8.2 assignment: law of cosines
score: 40/100 answered: 2/5
● question 3
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} $.
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Accepted Answer

# Explanation:
## Step1: Calculate side $a$ using the Law of Cosines
The Law of Cosines formula is $a^{2}=b^{2}+c^{2}-2bc\cos\alpha$.
Substitute $b = 10.9$, $c = 14.4$, and $\alpha=38.9^{\circ}$ (so $\cos\alpha=\cos(38.9^{\circ})\approx0.7794$) into the formula:
$$
\begin{align*}
a^{2}&=(10.9)^{2}+(14.4)^{2}-2\times10.9\times14.4\times0.7794\
a^{2}&=118.81 + 207.36-2\times10.9\times14.4\times0.7794\
a^{2}&=326.17-(314.88\times0.7794)\
a^{2}&=326.17 - 245.32\
a^{2}&=80.85\
a&\approx8.99
\end{align*}
$$
## Step2: Calculate $\cos\beta$ using the Law of Cosines
The Law of Cosines formula for $\cos\beta$ is $\cos\beta=\frac{a^{2}+c^{2}-b^{2}}{2ac}$.
Substitute $a\approx8.99$, $b = 10.9$, $c = 14.4$ into the formula:
$$
\begin{align*}
\cos\beta&=\frac{(8.99)^{2}+(14.4)^{2}-(10.9)^{2}}{2\times8.99\times14.4}\
&=\frac{80.82+207.36 - 118.81}{259.01}\
&=\frac{169.37}{259.01}\
&\approx0.6539
\end{align*}
$$
## Step3: Calculate $\beta$
Since $\beta=\cos^{-1}(0.6539)$, $\beta\approx49.00^{\circ}$
## Step4: Calculate $\cos\gamma$ using the Law of Cosines
The Law of Cosines formula for $\cos\gamma$ is $\cos\gamma=\frac{a^{2}+b^{2}-c^{2}}{2ab}$.
Substitute $a\approx8.99$, $b = 10.9$, $c = 14.4$ into the formula:
$$
\begin{align*}
\cos\gamma&=\frac{(8.99)^{2}+(10.9)^{2}-(14.4)^{2}}{2\times8.99\times10.9}\
&=\frac{80.82 + 118.81-207.36}{196.00}\
&=\frac{-7.73}{196.00}\
&\approx - 0.0394
\end{align*}
$$
## Step5: Calculate $\gamma$
Since $\gamma=\cos^{-1}(-0.0394)$, $\gamma\approx92.20^{\circ}$

# Answer:
$a\approx8.99$ km, $\cos\beta\approx0.6539$, $\beta\approx49.00^{\circ}$, $\cos\gamma\approx - 0.0394$, $\gamma\approx92.20^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate side \(a\) using the Law of Cosines

Step2: Calculate \(\cos\beta\) using the Law of Cosines

Step3: Calculate \(\beta\)

Step4: Calculate \(\cos\gamma\) using the Law of Cosines

Step5: Calculate \(\gamma\)

Answer: