a rover on mars leaves station a and travels 22 km due east to sample rocks at outcrop b. it then adjusts its course 51° southward. it travels 55 km in that direction until it reaches deposit c to sample soil. what angle θ with respect to due south could the rover have used to travel directly from station a to deposit c? see the figure below. (the figure is not drawn to scale.)
carry your intermediate computations to at least four decimal places.
round your answer to the nearest tenth of a degree.
θ = □°

Question

Accepted Answer

# Explanation:
## Step1: Use the Law of Cosines
The Law of Cosines formula is $c^{2}=a^{2}+b^{2}-2ab\cos C$. In $\triangle ABC$, $a = 22$, $b=55$, and $C = 180^{\circ}- 51^{\circ}=129^{\circ}$. First, find the length of $AC$ (let $AC = c$).
$$
\begin{align*}
c^{2}&=22^{2}+55^{2}-2\times22\times55\times\cos(129^{\circ})\
c^{2}&=484 + 3025- 2420\times(-0.6293)\
c^{2}&=484+3025 + 1522.906\
c^{2}&=5031.906\
c&\approx70.94
\end{align*}
$$
## Step2: Use the Law of Sines
The Law of Sines formula is $\frac{\sin A}{a}=\frac{\sin C}{c}$. We want to find $\theta$ (angle $A$), $a = 55$, $c\approx70.94$, $C = 129^{\circ}$.
$$
\begin{align*}
\frac{\sin\theta}{55}&=\frac{\sin(129^{\circ})}{70.94}\
\sin\theta&=\frac{55\times\sin(129^{\circ})}{70.94}\
\sin\theta&=\frac{55\times0.7771}{70.94}\
\sin\theta&=\frac{42.7405}{70.94}\
\sin\theta&\approx0.6025
\end{align*}
$$
## Step3: Find the angle $\theta$
Since $\sin\theta\approx0.6025$, then $\theta=\sin^{- 1}(0.6025)\approx37.1^{\circ}$

# Answer:
$37.1^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the Law of Cosines

Step2: Use the Law of Sines

Step3: Find the angle \(\theta\)

Answer: