Question

sine law and cosine law
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find each measurement indicated. round your answers to the nearest tenth.

1. find ac
2. find bc
3. find ac
4. find m∠a
5. find m∠b
6. find m∠c

solve each triangle. round your answers to the nearest tenth.
7)
8)
9)

Explanation:

- ## Step1: Find $\angle C$
- In a triangle, the sum of interior - angles is $180^{\circ}$. So, $\angle C=180^{\circ}-92^{\circ}-28^{\circ}=60^{\circ}$.
- ## Step2: Apply the Sine Law
- The Sine Law is $\frac{AC}{\sin B}=\frac{AB}{\sin C}$. We know $AB = 15$ yd, $\angle B = 28^{\circ}$, and $\angle C = 60^{\circ}$. So, $AC=\frac{15\times\sin28^{\circ}}{\sin60^{\circ}}$.
- $\sin28^{\circ}\approx0.4695$, $\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$. Then $AC=\frac{15\times0.4695}{0.866}\approx8.1$ yd.
- # Answer:
- $8.1$ yd
2. **2) Find BC**:
- # Explanation:
- ## Step1: Find $\angle C$
- $\angle C=180^{\circ}-15^{\circ}-59^{\circ}=106^{\circ}$.
- ## Step2: Apply the Sine Law
- By the Sine Law $\frac{BC}{\sin A}=\frac{AB}{\sin C}$. Given $AB = 10$ yd, $\angle A = 59^{\circ}$, $\angle C = 106^{\circ}$. So, $BC=\frac{10\times\sin59^{\circ}}{\sin106^{\circ}}$.
- $\sin59^{\circ}\approx0.8572$, $\sin106^{\circ}\approx0.9613$. Then $BC=\frac{10\times0.8572}{0.9613}\approx8.9$ yd.
- # Answer:
- $8.9$ yd
3. **3) Find AC**:
- # Explanation:
- ## Step1: Find $\angle C$
- $\angle C=180^{\circ}-83^{\circ}-38^{\circ}=59^{\circ}$.
- ## Step2: Apply the Sine Law
- Using the Sine Law $\frac{AC}{\sin B}=\frac{AB}{\sin C}$. With $AB = 25$ m, $\angle B = 83^{\circ}$, $\angle C = 59^{\circ}$. So, $AC=\frac{25\times\sin83^{\circ}}{\sin59^{\circ}}$.
- $\sin83^{\circ}\approx0.9925$, $\sin59^{\circ}\approx0.8572$. Then $AC=\frac{25\times0.9925}{0.8572}\approx29.0$ m.
- # Answer:
- $29.0$ m
4. **4) Find $m\angle A$**:
- # Explanation:
- ## Step1: Apply the Sine Law
- By the Sine Law $\frac{\sin A}{BC}=\frac{\sin B}{AC}$. Given $BC = 7$ yd, $AC = 28$ yd, $\angle B = 75^{\circ}$. So, $\sin A=\frac{7\times\sin75^{\circ}}{28}$.
- $\sin75^{\circ}=\sin(45^{\circ}+30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$.
- $\sin A=\frac{7\times0.9659}{28}\approx0.2415$. Then $m\angle A=\sin^{- 1}(0.2415)\approx14.0^{\circ}$.
- # Answer:
- $14.0^{\circ}$
5. **5) Find $m\angle B$**:
- # Explanation:
- ## Step1: Apply the Sine Law
- By the Sine Law $\frac{\sin B}{AC}=\frac{\sin C}{AB}$. Given $AC = 21$ mi, $AB = 32$ mi, $\angle C = 28^{\circ}$. So, $\sin B=\frac{21\times\sin28^{\circ}}{32}$.
- $\sin28^{\circ}\approx0.4695$. Then $\sin B=\frac{21\times0.4695}{32}\approx0.308$. So, $m\angle B=\sin^{-1}(0.308)\approx18.0^{\circ}$.
- # Answer:
- $18.0^{\circ}$
6. **6) Find $m\angle C$**:
- # Explanation:
- ## Step1: Apply the Cosine Law
- The Cosine Law is $c^{2}=a^{2}+b^{2}-2ab\cos C$. Here, $a = 11$ ft, $b = 19$ ft, $c$ is opposite $\angle C$. First, $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$.
- Let $c$ be the side opposite $\angle C$. We can also use the Sine Law. First, find $\sin C$ using the Sine Law $\frac{\sin C}{AB}=\frac{\sin A}{BC}$.
- $\sin A=\sin98^{\circ}\approx0.9903$. By the Sine Law $\sin C=\frac{11\times\sin98^{\circ}}{19}$.
- $\sin C=\frac{11\times0.9903}{19}\approx0.577$. Then $m\angle C=\sin^{-1}(0.577)\approx35.2^{\circ}$.
- # Answer:
- $35.2^{\circ}$
7. **7) Solve the triangle**:
- # Explanation:
- ## Step1: Find $\angle C$
- $\angle C=180^{\circ}-34^{\circ}-109^{\circ}=37^{\circ}$.
- ## Step2: Apply the Sine Law to find $AB$
- $\frac{AB}{\sin C}=\frac{BC}{\sin A}$. Given $BC = 22$ cm, $\angle A = 109^{\circ}$, $\angle C = 37^{\circ}$. So, $AB=\frac{22\times\sin37^{\circ}}{\sin109^{\circ}}$.
- $\sin37^{\circ}\approx0.6018$, $\sin109^{\circ}\approx0.9455$. Then $AB=\frac{22\times0.6018}{0.9455}\approx14.0$ cm.
- ## Step3: Apply the Sine Law to find $AC$
- $\frac{AC}{\sin B}=\frac{BC}{\sin A}$. $\angle B = 34^{\circ}$, so $AC=\frac{22\times\sin34^{\circ}}{\sin109^{\circ}}$.
- $\sin34^{\circ}\approx0.5592$. Then $AC=\frac{22\times0.5592}{0.9455}\approx12.9$ cm.
- # Answer:
- $\angle C = 37^{\circ}$, $AB\approx14.0$ cm, $AC\approx12.9$ cm
8. **8) Solve the triangle**:
- # Explanation:
- ## Step1: Find $\angle B$
- $\angle B=180^{\circ}-37^{\circ}-104^{\circ}=39^{\circ}$.
- ## Step2: Apply the Sine Law to find $AB$
- $\frac{AB}{\sin C}=\frac{AC}{\sin B}$. Given $AC = 24$ km, $\angle C = 104^{\circ}$, $\angle B = 39^{\circ}$. So, $AB=\frac{24\times\sin104^{\circ}}{\sin39^{\circ}}$.
- $\sin104^{\circ}\approx0.9703$, $\sin39^{\circ}\approx0.6293$. Then $AB=\frac{24\times0.9703}{0.6293}\approx37.0$ km.
- ## Step3: Apply the Sine Law to find $BC$
- $\frac{BC}{\sin A}=\frac{AC}{\sin B}$. $\angle A = 37^{\circ}$, so $BC=\frac{24\times\sin37^{\circ}}{\sin39^{\circ}}$.
- $\sin37^{\circ}\approx0.6018$. Then $BC=\frac{24\times0.6018}{0.6293}\approx23.0$ km.
- # Answer:
- $\angle B = 39^{\circ}$, $AB\approx37.0$ km, $BC\approx23.0$ km
9. **9) Solve the triangle**:
- # Explanation:
- ## Step1: Apply the Cosine Law to find $BC$
- By the Cosine Law $BC^{2}=AB^{2}+AC^{2}-2AB\cdot AC\cdot\cos A$. Given $AB = 34$ m, $AC = 20$ m, $\angle A = 127^{\circ}$, $\cos127^{\circ}\approx - 0.6018$.
- $BC^{2}=34^{2}+20^{2}-2\times34\times20\times(-0.6018)=1156 + 400+820.48=2376.48$. So, $BC=\sqrt{2376.48}\approx48.7$ m.
- ## Step2: Apply the Sine Law to find $\angle B$
- $\frac{\sin B}{AC}=\frac{\sin A}{BC}$. $\sin B=\frac{20\times\sin127^{\circ}}{48.7}$.
- $\sin127^{\circ}\approx0.7986$. Then $\sin B=\frac{20\times0.7986}{48.7}\approx0.327$. So, $m\angle B=\sin^{-1}(0.327)\approx19.1^{\circ}$.
- ## Step3: Find $\angle C$
- $\angle C=180^{\circ}-127^{\circ}-19.1^{\circ}=33.9^{\circ}$.
- # Answer:
- $BC\approx48.7$ m, $\angle B\approx19.1^{\circ}$, $\angle C\approx33.9^{\circ}$

Answer:

1. 1) Find AC:

• # Explanation:
• ## Step1: Find $\angle C$
• In a triangle, the sum of interior - angles is $180^{\circ}$. So, $\angle C=180^{\circ}-92^{\circ}-28^{\circ}=60^{\circ}$.
• ## Step2: Apply the Sine Law
• The Sine Law is $\frac{AC}{\sin B}=\frac{AB}{\sin C}$. We know $AB = 15$ yd, $\angle B = 28^{\circ}$, and $\angle C = 60^{\circ}$. So, $AC=\frac{15\times\sin28^{\circ}}{\sin60^{\circ}}$.
• $\sin28^{\circ}\approx0.4695$, $\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$. Then $AC=\frac{15\times0.4695}{0.866}\approx8.1$ yd.
• # Answer:
• $8.1$ yd

1. 2) Find BC:

• # Explanation:
• ## Step1: Find $\angle C$
• $\angle C=180^{\circ}-15^{\circ}-59^{\circ}=106^{\circ}$.
• ## Step2: Apply the Sine Law
• By the Sine Law $\frac{BC}{\sin A}=\frac{AB}{\sin C}$. Given $AB = 10$ yd, $\angle A = 59^{\circ}$, $\angle C = 106^{\circ}$. So, $BC=\frac{10\times\sin59^{\circ}}{\sin106^{\circ}}$.
• $\sin59^{\circ}\approx0.8572$, $\sin106^{\circ}\approx0.9613$. Then $BC=\frac{10\times0.8572}{0.9613}\approx8.9$ yd.
• # Answer:
• $8.9$ yd

1. 3) Find AC:

• # Explanation:
• ## Step1: Find $\angle C$
• $\angle C=180^{\circ}-83^{\circ}-38^{\circ}=59^{\circ}$.
• ## Step2: Apply the Sine Law
• Using the Sine Law $\frac{AC}{\sin B}=\frac{AB}{\sin C}$. With $AB = 25$ m, $\angle B = 83^{\circ}$, $\angle C = 59^{\circ}$. So, $AC=\frac{25\times\sin83^{\circ}}{\sin59^{\circ}}$.
• $\sin83^{\circ}\approx0.9925$, $\sin59^{\circ}\approx0.8572$. Then $AC=\frac{25\times0.9925}{0.8572}\approx29.0$ m.
• # Answer:
• $29.0$ m

1. 4) Find $m\angle A$:

• # Explanation:
• ## Step1: Apply the Sine Law
• By the Sine Law $\frac{\sin A}{BC}=\frac{\sin B}{AC}$. Given $BC = 7$ yd, $AC = 28$ yd, $\angle B = 75^{\circ}$. So, $\sin A=\frac{7\times\sin75^{\circ}}{28}$.
• $\sin75^{\circ}=\sin(45^{\circ}+30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$.
• $\sin A=\frac{7\times0.9659}{28}\approx0.2415$. Then $m\angle A=\sin^{- 1}(0.2415)\approx14.0^{\circ}$.
• # Answer:
• $14.0^{\circ}$

1. 5) Find $m\angle B$:

• # Explanation:
• ## Step1: Apply the Sine Law
• By the Sine Law $\frac{\sin B}{AC}=\frac{\sin C}{AB}$. Given $AC = 21$ mi, $AB = 32$ mi, $\angle C = 28^{\circ}$. So, $\sin B=\frac{21\times\sin28^{\circ}}{32}$.
• $\sin28^{\circ}\approx0.4695$. Then $\sin B=\frac{21\times0.4695}{32}\approx0.308$. So, $m\angle B=\sin^{-1}(0.308)\approx18.0^{\circ}$.
• # Answer:
• $18.0^{\circ}$

1. 6) Find $m\angle C$:

• # Explanation:
• ## Step1: Apply the Cosine Law
• The Cosine Law is $c^{2}=a^{2}+b^{2}-2ab\cos C$. Here, $a = 11$ ft, $b = 19$ ft, $c$ is opposite $\angle C$. First, $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$.
• Let $c$ be the side opposite $\angle C$. We can also use the Sine Law. First, find $\sin C$ using the Sine Law $\frac{\sin C}{AB}=\frac{\sin A}{BC}$.
• $\sin A=\sin98^{\circ}\approx0.9903$. By the Sine Law $\sin C=\frac{11\times\sin98^{\circ}}{19}$.
• $\sin C=\frac{11\times0.9903}{19}\approx0.577$. Then $m\angle C=\sin^{-1}(0.577)\approx35.2^{\circ}$.
• # Answer:
• $35.2^{\circ}$

