \(\gamma=112^\circ\): \(\beta+112^\circ=180^\circ\). 3. Therefore, \(\beta=68^\circ\).

\(\beta=68^\circ\). The two angles are supplementary.

- The two smaller angles are adjacent parts of one larger angle. - Add the two measures.

1. Adjacent angle measures add to the whole angle. 2. \(\gamma=25^\circ+45^\circ=70^\circ\).

\(\gamma=70^\circ\)

- How many equal parts make up the whole angle? - Divide the total by that number.

1. Divide the total measure equally among three parts. 2. \(114^\circ\div3=38^\circ\).

Each smaller angle measures \(38^\circ\).

- What is the sum of the three interior angles of a triangle? - Subtract the two known angle measures from that total.

1. The interior angles of a triangle sum to \(180^\circ\). 2. Therefore, \(\gamma=180^\circ-55^\circ-65^\circ=60^\circ\).

\(\gamma=60^\circ\)

- In which smaller triangle is the unknown angle located? - Which two angle measures are known in that triangle?

1. Consider triangle \(APC\). 2. Its two known angles measure \(30^\circ\) and \(20^\circ\). 3. Use the triangle angle sum: \(\angle APC=180^\circ-30^\circ-20^\circ=130^\circ\).

\(\angle APC=130^\circ\)

- What is the total measure of the three interior angles of a triangle? - Use the two labeled interior angles to determine how much of that total remains for \(\gamma\).

1. The interior angles of a triangle sum to \(180^\circ\). 2. Therefore, \(\gamma=180^\circ-33^\circ-112^\circ=35^\circ\).

\(\gamma=35^\circ\)

- What total must the two acute angles of a right triangle have? - Subtract the known acute angle from that total.

1. The two acute angles of a right triangle sum to \(90^\circ\). 2. Therefore, \(\beta=90^\circ-25^\circ=65^\circ\).

\(\beta=65^\circ\)

- What is the sum of the three interior angles of a triangle? - How many degrees remain after subtracting the known angle? - What measurements define acute, right, and obtuse angles?

1. The interior angles of a triangle have a sum of \(180^\circ\). 2. Since one angle measures \(110^\circ\), the other two angles have a sum of \(180^\circ - 110^\circ = 70^\circ\). 3. Each angle in a triangle must be greater than \(0^\circ\), so each of the two remaining angles must be less than \(70^\circ\). 4. Any angle less than \(90^\circ\) is acute. Therefore, both remaining angles must be acute.

Both remaining angles must be acute. Together they measure \(70^\circ\), and each one must be greater than \(0^\circ\), so each is less than \(70^\circ\) and therefore acute.

- What is the sum of the interior angles of a quadrilateral? - Add the given angle measures in each part. - Compare the total in part b with the required quadrilateral angle sum.

1. The interior angles of any quadrilateral total \(360^\circ\). 2. For part a, \(\delta = 360^\circ - (85^\circ + 110^\circ + 45^\circ) = 360^\circ - 240^\circ = 120^\circ\). 3. For part b, the claimed angles total \(100^\circ + 120^\circ + 80^\circ + 70^\circ = 370^\circ\). 4. Since \(370^\circ \ne 360^\circ\), those angles cannot form a quadrilateral.

a) \(\delta = 120^\circ\) b) No. The four angles total \(370^\circ\), not \(360^\circ\).

- What is the sum of the interior angles of a triangle? - What is true about the base angles of an isosceles triangle? - When is a triangle classified as obtuse? - Pay attention to whether the given angle is a vertex angle or a base angle.

1. In the first triangle, the two base angles total \(180^\circ - 40^\circ = 140^\circ\). 2. The base angles of an isosceles triangle are congruent, so each measures \(140^\circ \div 2 = 70^\circ\). The angles are \(40^\circ\), \(70^\circ\), and \(70^\circ\). 3. In the second triangle, both base angles measure \(40^\circ\). The vertex angle is \(180^\circ - (40^\circ + 40^\circ) = 100^\circ\). 4. The second triangle is obtuse because it has an angle greater than \(90^\circ\).

The first triangle has missing angles of \(70^\circ\) and \(70^\circ\). The second triangle has missing angles of \(40^\circ\) and \(100^\circ\). The second triangle is obtuse.

- What is the sum of the interior angles of a triangle? - How can angle measures show that a triangle is isosceles? - What angle measure makes a triangle a right triangle?

1. For part a, \(\gamma = 180^\circ - (42^\circ + 96^\circ) = 42^\circ\). Since \(\alpha = \gamma\), the triangle is isosceles. 2. For part b, \(\alpha = 180^\circ - (25^\circ + 130^\circ) = 25^\circ\). Since \(\alpha = \beta\), the triangle is isosceles. 3. For part c, \(\beta = 180^\circ - (45^\circ + 90^\circ) = 45^\circ\). Since \(\alpha = \beta\), the triangle is isosceles. Since \(\gamma = 90^\circ\), it is also a right triangle.

a) \(\gamma = 42^\circ\); isosceles b) \(\alpha = 25^\circ\); isosceles c) \(\beta = 45^\circ\); isosceles and right

- What is the sum of the interior angles of every triangle? - What is true about the base angles of an isosceles triangle? - What are the angle measures in an equilateral triangle? - A quick sketch may help you test each statement.

1. For a), two right angles have a sum of \(90^\circ + 90^\circ = 180^\circ\). That would leave \(0^\circ\) for the third angle, so the statement is false. 2. For b), if the vertex angle is \(120^\circ\), the two congruent base angles have a sum of \(180^\circ - 120^\circ = 60^\circ\). Each base angle is \(60^\circ \div 2 = 30^\circ\), so such a triangle exists. The statement is true. 3. For c), the three angles of an equilateral triangle are congruent. Each angle measures \(180^\circ \div 3 = 60^\circ\). Since all three angles are less than \(90^\circ\), the triangle is acute. The statement is true.

a) False b) True c) True

- What is the total turn after one complete trip around a convex polygon? - How are an exterior angle and its adjacent interior angle related? - Subtract the known turns from the full turn.

1. The exterior turns of any convex polygon total \(360^\circ\). 2. The fourth exterior angle is \(360^\circ - (85^\circ + 105^\circ + 90^\circ) = 80^\circ\). 3. An interior angle and its adjacent exterior angle are supplementary. 4. Therefore, the fourth interior angle is \(180^\circ - 80^\circ = 100^\circ\).

a) \(80^\circ\) b) \(100^\circ\)

- The two base angles of an isosceles triangle are congruent. - The three interior angles of a triangle sum to \(180^\circ\). - Express the vertex angle using one base-angle variable.

1. Let \(b\) be the measure of each base angle. 2. The vertex angle is \(b+15^\circ\). 3. The interior angles of a triangle sum to \(180^\circ\), so \(2b+(b+15^\circ)=180^\circ\). 4. Simplify: \(3b+15^\circ=180^\circ\), so \(b=55^\circ\). 5. The vertex angle is \(55^\circ+15^\circ=70^\circ\).

The two base angles are \(55^\circ\) each, and the vertex angle is \(70^\circ\).

- The two acute angles in a right triangle are complementary. - Express one acute angle in terms of the other. - Substitute that expression into the \(90^\circ\) angle sum.

1. The acute angles of a right triangle sum to \(90^\circ\). 2. Since \(\alpha=3\beta+12^\circ\), write \((3\beta+12^\circ)+\beta=90^\circ\). 3. Simplify: \(4\beta+12^\circ=90^\circ\), so \(\beta=19.5^\circ\). 4. Then \(\alpha=90^\circ-19.5^\circ=70.5^\circ\).

\(\alpha=70.5^\circ\) and \(\beta=19.5^\circ\).

- Subtract the given angle from \(360^\circ\). - A reflex angle is greater than \(180^\circ\) and less than \(360^\circ\).

1. A full circle measures \(360^\circ\). 2. The remaining angle is \(\alpha=360^\circ-75^\circ=285^\circ\). 3. Because \(180^\circ<285^\circ<360^\circ\), \(\alpha\) is a reflex angle.

\(\alpha=285^\circ\), a reflex angle

- A straight angle measures \(180^\circ\). - Add the measured angle to the straight angle.

1. A straight angle measures \(180^\circ\). 2. The reflex angle is the straight angle plus the measured acute angle: \(180^\circ+42^\circ=222^\circ\).

\(222^\circ\)

- Compare each obtuse angle with \(90^\circ\). - Decide how the total of four angles greater than \(90^\circ\) compares with \(360^\circ\).

1. If all four angles were obtuse, each would be greater than \(90^\circ\). 2. Their sum would therefore be greater than \(4 \cdot 90^\circ=360^\circ\). 3. This contradicts the quadrilateral angle sum, so the claim is false.

No. Four obtuse angles would have a sum greater than \(360^\circ\).

- The interior angles sum to \(360^\circ\). - Subtract the three known angles from the total. - Compare the result with \(90^\circ\).

1. The three known angles have a sum of \(90^\circ+90^\circ+90^\circ=270^\circ\). 2. The fourth angle is \(360^\circ-270^\circ=90^\circ\). 3. The fourth angle is a right angle, so all four angles are right angles. The quadrilateral cannot have an obtuse angle.

a) \(90^\circ\) b) Right angle c) No. All four angles must be right angles.

- Plot the points before deciding the direction of each ray. - Compare the horizontal and vertical coordinate changes from \(S\). - Identify horizontal, vertical, and equal-change diagonal directions.

1. Ray \(\overrightarrow{SA}\) is horizontal and points to the right. 2. Ray \(\overrightarrow{SB}\) is vertical and points upward, so it is perpendicular to \(\overrightarrow{SA}\). The angle is \(90^\circ\). 3. From \(S\) to \(C\), the horizontal and vertical changes are both \(4\), so \(\overrightarrow{SC}\) follows a \(45^\circ\) diagonal. The smaller angle is \(45^\circ\). 4. Ray \(\overrightarrow{SD}\) is vertical and points downward, so it is perpendicular to \(\overrightarrow{SA}\). The smaller angle is \(90^\circ\).

a) \(90^\circ\) b) \(45^\circ\) c) \(90^\circ\)

- What is the defining property of an angle bisector? - How are the two smaller angles related? - Add the two parts to find the full angle.

1. An angle bisector divides an angle into two congruent angles. 2. Therefore, the angle between \(w\) and \(h\) is equal to the given \(28^{\circ}\) angle between \(g\) and \(w\), so \(\alpha = 28^{\circ}\). 3. The full angle is \(\beta = 28^{\circ} + 28^{\circ} = 56^{\circ}\).

\(\alpha = 28^{\circ}\) and \(\beta = 56^{\circ}\)

- A full turn measures \(360^\circ\). - Subtract the interior angle from the full turn. - A reflex angle is between \(180^\circ\) and \(360^\circ\).

1. The interior angle and the reflex angle make a full turn of \(360^\circ\). 2. Therefore, \(\alpha=360^\circ-135^\circ=225^\circ\). 3. Since \(180^\circ<225^\circ<360^\circ\), \(\alpha\) is a reflex angle.

1. Reflex angle 2. \(\alpha=225^\circ\)

- What is the interior-angle sum of any quadrilateral? - Subtract the sum of the three known angles from that total.

1. The interior angles of a quadrilateral total \(360^\circ\). 2. The three known angles total \(270^\circ\). 3. Therefore, \(\delta=360^\circ-270^\circ=90^\circ\).

\(\delta=90^\circ\), by the quadrilateral interior-angle sum theorem.

- Use the quadrilateral interior-angle sum. - Add the three given angles before subtracting.

1. The interior angles of a quadrilateral total \(360^\circ\). 2. The known angles total \(90^\circ+90^\circ+115^\circ=295^\circ\). 3. Therefore, \(\alpha=65^\circ\).

\(\alpha=65^\circ\)

- Use the angle-bisector condition first. - Then combine the two equal halves. - Finish with the linear pair on the straight line.

1. Since \(w\) is an angle bisector, \(\alpha=55^\circ\). 2. The full bisected angle is \(110^\circ\). 3. That angle and \(\beta\) form a linear pair, so \(\beta=70^\circ\).

a) \(\alpha=55^\circ\) b) \(\beta=70^\circ\)

- What is the sum of the interior angles of a triangle? - Which two angle measures are already given in the diagram?

1. The two base angles of the triangle measure \(50^\circ\) and \(75^\circ\). 2. By the triangle angle sum, \(\gamma = 180^\circ - 50^\circ - 75^\circ = 55^\circ\).

\(\gamma = 55^\circ\)

- Add the two congruent base angles. - Subtract that sum from \(180^\circ\).

1. The two base angles total \(142^\circ\). 2. The triangle angle sum is \(180^\circ\), so \(\gamma=38^\circ\).

\(\gamma=38^\circ\)

- Add the two known adjacent parts. - Subtract that sum from the total angle.

1. The three angle measures total \(125^\circ\). 2. The known parts total \(80^\circ\). 3. Therefore, \(\gamma=125^\circ-80^\circ=45^\circ\).

\(\gamma=45^\circ\)

- Use the stated ray order to decide whether each target angle is a difference or a sum. - Compare the target endpoints with the given subangles.

1. The angle from \(a\) to \(b\) is \(75^\circ-25^\circ=50^\circ\). 2. The angle from \(b\) to \(d\) is \(25^\circ+35^\circ=60^\circ\).

The angle from \(a\) to \(b\) is \(50^\circ\), and the angle from \(b\) to \(d\) is \(60^\circ\).

- A straight angle measures \(180^\circ\). - Subtract the two known outer angles.

1. The three angles total \(180^\circ\). 2. The middle angle is \(180^\circ-(55^\circ+65^\circ)=60^\circ\).

The middle angle measures \(60^\circ\).

- Use the \(180^\circ\) straight-angle total. - Represent the two congruent outer angles with the same variable.

1. The three angles total \(180^\circ\). 2. Write \(\alpha+90^\circ+\alpha=180^\circ\). 3. Thus \(2\alpha=90^\circ\), so \(\alpha=45^\circ\).

\(\alpha=45^\circ\)

- Into how many triangles can a hexagon be divided from one vertex? - In a regular polygon, the interior-angle sum is divided equally among all vertices.

1. The sum of the interior angles of a hexagon is \((6 - 2) \cdot 180^\circ = 720^\circ\). 2. A regular hexagon has six congruent interior angles, so \(\alpha = 720^\circ \div 6 = 120^\circ\).

\(\alpha = 120^\circ\)

- What is the measure of each interior angle of a square? - What is the measure of each interior angle of an equilateral triangle? - How do the two angles at vertex \(B\) combine?

1. Each interior angle of a square measures \(90^\circ\), so \(\angle ABC = 90^\circ\). 2. Each interior angle of an equilateral triangle measures \(60^\circ\), so \(\angle ABE = 60^\circ\). 3. Therefore, \(\alpha = \angle ABC - \angle ABE = 90^\circ - 60^\circ = 30^\circ\).

\(\alpha = 30^\circ\)

- Recall the sum of all angles around a point. - Express the sum of the other three angles in terms of the unknown angle. - Translate “one-fourth of” into multiplication by a fraction.

1. Let \(x\) degrees be the angle measure. The four angles around the intersection add to \(360^\circ\). 2. The sum of the other three angles is \(360 - x\). 3. Translate the relationship into an equation: \(x = \frac{1}{4}(360 - x)\). 4. Multiply by \(4\): \(4x = 360 - x\). 5. Add \(x\): \(5x = 360\). Divide by \(5\): \(x = 72\).

The angle measures \(72^\circ\).

- Subtract the vertex angle from \(180^\circ\). - The two base angles are congruent, so split the remaining measure equally.

1. The base angles of an isosceles triangle are congruent, so \(\beta=\gamma\). 2. Subtract the vertex angle from the triangle angle sum: \(180^\circ-40^\circ=140^\circ\). 3. Divide the remaining measure equally: \(\beta=\gamma=140^\circ\div2=70^\circ\).

\(\beta=70^\circ\) and \(\gamma=70^\circ\)

- All adjacent angles on one side of a straight line form a \(180^\circ\) angle. - Combine the two known adjacent angles before finding the remaining part.

1. The adjacent angles on one side of a straight line sum to \(180^\circ\). 2. Write \(35^\circ+55^\circ+\gamma=180^\circ\). 3. Therefore, \(\gamma=180^\circ-90^\circ=90^\circ\).

\(\gamma=90^\circ\)

- How are an interior angle and its adjacent exterior angle related? - What does a pair of congruent angles imply about the opposite sides of a triangle?

1. The interior angle at \(C\) and its exterior angle form a linear pair, so \(\angle C=180^\circ-115^\circ=65^\circ\). 2. Since \(\angle A=\angle C=65^\circ\), the sides opposite those angles are congruent. Therefore, triangle \(ABC\) is isosceles. 3. The congruent legs are \(AB\) and \(BC\), and the base is \(AC\).

Triangle \(ABC\) is isosceles because \(\angle A=\angle C=65^\circ\). Its base is \(AC\), and its legs are \(AB\) and \(BC\).

- Identify the two congruent base angles from the equal legs. - Then use the triangle angle sum.

1. The base angles are congruent, so the other base angle also measures \(54^\circ\). 2. Use the triangle angle sum: \(\gamma=180^\circ-54^\circ-54^\circ=72^\circ\).

\(\gamma=72^\circ\)

- Find the angle measure left after the vertex angle. - Why must that remaining measure be split equally between the two base angles?

1. The two base angles are congruent. 2. Their total measure is \(180^\circ-104^\circ=76^\circ\). 3. Therefore, \(\alpha=\beta=76^\circ\div2=38^\circ\).

\(\alpha=38^\circ\) and \(\beta=38^\circ\)

- Which two smaller angles combine to form \(\angle AOC\)? - Write the corresponding decomposition for \(\angle BOD\). - Compare the two missing parts after subtracting the shared angle.

1. Since \(\angle AOC=\angle AOB+\angle BOC\), \(\angle AOB=80^\circ-35^\circ=45^\circ\). 2. Since \(\angle BOD=\angle BOC+\angle COD\), \(\angle COD=80^\circ-35^\circ=45^\circ\). 3. The two unknown angles are congruent.

\(\angle AOB=45^\circ\) and \(\angle COD=45^\circ\). The angles are congruent.

- What is the measure of a right angle? - How can you represent the larger acute angle in terms of the smaller one? - What is the sum of the two acute angles in a right triangle?

1. A right angle measures \(90^\circ\). 2. The two acute angles must total \(180^\circ - 90^\circ = 90^\circ\). 3. Let \(x\) be the measure of the smaller acute angle. Then the larger acute angle measures \(3x\). 4. Write and solve the equation \(x + 3x = 90^\circ\): \(4x = 90^\circ\), so \(x = 22.5^\circ\). 5. The larger acute angle is \(3 \cdot 22.5^\circ = 67.5^\circ\).

The three angle measures are \(22.5^\circ\), \(67.5^\circ\), and \(90^\circ\).

- Express all three angles in terms of one variable, such as \(\alpha\). - How can you write “\(\alpha\) is \(15^\circ\) less than \(\beta\)” as an equation? - Use the sum of the interior angles of a triangle.

1. Let \(\alpha = x\). Then \(\beta = x + 15^\circ\) and \(\gamma = 2x\). 2. Use the triangle angle sum: \(x + (x + 15^\circ) + 2x = 180^\circ\). 3. Combine like terms and solve: \(4x + 15^\circ = 180^\circ\), so \(4x = 165^\circ\) and \(x = 41.25^\circ\). 4. Therefore, \(\alpha = 41.25^\circ\), \(\beta = 41.25^\circ + 15^\circ = 56.25^\circ\), and \(\gamma = 2 \cdot 41.25^\circ = 82.5^\circ\).

\(\alpha = 41.25^\circ\), \(\beta = 56.25^\circ\), and \(\gamma = 82.5^\circ\).

- What is the sum of the interior angles of a quadrilateral? - After subtracting the two known angles, how much remains for the two equal angles? - What relationship do opposite angles have in a parallelogram?

1. The two given angles total \(72.5^\circ + 107.5^\circ = 180^\circ\). 2. The interior angles of a quadrilateral total \(360^\circ\), so \(\gamma + \delta = 360^\circ - 180^\circ = 180^\circ\). 3. Since \(\gamma = \delta\), each angle measures \(180^\circ \div 2 = 90^\circ\). 4. In a parallelogram, opposite angles are congruent. Here, \(\alpha \ne \gamma\) and \(\beta \ne \delta\), so the quadrilateral cannot be a parallelogram.

a) \(\gamma = 90^\circ\) and \(\delta = 90^\circ\) b) No. Its opposite angles are not congruent.

- Draw diagonals from one vertex so that the pentagon is divided into nonoverlapping triangles. - How many triangles are formed? - Subtract the sum of the four known angles from the pentagon’s total angle sum.

1. Drawing diagonals from one vertex divides a pentagon into \(5 - 2 = 3\) triangles. 2. Therefore, the pentagon’s interior angle sum is \(3 \cdot 180^\circ = 540^\circ\). 3. The four known angles total \(112^\circ + 98^\circ + 125^\circ + 105^\circ = 440^\circ\). 4. The fifth angle is \(540^\circ - 440^\circ = 100^\circ\).

a) \(540^\circ\) b) \(100^\circ\)

- Express each unknown angle in terms of one variable. - Translate “twice” and “three times” into algebraic expressions. - What is the measure of a right angle? - Use the interior angle sum of a quadrilateral.

1. The interior angles of a quadrilateral total \(360^\circ\), so \(\alpha + \beta + \gamma + \delta = 360^\circ\). 2. Substitute the given relationships: \(\alpha + 2\alpha + 3\alpha + 90^\circ = 360^\circ\). 3. Combine like terms: \(6\alpha + 90^\circ = 360^\circ\). 4. Solve: \(6\alpha = 270^\circ\), so \(\alpha = 45^\circ\). 5. Then \(\beta = 2 \cdot 45^\circ = 90^\circ\) and \(\gamma = 3 \cdot 45^\circ = 135^\circ\).

\(\alpha = 45^\circ\), \(\beta = 90^\circ\), and \(\gamma = 135^\circ\).

- Consider a triangle with a right angle, then a triangle with an obtuse angle. - What happens to the angle sum if two angles are each at least \(90^\circ\)? - How much angle measure would remain for the third angle?

1. An acute triangle has three acute angles, so it has at least two. 2. A right triangle has one \(90^\circ\) angle. The other two angles have a sum of \(180^\circ - 90^\circ = 90^\circ\). Since both are greater than \(0^\circ\), each is less than \(90^\circ\), so both are acute. 3. An obtuse triangle has one angle greater than \(90^\circ\). The other two angles have a sum less than \(90^\circ\), so both are acute. 4. If a triangle had only one acute angle, the other two angles would each be at least \(90^\circ\). Their sum would already be at least \(180^\circ\), leaving no positive measure for the third angle. That is impossible.

The statement is true. An acute triangle has three acute angles, while a right triangle and an obtuse triangle each have exactly two acute angles. A triangle cannot have fewer than two acute angles because two nonacute angles would have a sum of at least \(180^\circ\).

- What is true about the base angles of an isosceles triangle? - What would happen if both base angles were obtuse? - At which vertex could an obtuse angle occur without violating the angle sum?

1. In an isosceles triangle, the two base angles are congruent. 2. Suppose a base angle were obtuse, so it measured more than \(90^\circ\). The other base angle would also measure more than \(90^\circ\). 3. Those two angles alone would have a sum greater than \(180^\circ\), which is impossible because all three interior angles of a triangle have a sum of exactly \(180^\circ\). 4. Therefore, an obtuse angle in an isosceles triangle can only be the vertex angle between the congruent sides. For example, a triangle with angle measures \(120^\circ\), \(30^\circ\), and \(30^\circ\) is isosceles and obtuse.

a) Yes, an isosceles triangle can have an obtuse angle. b) No. An obtuse base angle would force both congruent base angles to be greater than \(90^\circ\), so their sum would exceed \(180^\circ\). The obtuse angle must be the vertex angle.

- Imagine drawing a segment from one vertex to the opposite vertex. What shapes are formed? - Once you know the total angle measure, how can you split it into four equal parts? - Which quadrilaterals have four right angles?

1. A triangle has an interior angle sum of \(180^\circ\). A diagonal divides a quadrilateral into two triangles, so the interior angle sum of the quadrilateral is \(2 \cdot 180^\circ = 360^\circ\). 2. If all four angles have the same measure, divide the total by \(4\): \(360^\circ \div 4 = 90^\circ\). 3. A rectangle and a square each have four right angles.

a) The interior angle sum is \(2 \cdot 180^\circ = 360^\circ\). b) Each angle measures \(90^\circ\). Two possible quadrilaterals are a rectangle and a square.

- Represent congruent angles with the same variable. - For part b, use the smallest angle as the variable. - In each part, use the \(360^\circ\) interior angle sum of a quadrilateral.

1. For part a, use the quadrilateral angle sum: \(112^\circ + 48^\circ + \beta + \delta = 360^\circ\). 2. Since \(\beta = \delta\), \(160^\circ + 2\beta = 360^\circ\). Thus \(2\beta = 200^\circ\), so \(\beta = 100^\circ\) and \(\delta = 100^\circ\). 3. For part b, write \(\alpha + (\alpha + 10^\circ) + (\alpha + 20^\circ) + (\alpha + 30^\circ) = 360^\circ\). 4. Combine like terms: \(4\alpha + 60^\circ = 360^\circ\). Then \(4\alpha = 300^\circ\), so \(\alpha = 75^\circ\). 5. Therefore, \(\beta = 85^\circ\), \(\gamma = 95^\circ\), and \(\delta = 105^\circ\).

a) \(\beta = 100^\circ\) and \(\delta = 100^\circ\) b) \(\alpha = 75^\circ\), \(\beta = 85^\circ\), \(\gamma = 95^\circ\), and \(\delta = 105^\circ\)

- What is the sum of the interior angles of a triangle? - Express all three angles in terms of \(\alpha\). - How is a triangle with one \(90^\circ\) angle classified?

1. Use the triangle angle sum: \(\alpha + \beta + \gamma = 180^\circ\). 2. Substitute the given relationships: \(\alpha + 2\alpha + 3\alpha = 180^\circ\). 3. Combine like terms and solve: \(6\alpha = 180^\circ\), so \(\alpha = 30^\circ\). 4. Then \(\beta = 2 \cdot 30^\circ = 60^\circ\) and \(\gamma = 3 \cdot 30^\circ = 90^\circ\). 5. Since one angle measures \(90^\circ\), the triangle is a right triangle.

\(\alpha = 30^\circ\), \(\beta = 60^\circ\), and \(\gamma = 90^\circ\). The triangle is a right triangle.

- Express every angle in terms of \(\alpha\). - How many copies of \(\alpha\) make up the full \(180^\circ\) angle sum? - What conditions must three angle measures satisfy to form a triangle?

1. The relationships are \(\beta = 3\alpha\) and \(\gamma = 2\beta = 6\alpha\). 2. Use the triangle angle sum: \(\alpha + \beta + \gamma = 180^\circ\). 3. Substitute: \(\alpha + 3\alpha + 6\alpha = 180^\circ\), so \(10\alpha = 180^\circ\). 4. Therefore, \(\alpha = 18^\circ\), \(\beta = 3 \cdot 18^\circ = 54^\circ\), and \(\gamma = 2 \cdot 54^\circ = 108^\circ\). 5. The angles are all positive and total \(180^\circ\), so such a triangle does exist.

\(\alpha = 18^\circ\), \(\beta = 54^\circ\), and \(\gamma = 108^\circ\). Yes, the triangle exists.

- Check the relevant angle sum in each statement. - Recall the definitions of right and obtuse angles. - What is true about the base angles of an isosceles triangle? - How are an interior angle and its adjacent exterior angle related?

1. For part a, \(35^\circ + 45^\circ + 110^\circ = 190^\circ\). Since the total is not \(180^\circ\), the statement is false. 2. For part b, a right angle measures \(90^\circ\), while an obtuse angle is greater than \(90^\circ\). A triangle cannot contain both because their sum would already exceed \(180^\circ\). The statement is true. 3. For part c, the two base angles measure \(70^\circ\) each. The vertex angle is \(180^\circ - 70^\circ - 70^\circ = 40^\circ\). The statement is true. 4. For part d, each interior angle of an equilateral triangle is \(60^\circ\). Its adjacent exterior angle is \(180^\circ - 60^\circ = 120^\circ\), which is twice \(60^\circ\). The statement is true.

a) False; the angles total \(190^\circ\). b) True; a right angle and an obtuse angle cannot both fit in a triangle. c) True; the vertex angle is \(40^\circ\). d) True; the interior angle is \(60^\circ\) and the adjacent exterior angle is \(120^\circ\).

- What is true about the base angles of an isosceles triangle? - Express the vertex angle in terms of one base angle. - How many copies of the base angle make up the full \(180^\circ\) angle sum?

1. The base angles of an isosceles triangle are congruent, so \(\alpha = \beta\). 2. The vertex angle satisfies \(\gamma = 4\alpha\). 3. Use the triangle angle sum: \(\alpha + \beta + \gamma = 180^\circ\). 4. Substitute the relationships: \(\alpha + \alpha + 4\alpha = 180^\circ\), so \(6\alpha = 180^\circ\). 5. Thus \(\alpha = 30^\circ\), \(\beta = 30^\circ\), and \(\gamma = 4 \cdot 30^\circ = 120^\circ\).

\(\alpha = 30^\circ\), \(\beta = 30^\circ\), and \(\gamma = 120^\circ\).

- What is the interior angle sum of a quadrilateral? - Write \(\delta\) in terms of \(\gamma\). - How much of the \(360^\circ\) total remains after subtracting the known angles?

1. The interior angles of a quadrilateral total \(360^\circ\). 2. Substitute the known angles into \(\alpha + \beta + \gamma + \delta = 360^\circ\): \(85^\circ + 110^\circ + \gamma + \delta = 360^\circ\). 3. Since \(\delta = \gamma - 15^\circ\), write \(195^\circ + \gamma + (\gamma - 15^\circ) = 360^\circ\). 4. Simplify: \(180^\circ + 2\gamma = 360^\circ\), so \(2\gamma = 180^\circ\) and \(\gamma = 90^\circ\). 5. Then \(\delta = 90^\circ - 15^\circ = 75^\circ\).

\(\gamma = 90^\circ\) and \(\delta = 75^\circ\).

- How is the interior angle sum related to the number of triangles formed from one vertex? - What does “regular” tell you about the angle measures? - Write an equation with the number of sides as the unknown.

1. Use the interior angle sum formula \(S = (n - 2) \cdot 180^\circ\): \(1440^\circ = (n - 2) \cdot 180^\circ\). 2. Divide by \(180^\circ\): \(8 = n - 2\), so \(n = 10\). The polygon is a decagon. 3. Since the polygon is regular, its interior angles are congruent. Each angle measures \(1440^\circ \div 10 = 144^\circ\).

a) The polygon has \(10\) sides. b) Each interior angle measures \(144^\circ\).

- How many degrees does the robot turn during one complete trip? - Divide the total turn by the turn at each vertex. - What kind of number must the number of sides be?

1. After one complete trip around the polygon, the robot has turned a total of \(360^\circ\). 2. For part a, the number of sides is \(360^\circ \div 45^\circ = 8\), so the robot traces a regular octagon. 3. For part b, \(360^\circ \div 50^\circ = 7.2\). 4. A polygon must have a whole-number count of sides, so a regular polygon with a \(50^\circ\) exterior angle is not possible.

a) A regular octagon b) No. The calculation gives \(7.2\) sides, which is not possible for a polygon.

- First find the interior angle sum of the second polygon. - Use the polygon interior angle sum formula. - How much does the angle sum increase when one side is added?

1. The second polygon has an interior angle sum of \(1260^\circ + 360^\circ = 1620^\circ\). 2. Use \((n - 2) \cdot 180^\circ = 1620^\circ\). 3. Divide by \(180^\circ\): \(n - 2 = 9\), so \(n = 11\). 4. Equivalently, increasing the angle sum by \(360^\circ\) adds \(360 \div 180 = 2\) sides.

The second polygon has \(11\) sides.

- Recall the definition of an obtuse angle. - Use the triangle angle sum to find a missing angle. - What must be true about two angles of an isosceles triangle? - Check each result against the \(180^\circ\) angle sum.

1. For a), a right angle measures \(90^\circ\), and an obtuse angle measures more than \(90^\circ\). Their sum would be greater than \(180^\circ\), which is impossible in a triangle. 2. For b), the third angle measures \(180^\circ - 85^\circ = 95^\circ\). Because \(95^\circ > 90^\circ\), the triangle is obtuse. 3. For c), the third angle would measure \(180^\circ - (90^\circ + 60^\circ) = 30^\circ\). The three angles would be \(30^\circ\), \(60^\circ\), and \(90^\circ\), with no two congruent angles. Therefore, the triangle could not be isosceles.

a) A right angle and an obtuse angle would have a sum greater than \(180^\circ\). b) The triangle is obtuse because the third angle measures \(95^\circ\). c) No. The angles would be \(30^\circ\), \(60^\circ\), and \(90^\circ\), so no two angles would be congruent.

- What does “regular” tell you about the exterior angles? - Divide the full \(360^\circ\) turn into \(8\) equal parts. - How are adjacent interior and exterior angles related? - Multiply one interior angle by the number of sides.

1. A regular octagon has \(8\) congruent exterior angles. One exterior angle measures \(360^\circ \div 8 = 45^\circ\). 2. An interior angle and its adjacent exterior angle are supplementary, so one interior angle measures \(180^\circ - 45^\circ = 135^\circ\). 3. The interior angle sum is \(8 \cdot 135^\circ = 1080^\circ\).

a) \(45^\circ\) b) \(135^\circ\) c) \(1080^\circ\)

- Which angles are congruent in an isosceles triangle? - How does knowing one angle help you find the other two? - Use the \(180^\circ\) triangle angle sum.

1. For a), the base angles have a sum of \(180^\circ - 110^\circ = 70^\circ\). Since they are congruent, \(\alpha = \beta = 70^\circ \div 2 = 35^\circ\). 2. For b), the other base angle is also \(\beta = 35^\circ\). The vertex angle is \(\gamma = 180^\circ - (35^\circ + 35^\circ) = 110^\circ\). 3. For c), a right vertex angle is possible. The remaining \(90^\circ\) is divided equally between the base angles, so \(\alpha = \beta = 90^\circ \div 2 = 45^\circ\).

a) \(\alpha = \beta = 35^\circ\) b) \(\beta = 35^\circ\) and \(\gamma = 110^\circ\) c) Yes. The base angles are each \(45^\circ\).

- Recall the sum of the interior angles of a triangle. - Express both \(\beta\) and \(\gamma\) in terms of \(\alpha\). - Substitute those expressions into the angle-sum equation. - Check that the three results total \(180^\circ\).

1. The angle sum is \(\alpha+\beta+\gamma=180^\circ\). 2. The given relationships are \(\beta=2\alpha\) and \(\gamma=\beta-30^\circ=2\alpha-30^\circ\). 3. Substitute into the angle-sum equation: \(\alpha+2\alpha+(2\alpha-30^\circ)=180^\circ\). 4. Simplify: \(5\alpha-30^\circ=180^\circ\), so \(5\alpha=210^\circ\) and \(\alpha=42^\circ\). 5. Then \(\beta=84^\circ\) and \(\gamma=54^\circ\).

\(\alpha=42^\circ\), \(\beta=84^\circ\), and \(\gamma=54^\circ\)

- Use the fact that the base angles of an isosceles triangle are congruent. - Express the vertex angle in terms of one base angle. - Use the sum of the interior angles of a triangle.

1. Because the triangle is isosceles, \(\alpha=\beta\). 2. The angle sum is \(2\alpha+\gamma=180^\circ\). 3. The given relationship is \(\gamma=1.5(\alpha+\beta)=1.5(2\alpha)=3\alpha\). 4. Substitute into the angle-sum equation: \(2\alpha+3\alpha=180^\circ\), so \(5\alpha=180^\circ\) and \(\alpha=36^\circ\). 5. Therefore, \(\beta=36^\circ\) and \(\gamma=108^\circ\).

\(\alpha=36^\circ\), \(\beta=36^\circ\), and \(\gamma=108^\circ\)

- How are an interior angle and its adjacent exterior angle related? - What is the sum of the three interior angles of a triangle? - A quick sketch may help you place the exterior angles.

1. An interior angle and its adjacent exterior angle form a linear pair and total \(180^\circ\). 2. Therefore, \(\alpha = 180^\circ - 125^\circ = 55^\circ\). 3. Similarly, \(\beta = 180^\circ - 110^\circ = 70^\circ\). 4. Use the triangle angle sum: \(\gamma = 180^\circ - 55^\circ - 70^\circ = 55^\circ\).

\(\alpha = 55^\circ\), \(\beta = 70^\circ\), and \(\gamma = 55^\circ\).

- Work with the two triangles separately first. - How are the full angles at \(A\) and \(C\) built from the smaller angles? - Relate the angle sums of the two triangles to the quadrilateral’s angle sum.

1. In triangle \(ABC\), \(\angle ABC = 180^\circ - 35^\circ - 45^\circ = 100^\circ\). 2. In triangle \(ADC\), \(\angle ADC = 180^\circ - 40^\circ - 60^\circ = 80^\circ\). 3. The full angle at \(A\) is \(35^\circ + 40^\circ = 75^\circ\), and the full angle at \(C\) is \(45^\circ + 60^\circ = 105^\circ\). 4. The quadrilateral angle sum is \(75^\circ + 100^\circ + 105^\circ + 80^\circ = 360^\circ\). Equivalently, the two triangles contribute \(180^\circ + 180^\circ = 360^\circ\).

a) \(\angle ABC = 100^\circ\) and \(\angle ADC = 80^\circ\) b) \(360^\circ\)

- What is the sum of the interior angles of a triangle? - What type of triangle has three congruent angles? - What is true about the side lengths of an equilateral triangle?

1. Use the triangle angle sum: \(\gamma = 180^\circ - 60^\circ - 60^\circ = 60^\circ\). 2. All three interior angles measure \(60^\circ\), so the triangle is equiangular and therefore equilateral. 3. All sides of an equilateral triangle are congruent. Since \(c = 8\,\text{cm}\), it follows that \(a = 8\,\text{cm}\) and \(b = 8\,\text{cm}\).

a) \(\gamma = 60^\circ\) b) \(a = 8\,\text{cm}\) and \(b = 8\,\text{cm}\). The triangle is equilateral because an equiangular triangle has three congruent sides.

- What is the interior angle sum of a quadrilateral? - Express every angle in terms of \(\alpha\). - Write one equation using the full angle sum. - Check that your four results total \(360^\circ\).

1. The interior angles of a quadrilateral total \(360^\circ\). 2. Let \(\alpha = x\). Then \(\beta = x\), \(\gamma = x + 30^\circ\), and \(\delta = 2x + 10^\circ\). 3. Write the equation \(x + x + (x + 30^\circ) + (2x + 10^\circ) = 360^\circ\). 4. Simplify: \(5x + 40^\circ = 360^\circ\). 5. Solve: \(5x = 320^\circ\), so \(x = 64^\circ\). 6. Therefore, \(\alpha = 64^\circ\), \(\beta = 64^\circ\), \(\gamma = 94^\circ\), and \(\delta = 2 \cdot 64^\circ + 10^\circ = 138^\circ\).

\(\alpha = 64^\circ\), \(\beta = 64^\circ\), \(\gamma = 94^\circ\), and \(\delta = 138^\circ\).

- Find how much of a full \(360^\circ\) turn is left. - Find how much the angle exceeds \(180^\circ\).

1. The remaining angle in a full turn is \(360^\circ-215^\circ=145^\circ\). 2. The amount beyond a straight angle is \(215^\circ-180^\circ=35^\circ\).

The remainder of a full turn is \(145^\circ\), and the amount beyond a straight angle is \(35^\circ\).

- Choose three angles less than \(90^\circ\). - Subtract their sum from \(360^\circ\). - Check whether the fourth angle is a valid interior angle.

1. Choose three acute angles of \(80^\circ\) each. Their sum is \(3 \cdot 80^\circ=240^\circ\). 2. The fourth angle must be \(360^\circ-240^\circ=120^\circ\). 3. The measures \(80^\circ\), \(80^\circ\), \(80^\circ\), and \(120^\circ\) sum to \(360^\circ\), so such a quadrilateral is possible.

No. One example has interior angles \(80^\circ\), \(80^\circ\), \(80^\circ\), and \(120^\circ\).

- At \({:}30\), the hour hand is halfway between two hour marks. - Each hour mark is \(30^\circ\) apart. - Find each hand’s position from \(12\), then subtract.

1. At half past an hour, the minute hand is at \(180^\circ\) from \(12\), and the hour hand is halfway between two hour marks. 2. At \(4{:}30\), the hour hand is at \(4 \cdot 30^\circ+15^\circ=135^\circ\). The angle is \(180^\circ-135^\circ=45^\circ\). 3. At \(10{:}30\), the hour hand is at \(10 \cdot 30^\circ+15^\circ=315^\circ\). The smaller angle is \(315^\circ-180^\circ=135^\circ\).

a) \(45^\circ\) b) \(135^\circ\)

- Each hour section measures \(30^\circ\). - Follow the specified counterclockwise direction. - Compare each angle measure with \(90^\circ\), \(180^\circ\), and \(360^\circ\).

1. Each hour section measures \(360^\circ \div 12=30^\circ\). 2. At \(2{:}00\), the counterclockwise angle is \(2 \cdot 30^\circ=60^\circ\), which is acute. 3. At \(3{:}00\), the angle is \(3 \cdot 30^\circ=90^\circ\), which is right. 4. At \(5{:}00\), the angle is \(5 \cdot 30^\circ=150^\circ\), which is obtuse. 5. At \(6{:}00\), the angle is \(6 \cdot 30^\circ=180^\circ\), which is straight. 6. At \(8{:}00\), the counterclockwise angle is \(8 \cdot 30^\circ=240^\circ\), which is reflex.

a) Acute b) Right c) Obtuse d) Straight e) Reflex

- Plot the points and draw the first ray before adding the second rays. - Opposite rays form a straight angle. - Compare horizontal, vertical, and equal-change diagonal directions.

1. Ray \(\overrightarrow{SA}\) is vertical and points upward. 2. The opposite ray points vertically downward, so the two rays form a straight angle. Thus, \(\beta=180^\circ\). 3. From \(S\) to \(Q\), the change is \(4\) units left and \(4\) units up. This equal-change diagonal makes a \(45^\circ\) angle with the vertical ray, so \(\beta=45^\circ\). 4. Ray \(\overrightarrow{SP}\) points horizontally left. A horizontal ray and a vertical ray are perpendicular, so \(\beta=90^\circ\).

a) \(180^\circ\) b) \(45^\circ\) c) \(90^\circ\)

- Plot each named point and draw the rays before calculating an angle. - Equal horizontal and vertical changes indicate a \(45^\circ\) diagonal direction. - Perpendicular rays form a right angle, and opposite rays form a straight angle.

1. Ray \(\overrightarrow{SA}\) is horizontal and points to the right. 2. From \(S\) to \(P\), the change is \(2\) units left and \(2\) units up. This northwest diagonal has direction \(135^\circ\) from the positive horizontal direction, so \(\gamma=135^\circ\). 3. A downward ray perpendicular to \(\overrightarrow{SA}\) is vertical. Therefore, \(\gamma=90^\circ\). It follows the line \(x=2\) and meets the x-axis at \((2, 0)\). 4. Ray \(\overrightarrow{SQ}\) points horizontally left, exactly opposite \(\overrightarrow{SA}\). Therefore, \(\gamma=180^\circ\).

a) \(135^\circ\) b) \(90^\circ\); the ray meets the x-axis at \((2, 0)\). c) \(180^\circ\)

- Substitute each number of sides into the given formula. - Solve the formula for \(n\) in part b. - What does “regular” tell you about how the total angle sum is divided?

1. For a hexagon, \(S = (6 - 2) \cdot 180^\circ = 720^\circ\). 2. For an octagon, \(S = (8 - 2) \cdot 180^\circ = 1080^\circ\). 3. For part b, solve \(1260^\circ = (n - 2) \cdot 180^\circ\). Dividing by \(180^\circ\) gives \(7 = n - 2\), so \(n = 9\). 4. Since the polygon is regular, one interior angle measures \(1260^\circ \div 9 = 140^\circ\).

a) Hexagon: \(720^\circ\); octagon: \(1080^\circ\) b) \(n = 9\) c) \(140^\circ\)

- How are an interior angle and its adjacent exterior angle related? - Use the \(360^\circ\) sum of the exterior angles. - The number of sides must be a whole number.

1. For part a, \(\alpha = 4\beta\) and \(\alpha + \beta = 180^\circ\). 2. Substitute: \(4\beta + \beta = 180^\circ\), so \(5\beta = 180^\circ\) and \(\beta = 36^\circ\). 3. The number of sides is \(n = 360^\circ \div 36^\circ = 10\). 4. For part b, an interior angle of \(175^\circ\) has an exterior angle of \(180^\circ - 175^\circ = 5^\circ\). 5. Then \(n = 360^\circ \div 5^\circ = 72\). Since \(72\) is a whole number, the polygon exists.

a) \(n = 10\) b) Yes. The polygon has \(72\) sides.

- Divide the full interior angle sum by the number of congruent angles. - Rearrange the equation in part b so that the terms containing \(n\) are on one side. - Can \(\frac{n - 2}{n}\) ever equal or exceed \(1\)?

1. For a regular hexagon, \(S = (6 - 2) \cdot 180^\circ = 720^\circ\), so one angle is \(720^\circ \div 6 = 120^\circ\). 2. For a regular dodecagon, \(S = (12 - 2) \cdot 180^\circ = 1800^\circ\), so one angle is \(1800^\circ \div 12 = 150^\circ\). 3. For part b, solve \(\frac{(n - 2) \cdot 180^\circ}{n} = 144^\circ\). This gives \(180n - 360 = 144n\), so \(36n = 360\) and \(n = 10\). 4. No. In \(\alpha = \frac{n - 2}{n} \cdot 180^\circ\), the factor \(\frac{n - 2}{n}\) is always less than \(1\), so \(\alpha < 180^\circ\). Geometrically, an angle of \(180^\circ\) would not form a vertex.

a) Hexagon: \(120^\circ\); dodecagon: \(150^\circ\) b) \(n = 10\) c) No. A regular polygon’s interior angle is always less than \(180^\circ\).

- What is true about the base angles of an isosceles triangle? - What is the sum of the interior angles of a triangle? - Express the vertex angle in terms of one base angle.

1. Let each base angle be \(\alpha\), and let the vertex angle be \(\gamma\). 2. The relationship is \(\gamma = 0.5\alpha\). 3. Use the triangle angle sum: \(\alpha + \alpha + \gamma = 180^\circ\). 4. Substitute: \(2\alpha + 0.5\alpha = 180^\circ\), so \(2.5\alpha = 180^\circ\). 5. Therefore, \(\alpha = 72^\circ\), and \(\gamma = 0.5 \cdot 72^\circ = 36^\circ\).

The base angles each measure \(72^\circ\), and the vertex angle measures \(36^\circ\).

- First find the interior angle at \(B\) from its adjacent exterior angle. - Then use the triangle angle sum.

1. The interior angle at \(B\) and the \(120^\circ\) exterior angle form a linear pair, so \(\angle B=60^\circ\). 2. The interior angles of a triangle total \(180^\circ\). 3. Therefore, \(\gamma=180^\circ-50^\circ-60^\circ=70^\circ\).

\(\gamma=70^\circ\)

- What is the measure of a straight angle? - How does an angle bisector divide \(\alpha\)? - What sum do angles in a linear pair have?

1. A straight angle measures \(180^\circ\). 2. Since \(w_\alpha\) bisects \(\alpha\), the angle between the rightward ray of \(g\) and \(w_\alpha\) is \(\frac{\alpha}{2}\). 3. That angle and \(\delta\) form a linear pair, so \(180^\circ-\frac{\alpha}{2}=145^\circ\). 4. Thus, \(\frac{\alpha}{2}=35^\circ\), so \(\alpha=70^\circ\). 5. Since \(\alpha\) and \(\beta\) form a linear pair, \(\beta=180^\circ-70^\circ=110^\circ\).

\(\alpha=70^\circ\) and \(\beta=110^\circ\).

- Express \(\beta\) in terms of \(\alpha\). - Use the stated total of the adjacent angles.

1. The measures satisfy \(\alpha+\beta=135^\circ\). 2. Since \(\beta=2\alpha\), \(\alpha+2\alpha=135^\circ\). 3. Thus \(\alpha=45^\circ\) and \(\beta=90^\circ\).

\(\alpha=45^\circ\) and \(\beta=90^\circ\)

- Apply the first angle bisector before the second. - Which two adjacent parts combine to make \(\beta\)?

1. Since \(w_1\) bisects \(74^\circ\), the angles on either side of \(w_1\) are \(37^\circ\). 2. Ray \(w_2\) bisects the \(37^\circ\) angle between \(w_1\) and \(b\), so \(\angle(w_1,w_2)=18.5^\circ\). 3. Therefore, \(\beta=37^\circ+18.5^\circ=55.5^\circ\).

\(\beta=55.5^\circ\)

- What relationship do vertical angles have? - What is the sum of a linear pair? - What is the sum of the interior angles of a triangle?

1. The \(65^\circ\) angle and \(\alpha\) are vertical angles, so \(\alpha = 65^\circ\). 2. The \(120^\circ\) angle and \(\beta\) form a linear pair, so \(\beta = 180^\circ - 120^\circ = 60^\circ\). 3. The three interior angles of the triangle total \(180^\circ\). Therefore, \(\gamma = 180^\circ - 65^\circ - 60^\circ = 55^\circ\).

\(\alpha = 65^\circ\), \(\beta = 60^\circ\), and \(\gamma = 55^\circ\).

- How are an interior angle and its adjacent exterior angle related? - What is the interior angle sum of a quadrilateral? - Find the angles connected to the given exterior angles before using the quadrilateral angle sum.

1. The \(70^\circ\) exterior angle and interior angle \(\gamma\) form a linear pair, so \(\gamma = 180^\circ - 70^\circ = 110^\circ\). 2. The \(75^\circ\) interior angle and exterior angle \(\beta\) form a linear pair, so \(\beta = 180^\circ - 75^\circ = 105^\circ\). 3. The interior angles of a quadrilateral total \(360^\circ\). Therefore, \(\alpha + 110^\circ + 75^\circ + 90^\circ = 360^\circ\). 4. Thus \(\alpha = 360^\circ - 275^\circ = 85^\circ\).

\(\alpha = 85^\circ\), \(\beta = 105^\circ\), and \(\gamma = 110^\circ\).

- Use the linear pair at \(D\) first. - Then work in triangle \(ABD\) before using the angle bisector. - Finish with the angle sum of triangle \(ABC\).

1. Since \(B,D,C\) are collinear, \(\angle ADB=180^\circ-75^\circ=105^\circ\). 2. In triangle \(ABD\), \(\angle DAB=180^\circ-105^\circ-40^\circ=35^\circ\). 3. Because \(AD\) bisects \(\angle BAC\), \(\alpha=35^\circ\) and \(\angle BAC=70^\circ\). 4. In triangle \(ABC\), \(\gamma=180^\circ-70^\circ-40^\circ=70^\circ\).

\(\alpha=35^\circ\) and \(\gamma=70^\circ\)

- Find the interior angle adjacent to the exterior angle. - Then use the triangle angle sum.

1. The exterior angle and \(\alpha\) form a linear pair, so \(\alpha=180^\circ-125^\circ=55^\circ\). 2. Then \(\gamma=180^\circ-55^\circ-65^\circ=60^\circ\).

\(\alpha=55^\circ\) and \(\gamma=60^\circ\)

- How are an interior angle and its adjacent exterior angle related? - What is the interior angle sum of a pentagon? - Subtract all known interior angles from the pentagon’s total.

1. At \(A\), \(\alpha\) and the \(75^\circ\) exterior angle form a linear pair, so \(\alpha = 180^\circ - 75^\circ = 105^\circ\). 2. At \(C\), \(\gamma\) and the \(85^\circ\) exterior angle form a linear pair, so \(\gamma = 180^\circ - 85^\circ = 95^\circ\). 3. The interior angle sum of a pentagon is \((5 - 2) \cdot 180^\circ = 540^\circ\). 4. Therefore, \(\epsilon = 540^\circ - 105^\circ - 115^\circ - 95^\circ - 130^\circ = 95^\circ\).

\(\alpha = 105^\circ\), \(\gamma = 95^\circ\), and \(\epsilon = 95^\circ\).

- Which two angles are congruent because of the kite's symmetry? - Use the \(360^\circ\) interior-angle sum of a quadrilateral.

1. Symmetry gives \(\beta=\delta\). 2. The quadrilateral angle sum gives \(38^\circ+82^\circ+\beta+\delta=360^\circ\). 3. Thus \(2\beta=240^\circ\), so \(\beta=\delta=120^\circ\).

\(\beta=120^\circ\) and \(\delta=120^\circ\)

- What relationship do vertical angles have? - What total do adjacent angles along a straight line form? - Find the vertical angles first.

1. The \(40^\circ\) angle and \(\alpha\) are vertical angles, so \(\alpha = 40^\circ\). 2. The \(75^\circ\) angle and \(\beta\) are vertical angles, so \(\beta = 75^\circ\). 3. The angles \(\alpha\), \(\beta\), and \(\gamma\) form a straight angle. Therefore, \(\gamma = 180^\circ - 40^\circ - 75^\circ = 65^\circ\).

\(\alpha = 40^\circ\), \(\beta = 75^\circ\), and \(\gamma = 65^\circ\).

- What is true about the base angles of an isosceles triangle? - Use the triangle angle sum.

1. Since \(AC=BC\), the base angles at \(A\) and \(B\) are congruent. 2. If each base angle is \(\alpha\), then \(2\alpha+54^\circ=180^\circ\). 3. Thus \(\alpha=63^\circ\).

\(\alpha=63^\circ\)

- Find the interior angle adjacent to the \(115^\circ\) exterior angle first. - Then use the triangle angle sum.

1. The exterior angle at \(C\) and \(\gamma\) form a linear pair, so \(\gamma=180^\circ-115^\circ=65^\circ\). 2. The triangle angle sum gives \(45^\circ+\beta+65^\circ=180^\circ\). 3. Therefore, \(\beta=70^\circ\).

\(\beta=70^\circ\) and \(\gamma=65^\circ\)

- Use the linear pair at \(D\). - Work in triangle \(ABD\) before doubling the bisected angle. - Finish with the angle sum of triangle \(ABC\).

1. Since \(B,D,C\) are collinear, \(\angle ADB=180^\circ-95^\circ=85^\circ\). 2. In triangle \(ABD\), \(\angle DAB=180^\circ-70^\circ-85^\circ=25^\circ\). 3. Because \(AD\) bisects \(\angle BAC\), \(\alpha=50^\circ\). 4. Then \(\gamma=180^\circ-50^\circ-70^\circ=60^\circ\).

\(\alpha=50^\circ\) and \(\gamma=60^\circ\)

- Convert the exterior angle at \(B\) to the adjacent interior angle. - Then use the quadrilateral interior-angle sum.

1. The exterior angle and \(\beta\) form a linear pair, so \(\beta=180^\circ-115^\circ=65^\circ\). 2. The interior angles of a quadrilateral total \(360^\circ\). 3. Thus \(75^\circ+65^\circ+\gamma+110^\circ=360^\circ\), giving \(\gamma=110^\circ\).

\(\beta=65^\circ\) and \(\gamma=110^\circ\)

- How is each exterior angle related to its adjacent interior angle? - Find the two unmarked interior angles first. - What is the sum of the interior angles of a triangle?

1. At \(A\), the \(130^\circ\) exterior angle and interior angle \(\alpha\) form a linear pair, so \(\alpha = 180^\circ - 130^\circ = 50^\circ\). 2. At \(C\), the \(120^\circ\) exterior angle and interior angle \(\gamma\) form a linear pair, so \(\gamma = 180^\circ - 120^\circ = 60^\circ\). 3. Use the triangle angle sum: \(50^\circ + \beta + 60^\circ = 180^\circ\). 4. Therefore, \(\beta = 70^\circ\).

\(\beta = 70^\circ\)

- How many sides does the polygon have? - What formula gives the sum of the interior angles of an \(n\)-gon? - After finding the total, how can you use the four given angles to find the missing angle?

1. The sum of the interior angles of an \(n\)-gon is \((n - 2) \cdot 180^\circ\). 2. A pentagon has \(5\) sides, so its interior angles total \((5 - 2) \cdot 180^\circ = 540^\circ\). 3. The four known angles total \(100^\circ + 110^\circ + 120^\circ + 95^\circ = 425^\circ\). 4. Therefore, \(\varphi = 540^\circ - 425^\circ = 115^\circ\).

\(\varphi = 115^\circ\)

- Find \(\angle A\) before using the angle bisector. - Work in triangle \(ABD\) next. - Use the straight line through \(B,D,C\) for the final angle.

1. \(\angle A=180^\circ-70^\circ-60^\circ=50^\circ\). 2. Since \(AD\) bisects \(\angle A\), \(\alpha=25^\circ\). 3. In triangle \(ABD\), \(\beta=180^\circ-70^\circ-25^\circ=85^\circ\). 4. Since \(B,D,C\) are collinear, \(\beta\) and \(\gamma\) form a linear pair, so \(\gamma=95^\circ\).

\(\alpha=25^\circ\), \(\beta=85^\circ\), and \(\gamma=95^\circ\)

- Which two vertices correspond across the symmetry axis? - Then use the quadrilateral interior-angle sum.

1. Symmetry gives \(\delta=120^\circ\). 2. The quadrilateral angle sum gives \(50^\circ+120^\circ+\gamma+120^\circ=360^\circ\). 3. Therefore, \(\gamma=70^\circ\).

\(\gamma=70^\circ\) and \(\delta=120^\circ\)

- What sum do the three adjacent angles have? - Build the angle between the bisectors from half of \(\alpha\), all of \(\beta\), and half of \(\gamma\). - For part b, use \(\alpha+\gamma=120^\circ\) with \(\gamma=2\alpha\).

1. Since \(\alpha+\beta+\gamma=180^\circ\) and \(\beta=60^\circ\), \(\alpha+\gamma=120^\circ\). 2. The angle between the two bisectors is \(\frac{\alpha}{2}+\beta+\frac{\gamma}{2}\). 3. Therefore, it measures \(\frac{\alpha+\gamma}{2}+60^\circ=\frac{120^\circ}{2}+60^\circ=120^\circ\). 4. For part b, substitute \(\gamma=2\alpha\) into \(\alpha+\gamma=120^\circ\): \(\alpha+2\alpha=120^\circ\). 5. Thus, \(3\alpha=120^\circ\), so \(\alpha=40^\circ\) and \(\gamma=80^\circ\).

a) \(120^\circ\). b) \(\alpha=40^\circ\) and \(\gamma=80^\circ\).

- Find the half-angle at \(A\) first. - Then use the angle sum in triangle \(ADC\).

1. Since \(AD\) bisects \(50^\circ\), \(\angle CAD=25^\circ\). 2. In triangle \(ADC\), \(25^\circ+70^\circ+\delta=180^\circ\). 3. Therefore, \(\delta=85^\circ\).

\(\delta=85^\circ\)

- Convert the exterior angle at \(C\) to the adjacent interior angle. - Then subtract the three known interior angles from \(360^\circ\).

1. The interior angle at \(C\) is \(180^\circ-105^\circ=75^\circ\). 2. The quadrilateral angle sum gives \(\beta=360^\circ-(85^\circ+75^\circ+90^\circ)=110^\circ\).

\(\beta=110^\circ\)

- Use the linear pair at \(D\) to get the angle in the left triangle. - Then solve the two smaller triangles separately.

1. Since \(A,D,B\) are collinear, \(\angle ADC=180^\circ-80^\circ=100^\circ\). 2. In triangle \(ADC\), \(\delta_1=180^\circ-55^\circ-100^\circ=25^\circ\). 3. In triangle \(BCD\), \(\delta_2=180^\circ-80^\circ-40^\circ=60^\circ\).

\(\delta_1=25^\circ\) and \(\delta_2=60^\circ\)

- Which angles are directly opposite each other? What is true about those angles? - Which adjacent angles form a straight angle, and what is their sum? - First find the angles determined directly by vertical angle relationships. - Then use a straight angle to find the remaining measure.

1. Angle \(\gamma\) is vertical to the given \(55^\circ\) angle, so \(\gamma = 55^\circ\). 2. Angle \(\alpha\) is vertical to the given \(70^\circ\) angle, so \(\alpha = 70^\circ\). 3. The angles \(55^\circ\), \(\alpha\), and \(\beta\) form a straight angle. Therefore, \(\beta = 180^\circ - 55^\circ - 70^\circ = 55^\circ\). 4. Angle \(\delta\) is vertical to \(\beta\), so \(\delta = 55^\circ\).

\(\alpha = 70^\circ\), \(\beta = 55^\circ\), \(\gamma = 55^\circ\), and \(\delta = 55^\circ\)

- Use the linear pair at the right base vertex first. - Then apply the triangle angle sum.

1. \(\beta=180^\circ-120^\circ=60^\circ\). 2. The triangle angle sum gives \(\gamma=180^\circ-65^\circ-60^\circ=55^\circ\).

\(\beta=60^\circ\) and \(\gamma=55^\circ\)

- Find the interior angles at \(A\), \(B\), and \(C\) first. - What is true about vertical angles? Use this at \(A\). - How can the exterior angle at \(C\) determine the interior angle there? - Use the interior angle sum of a quadrilateral to find \(\delta\).

1. The interior angle at \(A\) is vertical to the given \(120^\circ\) angle, so it measures \(120^\circ\). 2. The interior angle at \(B\) is \(70^\circ\). 3. The interior angle at \(C\) and the given \(75^\circ\) exterior angle form a linear pair, so the interior angle at \(C\) is \(180^\circ - 75^\circ = 105^\circ\). 4. The interior angles of a quadrilateral total \(360^\circ\). Therefore, \(\delta = 360^\circ - (120^\circ + 70^\circ + 105^\circ) = 65^\circ\).

\(\delta = 65^\circ\)

- Find the total measure left for the two unknown angles. - Express \(\delta\) as twice \(\gamma\).

1. The two unknown angles must total \(360^\circ-75^\circ-105^\circ=180^\circ\). 2. Since \(\delta=2\gamma\), \(\gamma+2\gamma=180^\circ\). 3. Thus \(\gamma=60^\circ\) and \(\delta=120^\circ\).

\(\gamma=60^\circ\) and \(\delta=120^\circ\)

- Solve the left triangle first. - Use the straight line through \(A,D,B\) to transfer to the right triangle. - Then use the triangle angle sum again.

1. In triangle \(ADC\), \(\delta=180^\circ-40^\circ-30^\circ=110^\circ\). 2. Since \(A,D,B\) are collinear, \(\angle BDC=70^\circ\). 3. In triangle \(BDC\), \(\gamma_2=180^\circ-70^\circ-55^\circ=55^\circ\).

\(\delta=110^\circ\) and \(\gamma_2=55^\circ\)

- Find the supplement of \(74^\circ\). - Bisect both angles. - Add the two half-angles that lie between the bisectors.

1. \(\beta=180^\circ-74^\circ=106^\circ\). 2. The half-angles are \(37^\circ\) and \(53^\circ\). 3. Therefore, \(\delta=37^\circ+53^\circ=90^\circ\).

\(\delta=90^\circ\)

- Find the interior angle at \(B\) from its exterior linear pair. - Then use the triangle angle sum.

1. The interior angle at \(B\) is \(180^\circ-130^\circ=50^\circ\). 2. Therefore, \(\gamma=180^\circ-40^\circ-50^\circ=90^\circ\).

\(\gamma=90^\circ\)

- Convert the exterior angle to the adjacent interior angle first. - Then use the triangle angle sum.

1. The right interior angle is \(180^\circ-135^\circ=45^\circ\). 2. The triangle angle sum gives \(\alpha=180^\circ-50^\circ-45^\circ=85^\circ\).

\(\alpha=85^\circ\)

- What is the sum of the interior angles of a triangle? - How are angles in a linear pair related? - What is true about vertical angles?

