Math / Trigonometry

QuestionUsing the Law of Sines to solve the all possible triangles if $\angle A=101^{\circ}, a=29, b=12$ . If no answer exists, enter DNE for all answers. $\angle B$ is $\square$ degrees; $\angle C$ is $\square$ degrees; $c=$ $\square$
Round to two decimal places as needed. Assume $\angle A$ is opposite side $a, \angle B$ is opposite side $b$ , and $\angle C$ is opposite side $c$ .

Studdy Solution

STEP 1

1. We are given $\angle A = 101^\circ$ , $a = 29$ , and $b = 12$ .
2. We need to find $\angle B$ , $\angle C$ , and $c$ .
3. The Law of Sines will be used to solve the triangle.
4. Round all answers to two decimal places.

STEP 2

1. Use the Law of Sines to find $\angle B$ .
2. Determine if a valid triangle exists.
3. Calculate $\angle C$ .
4. Use the Law of Sines to find side $c$ .

STEP 3

Use the Law of Sines to find $\angle B$ :
The Law of Sines states: $\frac{a}{\sin A} = \frac{b}{\sin B}$
Substitute the known values: $\frac{29}{\sin 101^\circ} = \frac{12}{\sin B}$
Solve for $\sin B$ : $\sin B = \frac{12 \cdot \sin 101^\circ}{29}$
Calculate $\sin B$ : $\sin B \approx \frac{12 \cdot 0.9816}{29} \approx 0.4068$

STEP 4

Determine if a valid triangle exists:
Check if $\sin B \leq 1$ . Since $\sin B \approx 0.4068$ , a valid triangle exists.
Calculate $\angle B$ : $\angle B = \sin^{-1}(0.4068) \approx 24.02^\circ$

STEP 5

Calculate $\angle C$ :
Use the angle sum property of triangles: $\angle C = 180^\circ - \angle A - \angle B$ $\angle C = 180^\circ - 101^\circ - 24.02^\circ \approx 54.98^\circ$

STEP 6

Use the Law of Sines to find side $c$ :
$\frac{a}{\sin A} = \frac{c}{\sin C}$
Substitute the known values: $\frac{29}{\sin 101^\circ} = \frac{c}{\sin 54.98^\circ}$
Solve for $c$ : $c = \frac{29 \cdot \sin 54.98^\circ}{\sin 101^\circ}$
Calculate $c$ : $c \approx \frac{29 \cdot 0.8192}{0.9816} \approx 24.20$
The solutions are: $\angle B \approx 24.02^\circ$ $\angle C \approx 54.98^\circ$ $c \approx 24.20$

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