1. 7) Solve the triangle:

• # Explanation:
• ## Step1: Find $\angle C$
• $\angle C=180^{\circ}-34^{\circ}-109^{\circ}=37^{\circ}$.
• ## Step2: Apply the Sine Law to find $AB$
• $\frac{AB}{\sin C}=\frac{BC}{\sin A}$. Given $BC = 22$ cm, $\angle A = 109^{\circ}$, $\angle C = 37^{\circ}$. So, $AB=\frac{22\times\sin37^{\circ}}{\sin109^{\circ}}$.
• $\sin37^{\circ}\approx0.6018$, $\sin109^{\circ}\approx0.9455$. Then $AB=\frac{22\times0.6018}{0.9455}\approx14.0$ cm.
• ## Step3: Apply the Sine Law to find $AC$
• $\frac{AC}{\sin B}=\frac{BC}{\sin A}$. $\angle B = 34^{\circ}$, so $AC=\frac{22\times\sin34^{\circ}}{\sin109^{\circ}}$.
• $\sin34^{\circ}\approx0.5592$. Then $AC=\frac{22\times0.5592}{0.9455}\approx12.9$ cm.
• # Answer:
• $\angle C = 37^{\circ}$, $AB\approx14.0$ cm, $AC\approx12.9$ cm

1. 8) Solve the triangle:

• # Explanation:
• ## Step1: Find $\angle B$
• $\angle B=180^{\circ}-37^{\circ}-104^{\circ}=39^{\circ}$.
• ## Step2: Apply the Sine Law to find $AB$
• $\frac{AB}{\sin C}=\frac{AC}{\sin B}$. Given $AC = 24$ km, $\angle C = 104^{\circ}$, $\angle B = 39^{\circ}$. So, $AB=\frac{24\times\sin104^{\circ}}{\sin39^{\circ}}$.
• $\sin104^{\circ}\approx0.9703$, $\sin39^{\circ}\approx0.6293$. Then $AB=\frac{24\times0.9703}{0.6293}\approx37.0$ km.
• ## Step3: Apply the Sine Law to find $BC$
• $\frac{BC}{\sin A}=\frac{AC}{\sin B}$. $\angle A = 37^{\circ}$, so $BC=\frac{24\times\sin37^{\circ}}{\sin39^{\circ}}$.
• $\sin37^{\circ}\approx0.6018$. Then $BC=\frac{24\times0.6018}{0.6293}\approx23.0$ km.
• # Answer:
• $\angle B = 39^{\circ}$, $AB\approx37.0$ km, $BC\approx23.0$ km

1. 9) Solve the triangle:

• # Explanation:
• ## Step1: Apply the Cosine Law to find $BC$
• By the Cosine Law $BC^{2}=AB^{2}+AC^{2}-2AB\cdot AC\cdot\cos A$. Given $AB = 34$ m, $AC = 20$ m, $\angle A = 127^{\circ}$, $\cos127^{\circ}\approx - 0.6018$.
• $BC^{2}=34^{2}+20^{2}-2\times34\times20\times(-0.6018)=1156 + 400+820.48=2376.48$. So, $BC=\sqrt{2376.48}\approx48.7$ m.
• ## Step2: Apply the Sine Law to find $\angle B$
• $\frac{\sin B}{AC}=\frac{\sin A}{BC}$. $\sin B=\frac{20\times\sin127^{\circ}}{48.7}$.
• $\sin127^{\circ}\approx0.7986$. Then $\sin B=\frac{20\times0.7986}{48.7}\approx0.327$. So, $m\angle B=\sin^{-1}(0.327)\approx19.1^{\circ}$.
• ## Step3: Find $\angle C$
• $\angle C=180^{\circ}-127^{\circ}-19.1^{\circ}=33.9^{\circ}$.
• # Answer:
• $BC\approx48.7$ m, $\angle B\approx19.1^{\circ}$, $\angle C\approx33.9^{\circ}$