1. The interior angle adjacent to the \(115^\circ\) exterior angle is \(180^\circ - 115^\circ = 65^\circ\). 2. The interior angle at the top vertex is vertical to the marked \(60^\circ\) angle, so it also measures \(60^\circ\). 3. Using the triangle angle sum, \(\alpha = 180^\circ - 65^\circ - 60^\circ = 55^\circ\).

\(\alpha = 55^\circ\)

- What is the sum of the interior angles of a quadrilateral? - Use linear-pair and vertical-angle relationships to find the interior angles at \(C\) and \(D\). - Subtract the three known interior angles from \(360^\circ\).

1. The interior angle at \(D\) and the marked \(110^\circ\) exterior angle form a linear pair, so the interior angle at \(D\) is \(70^\circ\). 2. The interior angle at \(C\) is vertical to the marked \(110^\circ\) angle, so it measures \(110^\circ\). 3. The interior angles of a quadrilateral total \(360^\circ\). Therefore, \(\alpha = 360^\circ - (80^\circ + 110^\circ + 70^\circ) = 100^\circ\).

\(\alpha = 100^\circ\)

- The \(360^\circ\) quadrilateral angle sum still applies to concave quadrilaterals. - The missing angle should be greater than \(180^\circ\).

1. The interior angles of any quadrilateral, including a concave quadrilateral, total \(360^\circ\). 2. The three given angles total \(100^\circ\). 3. Therefore, \(\delta=360^\circ-100^\circ=260^\circ\).

\(\delta=260^\circ\)

- Use one variable for the two congruent angles. - Subtract the two known angles from \(360^\circ\) before dividing by \(2\).

1. The quadrilateral angle sum is \(360^\circ\). 2. Write \(2\alpha+30^\circ+280^\circ=360^\circ\). 3. Then \(2\alpha=50^\circ\), so \(\alpha=25^\circ\).

\(\alpha=25^\circ\)

- Recover the whole angle at \(A\) from the bisected half. - Use the large triangle first, then triangle \(ADC\).

1. Since \(AD\) bisects \(\angle A\), the full angle at \(A\) is \(64^\circ\). 2. In triangle \(ABC\), \(\gamma=180^\circ-64^\circ-44^\circ=72^\circ\). 3. In triangle \(ADC\), \(\delta=180^\circ-32^\circ-72^\circ=76^\circ\).

\(\gamma=72^\circ\) and \(\delta=76^\circ\)

- First find each interior angle that forms a linear pair with a given exterior angle. - What does a \(90^\circ\) exterior angle imply about its adjacent interior angle? - Then use the interior-angle sum of a quadrilateral.

1. At \(A\), \(\alpha = 180^\circ - 110^\circ = 70^\circ\). 2. The interior angle at \(B\) is \(180^\circ - 85^\circ = 95^\circ\). 3. At \(C\), \(\gamma = 180^\circ - 90^\circ = 90^\circ\). 4. The interior angles of a quadrilateral total \(360^\circ\). Therefore, \(\delta = 360^\circ - (70^\circ + 95^\circ + 90^\circ) = 105^\circ\).

\(\alpha = 70^\circ\), \(\gamma = 90^\circ\), and \(\delta = 105^\circ\)

- What do matching arc marks tell you? - First use the given exterior angles to find adjacent interior angles. - Check that the four interior angles total \(360^\circ\).

1. Angles \(\alpha\) and \(\beta\) each form a linear pair with a \(70^\circ\) exterior angle. Therefore, \(\alpha = \beta = 180^\circ - 70^\circ = 110^\circ\). 2. Angle \(\gamma\) forms a linear pair with the \(110^\circ\) exterior angle, so \(\gamma = 70^\circ\). 3. The matching arc marks show that \(\delta = \gamma = 70^\circ\). 4. The check is \(110^\circ + 110^\circ + 70^\circ + 70^\circ = 360^\circ\).

\(\alpha = 110^\circ\), \(\beta = 110^\circ\), \(\gamma = 70^\circ\), and \(\delta = 70^\circ\)

- Use the isosceles-triangle relationships in the left triangle first. - Transfer across the straight line at \(E\). - Then use the isosceles relationship in triangle \(EBD\).

1. In isosceles triangle \(ADE\), \(\angle ADE=35^\circ\). 2. Therefore, \(\angle AED=180^\circ-35^\circ-35^\circ=110^\circ\). 3. Since \(A,E,B\) are collinear, \(\angle DEB=70^\circ\). 4. In isosceles triangle \(EBD\), the two base angles are equal, so \(70^\circ+2\varepsilon=180^\circ\), giving \(\varepsilon=55^\circ\).

\(\varepsilon=55^\circ\)

- Express both ray directions from the same side of line \(g\). - Then subtract the two direction angles.

1. Ray \(u\) forms \(180^\circ-115^\circ=65^\circ\) with the rightward ray of \(g\). 2. Ray \(v\) forms \(110^\circ\) with that same reference ray. 3. Therefore, \(\alpha=110^\circ-65^\circ=45^\circ\).

\(\alpha=45^\circ\)

- Apply the two angle bisectors in order. - Identify which two smaller angles combine to form the requested angle.

1. Ray \(d\) splits \(120^\circ\) into two \(60^\circ\) angles. 2. Ray \(e\) bisects one \(60^\circ\) angle, creating a \(30^\circ\) part. 3. The angle from \(c\) to \(e\) is \(60^\circ+30^\circ=90^\circ\).

The angle between \(c\) and \(e\) is \(90^\circ\).

- Find the supplement of \(126^\circ\) first. - Bisect that angle. - Use the straight line once more for the requested angle.

1. The angle between \(m\) and the rightward ray of \(g\) is \(180^\circ-126^\circ=54^\circ\). 2. Ray \(k\) bisects it, so each half is \(27^\circ\). 3. The requested angle is \(180^\circ-27^\circ=153^\circ\).

The angle measures \(153^\circ\).

- Find the interior-angle sum of a pentagon. - Use symmetry only after subtracting the three known angles.

1. A pentagon's interior angles total \(540^\circ\). 2. The three known angles total \(90^\circ+90^\circ+110^\circ=290^\circ\). 3. The two congruent angles total \(250^\circ\), so \(\gamma=125^\circ\).

\(\gamma=125^\circ\)

- Find the interior-angle sum of a hexagon. - Subtract the two right angles before dividing by \(4\).

1. A hexagon's interior angles total \(720^\circ\). 2. Subtract the two right angles: \(720^\circ-180^\circ=540^\circ\). 3. Divide the remaining total among four congruent angles: \(\gamma=135^\circ\).

\(\gamma=135^\circ\)

- Identify the upper and lower triangles. - What is the relationship between the opposite angles at the central intersection? - Use the triangle angle sum in each triangle. - Lines \(g\) and \(h\) do not need to be parallel.

1. In the lower triangle, the angle at the intersection of \(s\) and \(t\) is \(180^\circ - 45^\circ - 45^\circ = 90^\circ\). 2. The angle at the same intersection in the upper triangle is vertical to that angle, so it also measures \(90^\circ\). 3. Use the angle sum of the upper triangle: \(\delta = 180^\circ - 90^\circ - 35^\circ = 55^\circ\).

\(\delta = 55^\circ\)

- Start with the large triangle and use its angle sum. - Then use the smaller triangle that contains the obtuse angle at \(D\). - The two marked angles together form the right angle at \(C\).

1. In triangle \(ABC\), \(\angle B=180^\circ-90^\circ-25^\circ=65^\circ\). 2. In triangle \(BDC\), \(\alpha=180^\circ-65^\circ-100^\circ=15^\circ\). 3. The right angle at \(C\) is divided into \(\alpha\) and \(\beta\), so \(\beta=90^\circ-15^\circ=75^\circ\).

\(\alpha=15^\circ\) and \(\beta=75^\circ\)

- Start with the smaller triangle that already has two known angles. - The two angles at \(D\) form a linear pair. - Then use the angle sum in the other smaller triangle.

1. In triangle \(BDC\), \(\beta=180^\circ-50^\circ-35^\circ=95^\circ\). 2. The two angles at \(D\) form a linear pair, so \(\angle ADC=180^\circ-95^\circ=85^\circ\). 3. In triangle \(ADC\), \(\alpha=180^\circ-85^\circ-25^\circ=70^\circ\).

\(\alpha=70^\circ\) and \(\beta=95^\circ\)

- Begin with the smaller triangle that has two known angles. - Use the linear pair at \(D\) to move from one smaller triangle to the other. - Apply the triangle angle sum to the second smaller triangle.

1. In triangle \(ADC\), \(\alpha=180^\circ-45^\circ-85^\circ=50^\circ\). 2. The two angles at \(D\) form a linear pair, so \(\angle BDC=180^\circ-85^\circ=95^\circ\). 3. In triangle \(BDC\), \(\beta=180^\circ-95^\circ-55^\circ=30^\circ\).

\(\alpha=50^\circ\) and \(\beta=30^\circ\)

- Use each exterior angle and its adjacent interior angle as a linear pair. - Then use the triangle angle sum. - Classify the triangle from its equal angles and its right angle.

1. Each base interior angle forms a linear pair with a \(135^\circ\) exterior angle, so \(\angle A=\angle B=180^\circ-135^\circ=45^\circ\). 2. The third interior angle is \(\angle C=180^\circ-45^\circ-45^\circ=90^\circ\). 3. Because two angles are congruent, the triangle is isosceles. Because one angle is a right angle, it is also a right triangle.

The interior angles are \(45^\circ\), \(45^\circ\), and \(90^\circ\). The triangle is an isosceles right triangle.

- Use the equal side lengths to identify two isosceles triangles. - Find the exterior angle at \(D\) that connects the two smaller triangles. - The full angle \(\beta\) is made from two adjacent angles at \(B\).

1. Since \(AD=BD\), triangle \(ABD\) is isosceles, so \(\angle ABD=\alpha=25^\circ\). 2. The exterior angle \(\angle BDC\) of triangle \(ABD\) is \(25^\circ+25^\circ=50^\circ\). 3. Since \(BD=CD\), triangle \(BCD\) is isosceles. Its two base angles each measure \((180^\circ-50^\circ)\div2=65^\circ\), so \(\gamma=65^\circ\). 4. The full angle at \(B\) is \(\beta=25^\circ+65^\circ=90^\circ\).

\(\beta=90^\circ\) and \(\gamma=65^\circ\)

- Use the equal side lengths to identify the two isosceles triangles. - Connect the two triangles through the exterior angle at \(M\). - Then use the triangle angle sum in \(BCM\).

1. Since \(AM=MB\), triangle \(ABM\) is isosceles, so \(\angle ABM=\angle BAM=38^\circ\). 2. The exterior angle \(\angle BMC\) of triangle \(ABM\) is \(38^\circ+38^\circ=76^\circ\). 3. Since \(MB=MC\), triangle \(BCM\) is isosceles. Its base angles are congruent, so \(\gamma=(180^\circ-76^\circ)\div2=52^\circ\).

\(\gamma=52^\circ\)

- What angles of triangle \(ABI\) are created by the two angle bisectors? - Use the angle sum in triangle \(ABI\). - Then use the angle sum in triangle \(ABC\).

1. In triangle \(ABI\), the angles at \(A\) and \(B\) are half of the corresponding angles of triangle \(ABC\), so \(\frac{\angle A}{2}+\frac{\angle B}{2}+130^\circ=180^\circ\). 2. Thus, \(\frac{\angle A}{2}+\frac{\angle B}{2}=50^\circ\), so \(\angle A+\angle B=100^\circ\). 3. Using the angle sum of triangle \(ABC\), \(\angle C=180^\circ-100^\circ=80^\circ\).

\(\angle C=80^\circ\)

- The angles in a linear pair sum to \(180^\circ\). - Use the stated relationship between \(\alpha\) and \(\beta\) to write the supplementary-angle equation with one unknown.

1. A linear pair is supplementary, so \(\alpha + \beta = 180^\circ\). 2. Substitute \(\alpha = 90^\circ + \beta\): \((90^\circ + \beta) + \beta = 180^\circ\). 3. Solve: \(2\beta = 90^\circ\), so \(\beta = 45^\circ\). 4. Then \(\alpha = 90^\circ + 45^\circ = 135^\circ\).

\(\alpha = 135^\circ\) and \(\beta = 45^\circ\)

- Represent the smaller angle as one part and the larger angle as four equal parts of a \(180^\circ\) total. - Let one part be an unknown angle measure, then express the larger angle using that same unknown.

1. Let \(\beta\) be the smaller angle. Then \(\alpha = 4\beta\). 2. Since the angles form a linear pair, \(4\beta + \beta = 180^\circ\). 3. Solve: \(5\beta = 180^\circ\), so \(\beta = 36^\circ\). 4. Then \(\alpha = 4 \cdot 36^\circ = 144^\circ\).

The angle measures are \(36^\circ\) and \(144^\circ\).

- Into how many equal ratio parts is the \(180^\circ\) straight angle divided? - After finding the size of one ratio part, match one part to \(\alpha\) and five parts to \(\beta\).

