assume $ \angle \alpha $ is opposite side $ a, \angle \beta $ is opposite side $ b $, and $ \angle \gamma $ is opposite side $ c $.
solve triangle $ a b c $ if $ \angle a=\alpha=38.9^{\circ}, b = 10.9 \mathrm{~km} $, and $ c = 14.4 \mathrm{~km} $.
using the law of cosines,
$ a \approx $ km.
your answer should accurate to 2 decimal places.
using the law of cosines again,
$ \cos \angle b=\cos \beta \approx $
your answer should accurate to 5 decimal places.
thus, $ \beta \approx $
your answer should accurate to 2 decimal places.
$ \cos \angle c=\cos \gamma \approx $
your answer should accurate to 5 decimal places.
thus, $ \gamma \approx $
your answer should accurate to 2 decimal places.
check by seeing if the total of the three angles is $ 180^{circ} $.

Question

Accepted Answer

# Explanation:
## Step1: Recall the Law of Cosines formula
The Law of Cosines formula is $a^{2}=b^{2}+c^{2}-2bc\cos\alpha$. Given $\alpha = 38.9^{\circ}$, $b = 10.9$ km, and $c = 14.4$ km.
$$
\begin{align*}
a^{2}&=(10.9)^{2}+(14.4)^{2}-2\times10.9\times14.4\times\cos(38.9^{\circ})\
a^{2}&=118.81 + 207.36-314.88\times0.7794\
a^{2}&=326.17-245.33\
a^{2}&=80.84\
a&\approx9.05
\end{align*}
$$
## Step2: Use the Law of Cosines to find $\cos\beta$
The Law of Cosines formula for $\cos\beta$ is $\cos\beta=\frac{a^{2}+c^{2}-b^{2}}{2ac}$. Substitute $a = 9.05$, $b = 10.9$, $c = 14.4$
$$
\begin{align*}
\cos\beta&=\frac{(9.05)^{2}+(14.4)^{2}-(10.9)^{2}}{2\times9.05\times14.4}\
&=\frac{81.90 + 207.36-118.81}{259.92}\
&=\frac{170.45}{259.92}\
&\approx0.6558
\end{align*}
$$
## Step3: Use the Law of Cosines to find $\cos\gamma$
The Law of Cosines formula for $\cos\gamma$ is $\cos\gamma=\frac{a^{2}+b^{2}-c^{2}}{2ab}$. Substitute $a = 9.05$, $b = 10.9$, $c = 14.4$
$$
\begin{align*}
\cos\gamma&=\frac{(9.05)^{2}+(10.9)^{2}-(14.4)^{2}}{2\times9.05\times10.9}\
&=\frac{81.90+118.81 - 207.36}{197.29}\
&=\frac{-7.65}{197.29}\
&\approx - 0.0388
\end{align*}
$$

# Answer:
$\cos\beta\approx0.6558$, $\cos\gamma\approx - 0.0388$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the Law of Cosines formula

Step2: Use the Law of Cosines to find \(\cos\beta\)

Step3: Use the Law of Cosines to find \(\cos\gamma\)

Answer: