Free Math Worksheets

- Use the area formula for two sides and their included angle. - Make sure the calculator is in degree mode. - The sine area formula also works for an obtuse included angle.

1. Use the triangle area formula \(A = \frac{1}{2}bc\sin(\alpha)\). 2. Substitute: \(A = \frac{1}{2}\cdot42\cdot58\cdot\sin(125^{\circ})\). 3. \(A = 1218\sin(125^{\circ}) \approx 997.7\,\text{m}^2\).

\(A \approx 997.7\,\text{m}^2\)

- Which area formula uses two side lengths and their included angle? - Recall the exact value of \(\sin(30^{\circ})\).

1. Use the area formula with two sides and the included angle: \(A = \frac{1}{2}ab\sin(\gamma)\). 2. Substitute the values: \(A = \frac{1}{2}\cdot 8 \cdot 6 \cdot \sin(30^{\circ})\). 3. Since \(\sin(30^{\circ}) = \frac{1}{2}\), \(A = 12\,\text{cm}^2\).

\(12\,\text{cm}^2\)

- Adjacent angles in a parallelogram sum to \(180^{\circ}\). - Use a right triangle to express the height with sine. - You may also use \(A = ab\sin(\theta)\).

1. Adjacent angles in a parallelogram are supplementary, so \(\alpha = 180^{\circ} - 118^{\circ} = 62^{\circ}\). 2. The height relative to side \(a\) is \(h_a = 9.5\sin(62^{\circ}) \approx 8.388\,\text{cm}\). 3. The area is \(A = ah_a = 15.0\cdot8.388\ldots \approx 125.8\,\text{cm}^2\). 4. Equivalently, \(A = ab\sin(118^{\circ})\).

\(A \approx 125.8\,\text{cm}^2\)

- Express the altitude using sine in the right triangle. - Substitute that expression into \(A=\frac{1}{2}(\text{base})(\text{height})\). - Use the two given sides and their included angle.

1. Let altitude \(h_a\) from \(D\) to \(AB\) 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: \(A=\frac{1}{2}ah_a=\frac{1}{2}ab\sin(\alpha)\).

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

- Substitute the known values into the area formula for two sides and the included angle. - Rearrange the equation to isolate the unknown side. - Round only after evaluating the expression.

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 \(a\): \(a = \frac{24{,}000}{160\sin(40^{\circ})} \approx 233.36\,\text{ft}\). 3. Rounded to the nearest foot, \(a \approx 233\,\text{ft}\).

The side length is approximately \(233\,\text{ft}\).

- Use the triangle area formula that involves two side lengths and the included angle. - How does a diagonal divide a parallelogram? - Keep \(\sin(45^\circ)\) in exact form until the final decimal approximation.

1. For part a), use two sides and their included angle: \(A_{\triangle ABD}=\frac{1}{2}\cdot 8\cdot 5\cdot\sin(45^\circ)=10\sqrt{2}\,\text{cm}^2\approx 14.14\,\text{cm}^2\). 2. For part b), diagonal \(BD\) divides the parallelogram into two congruent triangles. Therefore, \(A_{ABCD}=2\cdot 10\sqrt{2}=20\sqrt{2}\,\text{cm}^2\approx 28.28\,\text{cm}^2\).

a) \(10\sqrt{2}\,\text{cm}^2\approx 14.14\,\text{cm}^2\) b) \(20\sqrt{2}\,\text{cm}^2\approx 28.28\,\text{cm}^2\)

- 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.

- Solve the sine area formula for \(\sin(\gamma)\). - A sine value between \(0^{\circ}\) and \(180^{\circ}\) can correspond to two supplementary angles. - Select the obtuse solution.

1. Use \(A = \frac{1}{2}ab\sin(\gamma)\). 2. Substitute: \(14 = \frac{1}{2}\cdot7.2\cdot5.5\cdot\sin(\gamma) = 19.8\sin(\gamma)\). 3. Thus, \(\sin(\gamma) = \frac{14}{19.8} \approx 0.70707\). 4. The inverse sine gives the acute solution \(45.0^{\circ}\). Since \(\gamma\) is obtuse, use the supplementary angle: \(\gamma = 180^{\circ} - 45.0^{\circ} \approx 135.0^{\circ}\).

\(\gamma \approx 135.0^{\circ}\)

- 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.