Free math worksheets

- Recall the cosine value at \(90^\circ\). - What happens to a product when one factor is zero? - Which side is opposite the right angle?

1. \(\cos(90^\circ)=0\). 2. Substituting gives \(c^2=a^2+b^2-2ab(0)=a^2+b^2\), which is the Pythagorean theorem. 3. Since \(c\) is opposite \(\gamma\), when \(\gamma=90^\circ\), side \(c\) is the hypotenuse. The resulting equation is exactly the right-triangle relationship between the hypotenuse and legs.

1. \(\cos(90^\circ)=0\) 2. The Pythagorean theorem results: \(c^2=a^2+b^2\). 3. With \(\gamma=90^\circ\), \(c\) is the hypotenuse, so the formula matches right-triangle geometry.

- The given angle is between sides \(a\) and \(b\). - Use the Law of Cosines to find the side opposite \(\gamma\).

1. Use the Law of Cosines: \(c^2=a^2+b^2-2ab\cos(\gamma)\). 2. Substitute the given values: \(c^2=(8.5)^2+(12.0)^2-2\cdot 8.5\cdot 12.0\cos(42^\circ)\). 3. Thus, \(c=\sqrt{(8.5)^2+(12.0)^2-2\cdot 8.5\cdot 12.0\cos(42^\circ)}\approx 8.04\,\text{cm}\).

\(c\approx 8.04\,\text{cm}\)

- Read the three side lengths and the target angle from the diagram. - Use the Law of Cosines because all three side lengths are known. - Put the side opposite \(\gamma\) on the left side before solving for \(\cos(\gamma)\).

1. Use the Law of Cosines: \(c^2=a^2+b^2-2ab\cos(\gamma)\). 2. Substitute the side lengths from the diagram: \(7^2=5^2+8^2-2\cdot5\cdot8\cos(\gamma)\). 3. Simplify: \(49=89-80\cos(\gamma)\), so \(80\cos(\gamma)=40\). 4. Therefore, \(\cos(\gamma)=\frac{1}{2}=0.5\).

\(\cos(\gamma)=0.5\)

- The diagram gives two sides and their included angle. - Use the Law of Cosines to find the side opposite the marked angle. - Remember that cosine is negative for an obtuse angle.

1. Use the Law of Cosines with the two known sides and their included angle: \(x^2=3^2+5^2-2\cdot3\cdot5\cos(120^\circ)\). 2. Since \(\cos(120^\circ)=-\frac{1}{2}\), \(x^2=9+25+15=49\). 3. Therefore, \(x=7\,\text{cm}\).

\(7\,\text{cm}\)

- All three side lengths are shown in the diagram. - Solve the Law of Cosines for the cosine of the target angle. - A negative cosine is consistent with an obtuse angle.

1. By the Law of Cosines, \(c^2=a^2+b^2-2ab\cos(x)\). 2. Substitute the side lengths from the diagram: \(7^2=3^2+5^2-2\cdot3\cdot5\cos(x)\). 3. Thus, \(\cos(x)=\frac{3^2+5^2-7^2}{2\cdot3\cdot5}=-\frac{1}{2}\). 4. Therefore, \(x=\cos^{-1}\left(-\frac{1}{2}\right)=120^\circ\).

\(x=120^\circ\)

- Solve the Law of Cosines for the cosine of the angle opposite side \(c\). - Recall the sign of cosine for acute, right, and obtuse angles. - Compare \(a^2+b^2\) with \(c^2\).

1. Solve the Law of Cosines for \(\cos(\gamma)\): \(\cos(\gamma)=\frac{a^2+b^2-c^2}{2ab}\). 2. Substitute: \(\cos(\gamma)=\frac{6^2+9^2-13^2}{2\cdot6\cdot9}=\frac{36+81-169}{108}\). 3. Therefore, \(\cos(\gamma)=\frac{-52}{108}=-\frac{13}{27}\approx-0.4815\). 4. Since \(\cos(\gamma)<0\) for an angle between \(0^\circ\) and \(180^\circ\), \(\gamma\) is obtuse.

\(\cos(\gamma)=-\frac{13}{27}\approx-0.4815\). Because the cosine is negative, \(\gamma\) is obtuse.

- In an isosceles triangle, the two legs have equal lengths. - Substitute \(b=c=s\) into the Law of Cosines before calculating. - Pay attention to the sign of \(\cos(110^\circ)\).

