- Find the vertex shared by both sides with length labels. - Name the angle made by the two rays from that vertex.

1. The two labeled sides meet at vertex \(A\). 2. Their included angle is \(\angle BAC\) (or \(\angle CAB\)).

\(\angle BAC\)

- Which angle is formed by the two legs? - What is the sine of that angle?

1. The included angle between the legs is \(C=90^\circ\), and \(\sin90^\circ=1\). 2. \(A=\frac{1}{2}\cdot6\cdot10\cdot1=30\,\text{cm}^2\). 3. This is the usual half-of-base-times-height formula because the two legs are perpendicular.

\(30\,\text{cm}^2\). Since \(\sin90^\circ=1\), the sine-area formula reduces to half the product of the legs.

- The formula uses the two sides touching the stated angle. - What is the exact value of \(\sin30^\circ\)?

1. The sides meet at the given angle, so \(A=\frac{1}{2}\cdot6\cdot8\cdot\sin30^\circ\). 2. As \(\sin30^\circ=\frac{1}{2}\), \(A=24\cdot\frac{1}{2}=12\,\text{cm}^2\).

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

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

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

- How does the sine of the included angle affect the area when both sides stay fixed? - How large can the sine of an angle be?

1. The area is \(A=\frac{1}{2}\cdot5\cdot12\sin C=30\sin C\). 2. Since \(\sin C\le1\), the greatest area is \(30\,\text{in}^2\), when \(C=90^\circ\).

\(A=30\sin C\,\text{in}^2\). The greatest area is \(30\,\text{in}^2\) at \(C=90^\circ\).

- Which two sides multiply in each area formula? - What relationship do you notice between the two included angles? - Does sine have the same value at supplementary angles?

1. Each triangle has two sides of \(10\,\text{cm}\) and \(6\,\text{cm}\), so its area is \(30\sin C\,\text{cm}^2\). 2. The included angles are \(35^\circ\) and \(145^\circ\), which are supplementary. 3. Because \(\sin145^\circ=\sin35^\circ\), the areas are equal even though the triangles have different shapes.

Both areas equal \(30\sin35^\circ\,\text{cm}^2\) because \(\sin145^\circ=\sin35^\circ\).

- Which factor in the sine-area formula is unknown? - Isolate the unknown side after replacing the known sine value.

1. \(30=\frac{1}{2}\cdot10\cdot x\cdot\sin30^\circ\). 2. As \(\sin30^\circ=\frac{1}{2}\), \(30=\frac{5x}{2}\). 3. Hence \(x=12\,\text{in}\).

\(12\,\text{in}\)

- Solve the sine-area equation for the sine of the angle. - Which calculator operation finds an angle from its sine? - How does the word acute select the appropriate solution?

1. \(25=\frac{1}{2}\cdot10\cdot8\sin C\), so \(\sin C=\frac{25}{40}=0.625\). 2. Since \(C\) is acute, use the acute inverse-sine solution: \(C=\sin^{-1}(0.625)\approx38.7^\circ\). 3. To the nearest degree, \(C\approx39^\circ\).

\(C\approx39^\circ\)

- Rearrange the area formula to find \(\sin C\). - Can two different angles between \(0^\circ\) and \(180^\circ\) have the same sine?

1. \(24=\frac{1}{2}\cdot8\cdot12\sin C\), so \(\sin C=\frac{1}{2}\). 2. An included angle of a triangle lies between \(0^\circ\) and \(180^\circ\). 3. In that interval, sine is \(\frac{1}{2}\) at \(30^\circ\) and \(150^\circ\). Both angles produce valid triangles.

\(30^\circ\) or \(150^\circ\)

- The sine-area formula needs two side lengths and the angle between them. - In each case, decide whether the stated angle sits at the shared vertex of the known sides.

1. In case a), both known sides meet at the given angle, so the formula can be used immediately. 2. In case b), the given angle is not between the two known sides. The included angle needed in the formula is not supplied. 3. Thus only case a) permits direct substitution into the sine-area formula.

a) Yes. The angle is included between the known sides. b) No. The included angle is not given, so the formula cannot be used directly.

- Which marked angle lies between the two measured sides? - The sine-area formula works even if neither side is horizontal.

1. The two shown lengths meet at \(A\), so the included angle is \(70^\circ\). 2. \(A_{\triangle ABC}=\frac{1}{2}\cdot8\cdot12\sin70^\circ\approx45.1\,\text{cm}^2\).

\(45.1\,\text{cm}^2\)

- Relate the sine of \(150^\circ\) to the sine of a familiar acute angle. - Does the sine-area formula change when the included angle is obtuse?

1. \(\sin150^\circ=\sin30^\circ=\frac{1}{2}\). 2. \(A=\frac{1}{2}\cdot8\cdot13\cdot\frac{1}{2}=26\,\text{cm}^2\).

\(26\,\text{cm}^2\)

- 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 them. - For each panel, ask whether the two sides meeting at a marked angle have known lengths. - Compute only after identifying a panel with enough direct information.

1. In panel a), the two given sides, \(8\) and \(11\), meet at the given \(47^\circ\) angle. 2. The area is \(\frac{1}{2}\cdot8\cdot11\sin(47^\circ)\approx32.18\), or \(32.2\) square units to the nearest tenth. 3. In panel b), the given \(47^\circ\) is not the angle between the two labeled sides, so the included angle is missing for direct substitution. 4. In panel c), only one side is given; another side forming a pair around a known included angle is missing.

Panel a) only: area \(\approx32.2\) square units. Panel b) lacks the included angle between the two known sides. Panel c) lacks a second known side for an included-angle pair.

- Find panel A's area before working on the unknown angle. - For panel B, use the fact that its area matches panel A's. - Which of the angles with the required sine is obtuse?

1. Panel A has area \(\frac{1}{2}\cdot6\cdot10\sin30^\circ=15\,\text{in}^2\). 2. For panel B, \(15=\frac{1}{2}\cdot5\cdot12\sin\theta=30\sin\theta\), so \(\sin\theta=\frac{1}{2}\). 3. The solutions between \(0^\circ\) and \(180^\circ\) are \(30^\circ\) and \(150^\circ\). Only \(150^\circ\) is obtuse.

\(\theta=150^\circ\)

- 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. - Which angles in the exterior right triangle are related to the given obtuse angle? - 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.