Solve the problem.
-A ship travels $74 \mathrm {~km}$ on a bearing of $38 ^ { \circ }$, and then travels on a bearing of $128 ^ { \circ }$ for $122 \mathrm {~km}$. Find the distance from the starting point to the end of the trip, to the nearest kilometer.
A) $143 \mathrm {~km}$
B) $196 \mathrm {~km}$
C) $58 \mathrm {~km}$
D) $46 \mathrm {~km}$

Question

Quizplus · Accepted Answer

To solve this problem, we can use the law of cosines since we have a non-right triangle formed by the ship's path. The change in bearing from $38^\circ$ to $128^\circ$ is $90^\circ$, making the angle between the two paths $90^\circ$. The law of cosines states that $c^2 = a^2 + b^2 - 2ab\cos(C)$, where $C$ is the angle opposite side $c$, and $a$ and $b$ are the other two sides. Here, $a = 74$, $b = 122$, and $C = 90^\circ$, so $\cos(90^\circ) = 0$, simplifying the equation to $c^2 = a^2 + b^2$. Plugging in the values, we get $c^2 = 74^2 + 122^2$, which calculates to $c \approx 143$. Therefore, the distance from the starting point to the end of the trip is approximately $143 \mathrm{km}$, rounded to the nearest kilometer.

Solve the Problem 74 km74 \mathrm {~km}74 km On a Bearing Of 38∘38 ^ { \circ }38∘

Solve the Problem $74 \mathrm {~km}$ On a Bearing Of $38 ^ { \circ }$