1. The ratio \(1:5\) divides the \(180^\circ\) total into \(1 + 5 = 6\) equal parts. 2. One part measures \(180^\circ \div 6 = 30^\circ\). 3. Therefore, \(\alpha = 30^\circ\) and \(\beta = 5 \cdot 30^\circ = 150^\circ\).

\(\alpha = 30^\circ\) and \(\beta = 150^\circ\)

- What is the total measure of all angles around a point? - Once you know one angle, which angle is congruent to it? - What relationship gives the two adjacent angle measures?

1. All four angles around the intersection total \(360^\circ\), so the omitted angle measures \(360^\circ - 305^\circ = 55^\circ\). 2. Its vertical angle also measures \(55^\circ\). 3. Each adjacent angle forms a linear pair with a \(55^\circ\) angle, so each measures \(180^\circ - 55^\circ = 125^\circ\).

Two angles measure \(55^\circ\), and two angles measure \(125^\circ\).

- The four angles around the intersection total \(360^\circ\). - Write an equation comparing the unknown angle with the sum of the other three.

1. Let \(\alpha\) be the unknown angle. The other three angles have a total measure of \(360^\circ - \alpha\). 2. Write the equation \(\alpha = \frac{1}{4}(360^\circ - \alpha)\). 3. Solve: \(4\alpha = 360^\circ - \alpha\), so \(5\alpha = 360^\circ\) and \(\alpha = 72^\circ\).

\(\alpha = 72^\circ\)

- First find the interior angle adjacent to the exterior angle at \(A\). - Then use the triangle angle sum and compare the three interior angles.

1. The interior angle at \(A\) is \(180^\circ-108^\circ=72^\circ\). 2. Use the triangle angle sum: \(\angle C=180^\circ-72^\circ-36^\circ=72^\circ\). 3. Since \(\angle A=\angle C=72^\circ\), the triangle is isosceles with base \(AC\).

\(\angle C=72^\circ\). The triangle is isosceles because \(\angle A=\angle C\).

- Convert each exterior angle to its adjacent interior angle. - What does equality of two interior angles tell you about the triangle?

1. Each corresponding interior angle forms a linear pair with a \(125^\circ\) exterior angle, so \(\angle A=\angle C=180^\circ-125^\circ=55^\circ\). 2. The remaining angle is \(\angle B=180^\circ-55^\circ-55^\circ=70^\circ\). 3. Since two interior angles are congruent, the triangle is isosceles.

\(\angle A=55^\circ\), \(\angle B=70^\circ\), and \(\angle C=55^\circ\). The triangle is isosceles.

- First find the interior angle adjacent to the exterior angle at \(B\). - Then use that angle with the interior angle at \(C\) and the triangle angle total.

1. The interior angle at \(B\) forms a linear pair with the exterior angle, so \(\angle B=180^\circ-100^\circ=80^\circ\). 2. Use the triangle angle sum: \(\alpha=180^\circ-80^\circ-38^\circ=62^\circ\).

\(\alpha=62^\circ\)

- First find the interior angles at \(A\) and \(B\). - Each shown exterior angle and its adjacent interior angle form a linear pair; then use the triangle angle total.

1. Convert the exterior angles to interior angles: \(\angle A=180^\circ-130^\circ=50^\circ\) and \(\angle B=180^\circ-110^\circ=70^\circ\). 2. Use the triangle angle sum: \(\gamma=180^\circ-50^\circ-70^\circ=60^\circ\).

\(\gamma=60^\circ\)

- First find the interior angle adjacent to the exterior angle at \(B\). - Then use the relationship between the two acute angles of a right triangle.

1. The interior angle at \(B\) is \(180^\circ-140^\circ=40^\circ\). 2. The two acute angles of a right triangle sum to \(90^\circ\), so \(\alpha=90^\circ-40^\circ=50^\circ\).

\(\alpha=50^\circ\)

- What is the relationship between vertical angles? - After finding \(\alpha\), use the triangle angle sum.

1. Vertical angles are congruent, so \(\alpha=42^\circ\). 2. Use the triangle angle sum: \(\beta=180^\circ-42^\circ-68^\circ=70^\circ\).

\(\beta=70^\circ\)

- Convert the exterior angle at \(A\) to its adjacent interior angle. - Use the equal legs to identify the congruent base angle, then finish with the triangle angle sum.

1. The interior angle at \(A\) is \(180^\circ-135^\circ=45^\circ\). 2. The base angles are congruent, so \(\angle B=45^\circ\). 3. The vertex angle is \(\angle C=180^\circ-45^\circ-45^\circ=90^\circ\).

\(\angle A=45^\circ\), \(\angle B=45^\circ\), and \(\angle C=90^\circ\)

- Use the marked \(65^\circ\) angle and first find the interior angle adjacent to \(110^\circ\). - Then apply the triangle angle sum.

1. The left interior angle of the triangle above the horizontal line measures \(65^\circ\). 2. The right interior angle forms a linear pair with the \(110^\circ\) angle, so it measures \(180^\circ - 110^\circ = 70^\circ\). 3. Use the triangle angle sum: \(x = 180^\circ - 65^\circ - 70^\circ = 45^\circ\).

\(x = 45^\circ\)

- Recall the sum of two consecutive angles in a parallelogram. - Represent the smaller angle with one variable. - Write an equation using the four-to-one relationship. - Recall the relationship between opposite angles in a parallelogram.

1. Consecutive angles in a parallelogram are supplementary, so \(\angle K + \angle L = 180^\circ\). 2. Let \(\angle L = x\). Then \(\angle K = 4x\), so \(x + 4x = 180^\circ\). 3. Solve: \(5x = 180^\circ\), so \(x = 36^\circ\). Therefore, \(\angle L = 36^\circ\) and \(\angle K = 4 \cdot 36^\circ = 144^\circ\). 4. Opposite angles in a parallelogram are congruent. Thus, \(\angle M = 144^\circ\) and \(\angle N = 36^\circ\).

\(\angle K = 144^\circ\), \(\angle L = 36^\circ\), \(\angle M = 144^\circ\), and \(\angle N = 36^\circ\)

- What is the sum of adjacent angles in a parallelogram? - What does an angle bisector do? - Use the angle sum of triangle \(ABP\).

1. Adjacent angles in a parallelogram are supplementary, so \(\alpha + \beta = 180^{\circ}\). 2. The angle bisectors create angles of \(\frac{\alpha}{2}\) and \(\frac{\beta}{2}\) in triangle \(ABP\). 3. Thus \(\frac{\alpha}{2} + \frac{\beta}{2} = \frac{180^{\circ}}{2} = 90^{\circ}\). 4. By the triangle angle-sum theorem, \(\angle APB = 180^{\circ} - 90^{\circ} = 90^{\circ}\).

\(\angle APB = 90^{\circ}\)

- Use the equal lengths \(AB=AM\) to identify an isosceles triangle. - Then apply the angle relationships in a parallelogram.

1. Since \(AB=AM\), triangle \(ABM\) is isosceles, so \(\angle ABM=\angle BMA=70^\circ\). 2. The angle at \(A\) is \(180^\circ-70^\circ-70^\circ=40^\circ\). 3. Opposite angles of a parallelogram are congruent, so \(\angle C=40^\circ\). 4. Consecutive angles are supplementary, so \(\angle B=\angle D=180^\circ-40^\circ=140^\circ\).

\(\angle A=40^\circ\), \(\angle B=140^\circ\), \(\angle C=40^\circ\), and \(\angle D=140^\circ\)

- Use the isosceles triangle formed by \(A\), \(D\), and \(C\). - Then connect the diagonal to the parallel bases.

1. Since \(AD=DC\), triangle \(ADC\) is isosceles. 2. Its base angles are congruent, so \(\angle DAC=\angle DCA=(180^\circ-110^\circ)\div2=35^\circ\). 3. Since \(AB\parallel CD\), \(\angle BAC\) and \(\angle DCA\) are alternate interior angles. Therefore, \(\angle BAC=35^\circ\).

\(\angle BAC=35^\circ\)

- First find the exterior angle corresponding to the \(140^\circ\) interior angle. - The exterior angles of any polygon total \(360^\circ\).

1. Each exterior angle is supplementary to its interior angle, so it measures \(180^\circ - 140^\circ = 40^\circ\). 2. The exterior angles of a polygon total \(360^\circ\), so the number of vertices is \(n = \frac{360^\circ}{40^\circ} = 9\).

The polygon has \(9\) vertices.

- Read the exact time from the clock first. - The minute hand moves \(6^\circ\) per minute, while the hour hand also moves between hour marks. - Find both positions from \(12\), then compare them.

1. The clock shows \(2{:}15\) p.m. At \(15\) minutes, the minute hand is \(15\cdot6^\circ=90^\circ\) clockwise from \(12\). 2. The hour hand is \(2\cdot30^\circ+15\cdot0.5^\circ=67.5^\circ\) clockwise from \(12\). 3. The smaller angle between the hands is \(|90^\circ-67.5^\circ|=22.5^\circ\).

\(22.5^\circ\)

- Read the exact time from the clock first. - Each hour mark represents \(30^\circ\), and the hour hand moves between hour marks as minutes pass. - Find each hand's position relative to \(12\), then compare the two positions.

1. The clock shows \(8{:}20\) p.m. At \(20\) minutes, the minute hand is \(20\cdot6^\circ=120^\circ\) clockwise from \(12\). 2. The hour hand is \(8\cdot30^\circ+20\cdot0.5^\circ=250^\circ\) clockwise from \(12\). 3. The difference is \(250^\circ-120^\circ=130^\circ\), which is the smaller angle.

\(130^\circ\)

- First find the measure of the supplementary angle. - Bisect each angle. - Add the two half-angles that form the requested angle.

1. Since \(\alpha\) and \(\beta\) are supplementary, \(\beta = 180^{\circ} - 124^{\circ} = 56^{\circ}\). 2. Their half-angles are \(124^{\circ} \div 2 = 62^{\circ}\) and \(56^{\circ} \div 2 = 28^{\circ}\). 3. The angle between the bisectors is \(\delta = 62^{\circ} + 28^{\circ} = 90^{\circ}\).

The angle between the two angle bisectors is \(90^{\circ}\).

- If you know half of \(\alpha\), how can you find the full angle? - What is the sum of a linear pair? - How does \(w_\beta\) divide \(\beta\)?

1. Since \(w_\alpha\) bisects \(\alpha\), \(\alpha=2\cdot37^\circ=74^\circ\). 2. The angles form a linear pair, so \(\beta=180^\circ-74^\circ=106^\circ\). 3. Since \(w_\beta\) bisects \(\beta\), the angle between \(s\) and \(w_\beta\) is \(106^\circ\div2=53^\circ\).

\(\alpha=74^\circ\), \(\beta=106^\circ\), and the angle between \(s\) and \(w_\beta\) is \(53^\circ\).

- What is the sum of adjacent angles formed by two intersecting lines? - Express the smaller angle in terms of the larger angle. - After finding one adjacent pair, which opposite angles must match them?

1. Let \(\alpha\) be the smaller angle and \(\beta\) the adjacent larger angle. Then \(\alpha + \beta = 180^\circ\) and \(\alpha = \beta - 60^\circ\). 2. Substitute: \((\beta - 60^\circ) + \beta = 180^\circ\). 3. Solve: \(2\beta = 240^\circ\), so \(\beta = 120^\circ\) and \(\alpha = 60^\circ\). 4. Vertical angles are congruent, so the four angle measures alternate between \(60^\circ\) and \(120^\circ\).

The four angles measure \(60^\circ\), \(120^\circ\), \(60^\circ\), and \(120^\circ\).

- What is true about the two angles not on the line of symmetry? - What is the sum of the interior angles of a quadrilateral? - Write an equation with the unknown angle.

1. In a kite with symmetry line \(AC\), the opposite angles at \(B\) and \(D\) are congruent, so \(\beta = \delta\). 2. The interior angles of a quadrilateral sum to \(360^{\circ}\), so \(80^{\circ} + \beta + 40^{\circ} + \beta = 360^{\circ}\). 3. Combine terms: \(120^{\circ} + 2\beta = 360^{\circ}\). 4. Solve: \(2\beta = 240^{\circ}\), so \(\beta = 120^{\circ}\).

The measure of \(\beta\) is \(120^{\circ}\).

- Use the angle sum of triangle \(ABD\) to find \(\angle DAB\). - Recall the relationship between opposite angles of a parallelogram.

1. The angle sum of triangle \(ABD\) is \(180^\circ\), so \(\angle DAB=180^\circ-42^\circ-38^\circ=100^\circ\). 2. Opposite angles of a parallelogram are congruent, so \(\angle BCD=\angle DAB=100^\circ\).

\(\angle BCD=100^\circ\)

- Use the equal side lengths in the square and the equilateral triangle. - Identify the isosceles triangles formed by point \(M\). - Write \(\angle AMC\) as a sum of two angles at \(M\).

1. Since \(AD = AM\), triangle \(ADM\) is isosceles. Also, \(\angle DAM = 90^\circ - 60^\circ = 30^\circ\), so \(\angle AMD = 75^\circ\). 2. Similarly, \(BC = BM\) and \(\angle CBM = 30^\circ\), so \(\angle BMC = 75^\circ\). 3. Therefore, \(\angle AMC = \angle AMB + \angle BMC = 60^\circ + 75^\circ = 135^\circ\).

\(\angle AMC = 135^\circ\)

- Sketch the square and attach an equilateral triangle along one side. - What are the angle measures in a square and an equilateral triangle? - Which sides must be congruent because of the definitions of a square and an equilateral triangle? - What is true about the base angles of an isosceles triangle?

1. At \(C\), the square contributes a \(90^\circ\) angle and the equilateral triangle contributes a \(60^\circ\) angle. Therefore, \(m\angle DCE = 90^\circ + 60^\circ = 150^\circ\). 2. In the square, \(DC = BC\). In the equilateral triangle, \(BC = CE\). Therefore, \(DC = CE\), so \(\triangle DCE\) is isosceles. 3. The two base angles of \(\triangle DCE\) are congruent. Their sum is \(180^\circ - 150^\circ = 30^\circ\), so \(m\angle EDC = 30^\circ \div 2 = 15^\circ\).

a) \(m\angle DCE = 150^\circ\) b) Since \(DC = BC\) and \(BC = CE\), it follows that \(DC = CE\). Thus, \(\triangle DCE\) is isosceles. c) \(m\angle EDC = 15^\circ\)

- Substitute the given angle into the formula for one interior angle of a regular polygon. - Double the number of sides found in part a. - Calculate the new angle and compare it with twice \(150^\circ\).

1. For a regular polygon, \(\alpha = \frac{(n - 2) \cdot 180^\circ}{n}\). 2. Substitute \(\alpha = 150^\circ\): \(150n = 180(n - 2)\). 3. Solve: \(150n = 180n - 360\), so \(30n = 360\) and \(n = 12\). 4. Doubling the number of sides gives \(24\) sides. The new interior angle is \(\frac{(24 - 2) \cdot 180^\circ}{24} = 165^\circ\). 5. Since \(165^\circ \ne 2 \cdot 150^\circ\), the claim is false.

a) The polygon has \(12\) sides. b) The claim is false. A regular polygon with \(24\) sides has interior angles of \(165^\circ\), not \(300^\circ\).

- How are an interior angle and its adjacent exterior angle related? - Express the vertex angle in terms of a base angle. - Use one variable for both congruent base angles.

1. Let each base angle be \(\alpha\), and let the vertex angle be \(\gamma\). 2. The exterior angle at a base vertex is \(180^\circ - \alpha\). 3. The given relationship is \(180^\circ - \alpha = \gamma + 45^\circ\). 4. Since the triangle is isosceles, \(\gamma = 180^\circ - 2\alpha\). 5. Substitute: \(180^\circ - \alpha = 180^\circ - 2\alpha + 45^\circ\). 6. Solving gives \(\alpha = 45^\circ\). 7. Then \(\gamma = 180^\circ - 2 \cdot 45^\circ = 90^\circ\).

The interior angles are \(45^\circ\), \(45^\circ\), and \(90^\circ\).

- The hour hand does not remain fixed between whole hours. - Find how far the hour hand moves in \(15\) minutes. - Subtract that movement from \(90^\circ\).

1. During the first \(15\) minutes after \(12{:}00\), the hour hand moves partway from \(12\) toward \(1\), so it is no longer at \(12\). 2. The minute hand is at \(90^\circ\) from \(12\). 3. The hour hand moves \(30^\circ\) in \(60\) minutes, so in \(15\) minutes it moves \(30^\circ \cdot \frac{15}{60}=7.5^\circ\). 4. The angle between the hands is \(90^\circ-7.5^\circ=82.5^\circ\).

a) The hour hand moves continuously and has already moved toward \(1\). b) \(82.5^\circ\)

- Which adjacent angles form a straight angle? - What is the sum of the interior angles of a triangle? - Which opposite angles are congruent when two lines intersect? - Find the missing angle in the left triangle before working with the right triangle.

1. The \(75^\circ\) angle and \(\alpha\) form a linear pair, so \(\alpha = 180^\circ - 75^\circ = 105^\circ\). 2. The interior angles of triangle \(ABS\) total \(180^\circ\). Therefore, \(\beta = 180^\circ - 75^\circ - 65^\circ = 40^\circ\). 3. The angles \(\angle ASB\) and \(\angle CSD\) are vertical angles, so \(\angle CSD = \beta = 40^\circ\). 4. The interior angles of triangle \(CDS\) total \(180^\circ\). Therefore, \(\gamma = 180^\circ - 40^\circ - 50^\circ = 90^\circ\).

\(\alpha = 105^\circ\), \(\beta = 40^\circ\), and \(\gamma = 90^\circ\)

- What is true about vertical angles and about adjacent angles that form a straight line? - Consider each triangle separately. What is the sum of its interior angles? - How can the angles at \(S\) connect the two triangles? - Look for angles that form a linear pair.

1. The given \(110^\circ\) angle and the interior angle of the left triangle at \(S\) form a linear pair. That interior angle is \(180^\circ - 110^\circ = 70^\circ\). 2. In the left triangle, \(\alpha = 180^\circ - 45^\circ - 70^\circ = 65^\circ\). 3. The interior angles at \(S\) in the two triangles are vertical angles, so the angle at \(S\) in the right triangle is also \(70^\circ\). 4. In the right triangle, \(\beta = 180^\circ - 70^\circ - 45^\circ = 65^\circ\). 5. The \(45^\circ\) angle and \(\gamma\) form a linear pair, so \(\gamma = 180^\circ - 45^\circ = 135^\circ\).

\(\alpha = 65^\circ\), \(\beta = 65^\circ\), and \(\gamma = 135^\circ\)

- Which triangles can you identify in the diagram? - What is the sum of the interior angles of a triangle? - What angle relationships occur where two lines intersect? - What is the sum of adjacent angles that form a straight line? - Which unknown can you find first using one triangle?

1. In the large triangle, \(\alpha = 180^\circ - 60^\circ - 55^\circ = 65^\circ\). 2. In the upper small triangle, the angle adjacent to the given \(135^\circ\) angle is \(180^\circ - 135^\circ = 45^\circ\). 3. Therefore, \(\beta = 180^\circ - 55^\circ - 45^\circ = 80^\circ\). 4. Angles \(\beta\) and \(\gamma\) form a linear pair, so \(\gamma = 180^\circ - 80^\circ = 100^\circ\).

\(\alpha = 65^\circ\), \(\beta = 80^\circ\), and \(\gamma = 100^\circ\)

- Use the converse of the isosceles triangle theorem. - Find the missing angle in triangle \(ABD\). - For part b, first find \(\angle C\) in the large triangle.

1. In triangle \(ABD\), \(\angle ADB=180^\circ-36^\circ-72^\circ=72^\circ\). Since \(\angle ABD=\angle ADB\), the opposite sides satisfy \(AD=AB\). 2. In triangle \(ABC\), \(\angle C=180^\circ-72^\circ-72^\circ=36^\circ\). Thus in triangle \(ADC\), \(\angle DAC=\angle ACD=36^\circ\), so \(AD=CD\).

a) Triangle \(ABD\) is isosceles with \(AB=AD\). b) Triangle \(ADC\) is isosceles with \(AD=CD\).

- Find \(\beta\) first. - Take half of each smaller angle. - Compare their sum with the original \(120^\circ\) angle.

1. \(\beta=120^\circ-40^\circ=80^\circ\). 2. The bisected parts next to the dividing ray are \(20^\circ\) and \(40^\circ\). 3. Thus \(\gamma=60^\circ\). 4. In general, the angle between the internal bisectors of two adjacent angles is half their combined angle.

\(\gamma=60^\circ\). In general, \(\gamma\) is one-half of the original angle.

- Start with isosceles triangle \(BDC\). - Use the linear pair at \(D\) to move to triangle \(ABD\). - Add the two parts of angle \(B\) only after solving both small triangles.

1. In isosceles triangle \(BDC\), \(\gamma=\frac{180^\circ-116^\circ}{2}=32^\circ\). 2. Since \(A,D,C\) are collinear, \(\angle ADB=64^\circ\). 3. In isosceles triangle \(ABD\), \(\alpha=\frac{180^\circ-64^\circ}{2}=58^\circ\). 4. The full angle at \(B\) is \(58^\circ+32^\circ=90^\circ\), so triangle \(ABC\) is right.

a) \(\alpha=58^\circ\) and \(\gamma=32^\circ\) b) \(\angle ABC=90^\circ\), so \(ABC\) is a right triangle.

- Start with the linear pair containing \(\beta\). - Use the angle sum of the left triangle. - Then use a linear pair and vertical angles to transfer the information to the right triangle. - Finish with the angle sum of the right triangle.

1. The interior angle adjacent to \(\beta\) in the left triangle is \(180^\circ - 115^\circ = 65^\circ\). 2. The third angle of the left triangle is \(180^\circ - 65^\circ - 45^\circ = 70^\circ\). 3. The adjacent angle in the right triangle forms a linear pair with that \(70^\circ\) angle, so it measures \(180^\circ - 70^\circ = 110^\circ\). 4. The top angle of the right triangle is vertical to \(\gamma\), so it measures \(40^\circ\). 5. Use the triangle angle sum: \(x = 180^\circ - 110^\circ - 40^\circ = 30^\circ\).

\(x = 30^\circ\)

- Use \(AD=DC\) to relate angles in triangle \(ADC\). - Use the parallel bases to connect an angle along diagonal \(AC\). - Use the right triangle formed by \(A\), \(B\), and \(C\).

1. Let \(\angle BAC=\alpha\). Since \(AB\parallel CD\), alternate interior angles give \(\angle DCA=\alpha\). 2. Since \(AD=DC\), triangle \(ADC\) is isosceles, so \(\angle DAC=\angle DCA=\alpha\). Therefore, \(\angle A=2\alpha\). 3. An isosceles trapezoid has congruent base angles, so \(\angle B=2\alpha\). 4. Triangle \(ABC\) is right at \(C\), so \(\alpha+2\alpha=90^\circ\). Thus, \(\alpha=30^\circ\). 5. Therefore, \(\angle A=\angle B=60^\circ\), and the supplementary upper angles satisfy \(\angle C=\angle D=120^\circ\).

\(\angle A=60^\circ\), \(\angle B=60^\circ\), \(\angle C=120^\circ\), and \(\angle D=120^\circ\)

- Use the side lengths of the square and the equilateral triangle to identify isosceles triangles. - Find the angles at \(P\) in triangles \(ABP\) and \(DCP\). - The angles around point \(P\) total \(360^\circ\).

1. Because \(ABCD\) is a square and triangle \(BCP\) is equilateral, \(AB = BP\) and \(CD = CP\). 2. At \(B\), \(\angle ABP = 90^\circ - 60^\circ = 30^\circ\). 3. Triangle \(ABP\) is isosceles, so \(\angle BPA = \frac{180^\circ - 30^\circ}{2} = 75^\circ\). 4. In the same way, \(\angle CPD = 75^\circ\). 5. The angles around \(P\) total \(360^\circ\), so \(\angle APD = 360^\circ - 75^\circ - 60^\circ - 75^\circ = 150^\circ\).

\(\angle APD = 150^\circ\)

- Compare the side lengths of the square and the equilateral triangle. - Find the vertex angle of isosceles triangle \(ABP\). - Use the symmetry of the figure and the \(60^\circ\) angle at \(P\).

1. Since \(AB = BC = BP\), triangle \(ABP\) is isosceles. 2. Its vertex angle is \(\angle ABP = 90^\circ + 60^\circ = 150^\circ\), so \(\angle BPA = \frac{180^\circ - 150^\circ}{2} = 15^\circ\). 3. Similarly, triangle \(DCP\) is isosceles and \(\angle DPC = 15^\circ\). 4. Since \(\angle BPC = 60^\circ\), \(\angle APD = 60^\circ - 15^\circ - 15^\circ = 30^\circ\).

\(\angle APD = 30^\circ\)

- A rhombus has four congruent sides. - Recall how the diagonals of a rhombus meet and how they divide vertex angles. - Use triangle angle sums after identifying the equal sides.

1. In a rhombus, all four sides are congruent. In \(\triangle PQS\), \(PQ=PS\), and the included angle at \(P\) is \(60^\circ\). The other two angles are each \((180^\circ-60^\circ)\div2=60^\circ\), so \(\triangle PQS\) is equilateral and acute. 2. In \(\triangle PQR\), \(PQ=QR\), so it is isosceles. The angle at \(Q\) is supplementary to the \(60^\circ\) angle at \(P\), so it is \(120^\circ\). Thus \(\triangle PQR\) is obtuse and isosceles. 3. The diagonals of a rhombus are perpendicular and bisect the vertex angles. In \(\triangle PQZ\), the angles are \(30^\circ\) at \(P\), \(60^\circ\) at \(Q\), and \(90^\circ\) at \(Z\). It is a scalene right triangle.

a) \(\triangle PQS\) is equilateral and acute. b) \(\triangle PQR\) is isosceles and obtuse. c) The angles are \(30^\circ\), \(60^\circ\), and \(90^\circ\); the triangle is scalene and right.