solve the triangle.
what is the length of side a?
a = \square
(round to the nearest whole number as needed.)
what is the measure of angle b?
b = \square°
(round to the nearest degree as needed.)
what is the measure of angle c?
c = \square°
(round to the nearest degree as needed.)

Question

Accepted Answer

# Explanation:
## Step1: Use the Law of Cosines to find side $ a $
The Law of Cosines states that for a triangle with sides $ b $, $ c $, and $ a $ opposite angles $ B $, $ C $, and $ A $ respectively, $ a^{2}=b^{2}+c^{2}-2bc\cos(A) $. Here, $ b = 8 $, $ c = 24 $, and $ A=34^{\circ} $.
$$
\begin{align*}
a^{2}&=8^{2}+24^{2}-2\times8\times24\times\cos(34^{\circ})\
a^{2}&=64 + 576- 384\times\cos(34^{\circ})\
\cos(34^{\circ})&\approx0.8290\
a^{2}&=640-384\times0.8290\
a^{2}&=640 - 318.336\
a^{2}&=321.664\
a&\approx\sqrt{321.664}\approx17.94\approx18
\end{align*}
$$
## Step2: Use the Law of Sines to find angle $ B $
The Law of Sines is $ \frac{\sin(B)}{b}=\frac{\sin(A)}{a} $. We know $ b = 8 $, $ a\approx18 $, and $ A = 34^{\circ} $.
$$
\begin{align*}
\sin(B)&=\frac{b\times\sin(A)}{a}\
\sin(B)&=\frac{8\times\sin(34^{\circ})}{18}\
\sin(34^{\circ})&\approx0.5592\
\sin(B)&=\frac{8\times0.5592}{18}\
\sin(B)&=\frac{4.4736}{18}\approx0.2485\
B&\approx\arcsin(0.2485)\approx14^{\circ}
\end{align*}
$$
## Step3: Find angle $ C $
The sum of angles in a triangle is $ 180^{\circ} $, so $ C=180^{\circ}-A - B $.
$$
\begin{align*}
C&=180^{\circ}-34^{\circ}-14^{\circ}\
C&=132^{\circ}
\end{align*}
$$
# Answer:
Length of side $ a $: $ \boldsymbol{18} $
Measure of angle $ B $: $ \boldsymbol{14^{\circ}} $
Measure of angle $ C $: $ \boldsymbol{132^{\circ}} $

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QUESTION IMAGE

Question

Explanation:

Step1: Use the Law of Cosines to find side \( a \)

Step2: Use the Law of Sines to find angle \( B \)

Step3: Find angle \( C \)

Answer: