Extra Practice - Sine Law and Cosine Law.pdf - RHHS - Math - Free Printable

Here are the step-by-step solutions for each problem on the worksheet. I have used the Sine Law and Cosine Law as appropriate, rounding all answers to the nearest tenth.

Part 1: Find each measurement indicated

1) Find AC
* Identify: We have angles $A=76^\circ$, $B=28^\circ$, and side $c (AB) = 15$. We need side $b (AC)$.
* Step 1: Find angle $C$.
$$180^\circ - 76^\circ - 28^\circ = 76^\circ$$
* Step 2: Use Sine Law. Since angle $A$ ($76^\circ$) equals angle $C$ ($76^\circ$), this is an isosceles triangle. The sides opposite these angles must be equal.
Side $AC$ is opposite angle $B$? No, side $AC$ is opposite angle $B$ is incorrect notation. Side $AC$ is side $b$, opposite angle $B$. Side $AB$ is side $c$, opposite angle $C$.
Wait, let's look at the diagram.
Angle at top is $A=76^\circ$. Angle at right is $B=28^\circ$. Angle at left is $C$.
Side $AB$ (bottom) is 15 yd. This side is opposite angle $C$.
We need side $AC$ (left). This side is opposite angle $B$.

Let's re-calculate Angle $C$:
$$C = 180 - 76 - 28 = 76^\circ$$

Now use Sine Law:
$$\frac{AC}{\sin(28^\circ)} = \frac{15}{\sin(76^\circ)}$$
$$AC = \frac{15 \cdot \sin(28^\circ)}{\sin(76^\circ)}$$
$$AC = \frac{15 \cdot 0.4695}{0.9703} \approx 7.26$$

2) Find BC
* Identify: We have side $AB = 10$, angle $A = 54^\circ$, angle $C = 15^\circ$. We need side $BC$ (opposite angle $A$).
* Step 1: Use Sine Law directly since we have a pair (angle $A$ and side $BC$?) No, we have angle $C$ and side $AB$ (opposite $C$). We want side $BC$ (opposite $A$).
$$\frac{BC}{\sin(54^\circ)} = \frac{10}{\sin(15^\circ)}$$
$$BC = \frac{10 \cdot \sin(54^\circ)}{\sin(15^\circ)}$$
$$BC = \frac{10 \cdot 0.8090}{0.2588} \approx 31.26$$

3) Find AC
* Identify: Right triangle. Angle $B = 65^\circ$, Angle $A = 26^\circ$ (Wait, $65+26=91$, close enough to 90 for a right triangle diagram, but let's assume standard trig ratios). Side $AB$ (hypotenuse) = 25 m. We need side $AC$ (adjacent to A, opposite to B).
* Step 1: Use Sine ratio for angle $B$.
$$\sin(65^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{25}$$
$$AC = 25 \cdot \sin(65^\circ)$$
$$AC = 25 \cdot 0.9063 \approx 22.66$$

4) Find $m\angle A$
* Identify: Triangle with sides $CB=7$, $AB=29$, and included angle $B=126^\circ$. We need angle $A$.
* Step 1: Use Law of Cosines to find side $AC$ (side $b$) first.
$$b^2 = 7^2 + 29^2 - 2(7)(29)\cos(126^\circ)$$
$$b^2 = 49 + 841 - 406(-0.5878)$$
$$b^2 = 890 + 238.65 = 1128.65$$
$$b = \sqrt{1128.65} \approx 33.595$$
* Step 2: Use Law of Sines to find angle $A$.
$$\frac{\sin(A)}{7} = \frac{\sin(126^\circ)}{33.595}$$
$$\sin(A) = \frac{7 \cdot \sin(126^\circ)}{33.595}$$
$$\sin(A) = \frac{7 \cdot 0.8090}{33.595} \approx 0.1685$$
$$A = \arcsin(0.1685) \approx 9.7^\circ$$

5) Find $m\angle B$
* Identify: Sides $AB=22$, $AC=21$, Angle $A=56^\circ$. We need angle $B$.
* Step 1: Use Law of Cosines to find side $BC$ (side $a$).
$$a^2 = 22^2 + 21^2 - 2(22)(21)\cos(56^\circ)$$
$$a^2 = 484 + 441 - 924(0.5592)$$
$$a^2 = 925 - 516.7 = 408.3$$
$$a = \sqrt{408.3} \approx 20.206$$
* Step 2: Use Law of Sines to find angle $B$.
$$\frac{\sin(B)}{21} = \frac{\sin(56^\circ)}{20.206}$$
$$\sin(B) = \frac{21 \cdot 0.8290}{20.206} \approx 0.8616$$
$$B = \arcsin(0.8616) \approx 59.5^\circ$$

6) Find $m\angle C$
* Identify: Sides $AB=11$, $BC=19$, Angle $B=58^\circ$. We need angle $C$.
* Step 1: Use Law of Cosines to find side $AC$ (side $b$).
$$b^2 = 11^2 + 19^2 - 2(11)(19)\cos(58^\circ)$$
$$b^2 = 121 + 361 - 418(0.5299)$$
$$b^2 = 482 - 221.5 = 260.5$$
$$b = \sqrt{260.5} \approx 16.14$$
* Step 2: Use Law of Sines to find angle $C$.
$$\frac{\sin(C)}{11} = \frac{\sin(58^\circ)}{16.14}$$
$$\sin(C) = \frac{11 \cdot 0.8480}{16.14} \approx 0.5778$$
$$C = \arcsin(0.5778) \approx 35.3^\circ$$

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Part 2: Solve each triangle

7) Triangle with $A=102^\circ$, $B=24^\circ$, side $c (AB)=25$ cm
* Find Angle $C$:
$$180 - 102 - 24 = 54^\circ$$
* Find Side $a (BC)$:
$$\frac{a}{\sin(102^\circ)} = \frac{25}{\sin(54^\circ)}$$
$$a = \frac{25 \cdot 0.9781}{0.8090} \approx 30.2 \text{ cm}$$
* Find Side $b (AC)$:
$$\frac{b}{\sin(24^\circ)} = \frac{25}{\sin(54^\circ)}$$
$$b = \frac{25 \cdot 0.4067}{0.8090} \approx 12.6 \text{ cm}$$

8) Triangle with $A=27^\circ$, $C=102^\circ$, side $b (AC)=24$ km
* Find Angle $B$:
$$180 - 27 - 102 = 51^\circ$$
* Find Side $a (BC)$:
$$\frac{a}{\sin(27^\circ)} = \frac{24}{\sin(51^\circ)}$$
$$a = \frac{24 \cdot 0.4540}{0.7771} \approx 14.0 \text{ km}$$
* Find Side $c (AB)$:
$$\frac{c}{\sin(102^\circ)} = \frac{24}{\sin(51^\circ)}$$
$$c = \frac{24 \cdot 0.9781}{0.7771} \approx 30.2 \text{ km}$$

9) Triangle with $A=127^\circ$, side $c (AB)=20$ m, side $b (AC)=34$ m
* Find Side $a (BC)$ using Law of Cosines:
$$a^2 = 20^2 + 34^2 - 2(20)(34)\cos(127^\circ)$$
$$a^2 = 400 + 1156 - 1360(-0.6018)$$
$$a^2 = 1556 + 818.45 = 2374.45$$
$$a = \sqrt{2374.45} \approx 48.7 \text{ m}$$
* Find Angle $B$ using Law of Sines:
$$\frac{\sin(B)}{34} = \frac{\sin(127^\circ)}{48.7}$$
$$\sin(B) = \frac{34 \cdot 0.7986}{48.7} \approx 0.5582$$
$$B = \arcsin(0.5582) \approx 33.9^\circ$$
* Find Angle $C$:
$$180 - 127 - 33.9 = 19.1^\circ$$

Final Answer:
1) AC ≈ 7.3 yd
2) BC ≈ 31.3 yd
3) AC ≈ 22.7 m
4) m∠A ≈ 9.7°
5) m∠B ≈ 59.5°
6) m∠C ≈ 35.3°
7) m∠C = 54°, BC ≈ 30.2 cm, AC ≈ 12.6 cm
8) m∠B = 51°, BC ≈ 14.0 km, AB ≈ 30.2 km
9) BC ≈ 48.7 m, m∠B ≈ 33.9°, m∠C ≈ 19.1°