solve the triangle.
a ≈ □° (round to one decimal place as needed.)
b ≈ □° (round to one decimal place as needed.)
c ≈ □ (round to two decimal places as needed.)

Question

Accepted Answer

# Explanation:
## Step1: Use the Law of Cosines
The Law of Cosines formula is $c^{2}=a^{2}+b^{2}-2ab\cos C$. Here, $a = 6$, $b = 9$, $C=105^{\circ}$.
$$
\begin{align*}
c^{2}&=6^{2}+9^{2}-2\times6\times9\times\cos(105^{\circ})\
&=36 + 81-108\times(-0.2588)\
&=117+27.9504\
&=144.9504\
c&=\sqrt{144.9504}\approx12.04
\end{align*}
$$
## Step2: Use the Law of Sines
The Law of Sines formula is $\frac{\sin A}{a}=\frac{\sin C}{c}$.
$$
\begin{align*}
\sin A&=\frac{a\sin C}{c}\
&=\frac{6\times\sin(105^{\circ})}{12.04}\
&=\frac{6\times0.9659}{12.04}\
&=\frac{5.7954}{12.04}\approx0.4814\
A&=\sin^{-1}(0.4814)\approx28.8^{\circ}
\end{align*}
$$
## Step3: Find angle B
Since the sum of angles in a triangle is $180^{\circ}$, $B = 180^{\circ}-A - C$.
$$
\begin{align*}
B&=180^{\circ}-28.8^{\circ}-105^{\circ}\
&=46.2^{\circ}
\end{align*}
$$

# Answer:
$c\approx12.04$, $A\approx28.8^{\circ}$, $B\approx46.2^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the Law of Cosines

Step2: Use the Law of Sines

Step3: Find angle B

Answer: