Free math worksheets

- The diagram shows two sides and their included angle. - Use the sine area formula for a triangle. - The formula works with an obtuse included angle as well as an acute one.

1. Use the triangle area formula with two sides and their included angle: \(A=\frac{1}{2}bc\sin(\alpha)\). 2. Substitute the diagram values: \(A=\frac{1}{2}\cdot42\cdot58\cdot\sin(125^\circ)\). 3. Therefore, \(A=1218\sin(125^\circ)\approx997.7\,\text{m}^2\).

\(A\approx997.7\,\text{m}^2\)

- Identify the two sides that form the marked angle. - Substitute those two side lengths and the included angle into the required formula. - Round only after evaluating the complete sine-area expression.

1. The two displayed sides adjacent to the included angle have lengths \(8\,\text{cm}\) and \(6\,\text{cm}\), and the included angle is \(37^\circ\). 2. Apply the required formula: \(A=\frac{1}{2}\cdot8\cdot6\cdot\sin(37^\circ)\). 3. Thus, \(A\approx14.4436\,\text{cm}^2\), which rounds to \(14.4\,\text{cm}^2\).

\(14.4\,\text{cm}^2\)

- Read the two adjacent side lengths and the marked interior angle from the diagram. - The sine-area relationship can use either an acute angle or its supplementary obtuse angle. - A parallelogram has twice the area of a triangle with the same two adjacent sides and included angle.

1. Adjacent angles in a parallelogram are supplementary, so the acute included angle is \(180^\circ-118^\circ=62^\circ\). 2. The height relative to the \(15.0\,\text{cm}\) side is \(9.5\sin(62^\circ)\approx8.388\,\text{cm}\). 3. The area is \(15.0\cdot8.388\ldots\approx125.8\,\text{cm}^2\). 4. Equivalently, \(A=15.0\cdot9.5\sin(118^\circ)\).

\(A\approx125.8\,\text{cm}^2\)

- Focus on the right triangle formed by the altitude in the diagram. - Express the altitude in terms of \(b\) and \(\alpha\). - Substitute that expression into \(A=\frac{1}{2}(\text{base})(\text{height})\).

1. Let the shown altitude \(h_a\) from \(D\) meet \(AB\) at \(F\). 2. In right triangle \(AFD\), \(\sin(\alpha)=\frac{h_a}{b}\), so \(h_a=b\sin(\alpha)\). 3. Substitute into the triangle area formula: \(A=\frac{1}{2}ah_a=\frac{1}{2}ab\sin(\alpha)\).

\(A=\frac{1}{2}ab\sin(\alpha)\)

- Read the known side and included angle from the diagram. - Substitute the area and diagram values into the sine area formula. - Isolate \(a\) before rounding.

1. Use the triangle area formula with two sides and the included angle: \(12{,}000=\frac{1}{2}\cdot a\cdot160\cdot\sin(40^\circ)\). 2. Solve for \(a\): \(a=\frac{24{,}000}{160\sin(40^\circ)}\approx233.36\,\text{ft}\). 3. Rounded to the nearest foot, \(a\approx233\,\text{ft}\).

\(a\approx233\,\text{ft}\)

- Read the two sides of \(\triangle ABD\) and their included angle from the diagram. - Use the triangle sine area formula for part a). - A diagonal divides a parallelogram into two congruent triangles.

1. For part a), use the two shown sides and their included angle: \(A_{\triangle ABD}=\frac{1}{2}\cdot8\cdot5\cdot\sin(45^\circ)=10\sqrt{2}\,\text{cm}^2\approx14.14\,\text{cm}^2\). 2. For part b), diagonal \(BD\) divides the parallelogram into two congruent triangles. Therefore, \(A_{ABCD}=2\cdot10\sqrt{2}=20\sqrt{2}\,\text{cm}^2\approx28.28\,\text{cm}^2\).

a) \(10\sqrt{2}\,\text{cm}^2\approx14.14\,\text{cm}^2\) b) \(20\sqrt{2}\,\text{cm}^2\approx28.28\,\text{cm}^2\)

- The sine-area formula needs two side lengths and the angle between those two sides. - In each panel, locate the displayed angle and ask whether both sides that form that angle have known lengths. - Only after selecting the applicable panel should you substitute into the formula.

1. In panel a), the two displayed side lengths, \(8\) and \(11\), meet at the displayed \(47^\circ\) angle. This is exactly the side-angle-side information needed for the sine-area formula. 2. The area is \(\frac{1}{2}\cdot8\cdot11\sin(47^\circ)\approx32.1796\), so the area is about \(32.2\) square units. 3. In panel b), two sides are given, but the displayed \(47^\circ\) angle is not the included angle between them, so the included angle is missing for an immediate use of the formula. 4. In panel c), angles are given with only one side, so a second side forming an included-angle pair is missing.

Panel a) only. Its area is \(\approx32.2\) square units. Panel b) is missing the included angle between the two given sides, and panel c) is missing a second side needed with an included angle.

- Use the formula for the area of a triangle when two sides and their included angle are known. - What equation does the given area create for \(\sin C\)? - Which two angles between \(0^\circ\) and \(180^\circ\) have the same sine? - Would triangles with different included angles be congruent?

1. Let \(C\) be the included angle between the sides of lengths \(4\,\text{cm}\) and \(10\,\text{cm}\). 2. Use the area formula \(A = \frac{1}{2}ab\sin C\): \(10 = \frac{1}{2}(4)(10)\sin C\). 3. This gives \(\sin C = \frac{1}{2}\), so the included angle can be \(30^\circ\) or \(150^\circ\). 4. These choices produce triangles with different included angles and therefore different third-side lengths. The triangles do not have to be congruent.

No. The area condition gives \(\sin C = \frac{1}{2}\), so the included angle can be either \(30^\circ\) or \(150^\circ\). These choices produce noncongruent triangles.

- Read the two side lengths adjacent to \(\gamma\) from the diagram. - Solve the sine-area formula for \(\sin(\gamma)\). - Two supplementary angles have the same sine; use the condition that \(\gamma\) is obtuse.

1. Use \(A=\frac{1}{2}ab\sin(\gamma)\). 2. Substitute the area and the two side lengths from the diagram: \(14=\frac{1}{2}\cdot7.2\cdot5.5\sin(\gamma)=19.8\sin(\gamma)\). 3. Thus, \(\sin(\gamma)=\frac{14}{19.8}\approx0.70707\). 4. The inverse sine gives the acute solution \(45.0^\circ\). The requested angle is obtuse, so \(\gamma=180^\circ-45.0^\circ\approx135.0^\circ\).

\(\gamma\approx135.0^\circ\)

- Consider the altitude to the extension of the base. - Identify the acute supplementary angle in the exterior right triangle. - Use the given sine relationship to express the altitude.

1. Let the altitude from \(A\) meet the extension of side \(BC\) at \(F\). 2. In right triangle \(ACF\), the acute angle at \(C\) is \(180^\circ-\gamma\). 3. Thus, \(\sin(180^\circ-\gamma)=\frac{h_a}{b}\). 4. Since \(\sin(180^\circ-\gamma)=\sin(\gamma)\), \(h_a=b\sin(\gamma)\). 5. Substituting into \(A=\frac{1}{2}ah_a\) gives \(A=\frac{1}{2}ab\sin(\gamma)\).

The exterior right triangle gives \(h_a=b\sin(180^\circ-\gamma)=b\sin(\gamma)\). Therefore, \(A=\frac{1}{2}ah_a=\frac{1}{2}ab\sin(\gamma)\), even when \(\gamma\) is obtuse.