1. Apply the Law of Cosines to the base: \(a^2=s^2+s^2-2s^2\cos(\alpha)\). 2. Factor the expression: \(a^2=2s^2(1-\cos(\alpha))\). 3. Substitute \(s=10\) and \(\alpha=110^\circ\): \(a=\sqrt{2(10)^2(1-\cos(110^\circ))}\approx16.38\,\text{cm}\). 4. In general, the simplified formula is \(a^2=2s^2(1-\cos(\alpha))\), or equivalently, \(a=s\sqrt{2(1-\cos(\alpha))}\).

a) \(a\approx16.38\,\text{cm}\) b) \(a^2=2s^2(1-\cos(\alpha))\), so \(a=s\sqrt{2(1-\cos(\alpha))}\).

- The largest angle is opposite the longest side. - Use the Law of Cosines because all three side lengths are known. - The sign of the cosine can help you classify the angle before finding its measure. - A triangle is obtuse when one interior angle is greater than \(90^\circ\).

1. The largest angle is \(\gamma\) because it is opposite the longest side, \(c=20\,\text{cm}\). 2. Solve the Law of Cosines for \(\cos(\gamma)\): \(\cos(\gamma)=\frac{a^2+b^2-c^2}{2ab}\). 3. Substitute: \(\cos(\gamma)=\frac{12^2+15^2-20^2}{2\cdot12\cdot15}=\frac{-31}{360}\). 4. Therefore, \(\gamma=\cos^{-1}\left(-\frac{31}{360}\right)\approx94.94^\circ\). 5. Since the largest angle is greater than \(90^\circ\), the triangle is obtuse.

The largest angle is \(\gamma\approx94.94^\circ\), so the triangle is obtuse.

- Read the three vertex coordinates from the graph. - Use the distance formula to find the three side lengths. - Apply the Law of Cosines to the side opposite angle \(\alpha\), then solve for the cosine.

1. Read the vertices as \(A(0, 0)\), \(B(5, 0)\), and \(C(3, 4)\). 2. The side lengths are \(AB=5\), \(AC=\sqrt{3^2+4^2}=5\), and \(BC=\sqrt{(3-5)^2+(4-0)^2}=\sqrt{20}\). 3. Apply the Law of Cosines with \(BC\) opposite \(\alpha\): \((\sqrt{20})^2=5^2+5^2-2\cdot5\cdot5\cos(\alpha)\). 4. Thus, \(20=50-50\cos(\alpha)\), so \(\cos(\alpha)=\frac{30}{50}=0.6\).

\(\cos(\alpha)=0.6\)

- The shorter diagonal and two adjacent sides form a triangle. - Read the two side lengths and their included angle from the diagram. - Use the Law of Cosines on that triangle.

1. The shorter diagonal forms a triangle with the two side lengths and the included \(60^\circ\) angle. 2. By the Law of Cosines, \(f^2=8^2+5^2-2\cdot8\cdot5\cos(60^\circ)\). 3. Since \(\cos(60^\circ)=\frac{1}{2}\), \(f^2=64+25-40=49\). 4. Therefore, \(f=7\,\text{cm}\).

\(7\,\text{cm}\)

- Identify the information pattern formed by the two known sides and the marked angle. - The unknown side lies opposite the marked included angle. - Substitute the three displayed values into the general-triangle relation that connects an SAS configuration to the opposite side.

1. The diagram gives two sides and their included angle, so apply the Law of Cosines to the side \(x\) opposite \(67^\circ\). 2. \(x^2=8^2+11^2-2\cdot8\cdot11\cos(67^\circ)\). 3. Therefore, \(x\approx10.7811\), so \(x\approx10.8\).

\(x\approx10.8\)

- The diagram gives SAS information, so first find the third side with the Law of Cosines. - Compare the resulting side length with the side labeled \(a\). - Equal sides in a triangle have equal opposite angles.

1. Use the Law of Cosines to find side \(b\): \(b^2=3^2+(3\sqrt{3})^2-2\cdot3\cdot3\sqrt{3}\cos(30^\circ)\). 2. Since \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\), \(b^2=9+27-27=9\), so \(b=3\). 3. Because \(a=b\), the opposite angles are congruent. Therefore, \(x=30^\circ\).

\(x=30^\circ\)

- Inventory the given information before naming a method: note whether the triangle is right, whether any known side has its opposite angle, and whether two known sides meet at a known angle. - Eliminate any method that cannot use the displayed givens directly. - After choosing the method, write an equation that uses exactly the two known sides, the known angle, and the unknown side.

1. The diagram gives two sides and the included angle: \(AB=13\), \(AC=9\), and \(\angle A=58^\circ\). 2. This is an SAS configuration, so the Law of Cosines is the most direct method. 3. Let \(a=BC\). Then \(a^2=9^2+13^2-2\cdot9\cdot13\cos(58^\circ)\). 4. Therefore, \(a\approx11.2249\), so \(BC\approx11.2\) to the nearest tenth.

Law of Cosines: \(BC^2=9^2+13^2-2\cdot9\cdot13\cos(58^\circ)\), so \(BC\approx11.2\).

- The diagonal divides the quadrilateral into two triangles whose three side lengths are shown. - Apply the Law of Cosines separately in the two triangles. - At \(B\) and \(D\), add the two smaller triangle angles that meet at the same vertex.

1. In \(\triangle ABD\), the Law of Cosines gives \(\cos(\alpha)=\frac{15^2+13^2-14^2}{2\cdot15\cdot13}\), so \(\alpha\approx59.5^\circ\). 2. In \(\triangle BCD\), \(\cos(\gamma)=\frac{12^2+10^2-14^2}{2\cdot12\cdot10}\), so \(\gamma\approx78.5^\circ\). 3. At \(B\), \(\angle ABD\approx53.1^\circ\) and \(\angle DBC\approx44.4^\circ\). Therefore, \(\beta\approx97.5^\circ\). 4. At \(D\), the two triangle angles are approximately \(67.4^\circ\) and \(57.1^\circ\). Therefore, \(\delta\approx124.5^\circ\). 5. The four angles total \(360.0^\circ\), which checks the result.

\(\alpha\approx59.5^\circ\) \(\beta\approx97.5^\circ\) \(\gamma\approx78.5^\circ\) \(\delta\approx124.5^\circ\)

- The diagram gives two sides with an included angle and the side opposite that angle. - Applying the Law of Cosines produces a quadratic equation in \(x\). - Use the condition \(x>4\) only after solving the quadratic.

1. Apply the Law of Cosines to the side labeled \(7\): \(7^2=x^2+8^2-2\cdot x\cdot8\cos(60^\circ)\). 2. Simplify: \(49=x^2+64-8x\), so \(x^2-8x+15=0\). 3. Factor: \((x-3)(x-5)=0\), giving \(x=3\) or \(x=5\). 4. Because \(x>4\), the required length is \(x=5\).

\(x=5\)

- The given information is SAS, so first find the third side. - After that, a side-angle opposite pair is available for the Law of Sines. - In the final inverse-sine step, compare side lengths to decide which angle is possible.

1. First use the Law of Cosines to find \(c\): \(c^2=2^2+(1+\sqrt{3})^2-2\cdot2(1+\sqrt{3})\cos(60^\circ)=6\). Thus, \(c=\sqrt{6}\). 2. By the Law of Sines, \(\frac{\sin(x)}{2}=\frac{\sin(60^\circ)}{\sqrt{6}}\), so \(\sin(x)=\frac{1}{\sqrt{2}}\). 3. This gives \(x=45^\circ\) or \(135^\circ\). Since the side opposite \(x\) is shorter than \(c=\sqrt{6}\), the opposite angles satisfy \(x<60^\circ\). 4. Therefore, \(x=45^\circ\).

\(x=45^\circ\)

- Treat the acute and obtuse diagrams separately before trying to combine the algebra. - In the obtuse panel, relate the acute exterior angle to \(A\) using a supplementary-angle relationship. - Express the horizontal leg adjacent to vertex \(B\) in each panel using \(b\), \(c\), and \(\cos A\). - After the Pythagorean step, simplify the squared sine and cosine terms together.

1. In panel a), \(AD=b\cos A\) and \(CD=b\sin A\). Since \(AB=c\), \(BD=c-b\cos A\). 2. Apply the Pythagorean theorem to \(\triangle BCD\): \(a^2=(c-b\cos A)^2+(b\sin A)^2\). 3. Expanding and using \(\sin^2 A+\cos^2 A=1\) gives \(a^2=b^2+c^2-2bc\cos A\). 4. In panel b), let \(E\) be the foot of the exterior altitude. The acute angle between \(AC\) and \(AE\) is \(180^\circ-A\). 5. Thus, \(AE=b\cos(180^\circ-A)=-b\cos A\) and \(CE=b\sin(180^\circ-A)=b\sin A\). 6. Because \(BE=BA+AE\), \(BE=c-b\cos A\). Applying the Pythagorean theorem to \(\triangle BCE\) gives the same expression \(a^2=(c-b\cos A)^2+(b\sin A)^2\), which simplifies to the Law of Cosines. 7. If \(A=90^\circ\), then \(\cos A=0\), so the formula becomes \(a^2=b^2+c^2\), exactly the Pythagorean theorem.

a) For acute \(A\), \(BD=c-b\cos A\) and \(CD=b\sin A\), giving \(a^2=b^2+c^2-2bc\cos A\). b) For obtuse \(A\), \(AE=-b\cos A\), so \(BE=c-b\cos A\) and \(CE=b\sin A\); the same formula follows. c) For \(A=90^\circ\), \(\cos A=0\), so the formula becomes the Pythagorean theorem.