Solve the problem.
-A person is watching a boat from the top of a lighthouse. The boat is approaching the lighthouse directly. When first noticed the angle of depression to the boat is $15 ^ { \circ } 57 ^ { \prime }$. When the boat stops, the angle of depression is $55 ^ { \circ } 5 ^ { \prime }$ . The lighthouse is 200 feet tall. How far did the boat travel from when it was first noticed until it stopped? Round your answer to the hundredths place.
A) $560.18 \mathrm { ft }$
B) $605.38 \mathrm { ft }$
C) $544.88 \mathrm { ft }$
D) $582.68 \mathrm { ft }$

Question

Quizplus · Accepted Answer

To solve this problem, we use trigonometry, specifically the tangent of the angles of depression, which gives us the ratio of the opposite side (distance traveled by the boat) to the adjacent side (height of the lighthouse). Let $d_1$ and $d_2$ be the distances of the boat from the base of the lighthouse when first noticed and when it stopped, respectively.For the first angle of depression $15^\circ 57'$:$$ \tan(15^\circ 57') = \frac{d_1}{200} $$$$ d_1 = 200 \cdot \tan(15^\circ 57') $$For the second angle of depression $55^\circ 5'$:$$ \tan(55^\circ 5') = \frac{d_2}{200} $$$$ d_2 = 200 \cdot \tan(55^\circ 5') $$The distance traveled by the boat is the difference between $d_1$ and $d_2$:$$ \text{Distance traveled} = d_2 - d_1 $$Calculating $d_1$ and $d_2$ with the given angles:$$ d_1 = 200 \cdot \tan(15^\circ 57') \approx 57.29 \, \text{ft} $$$$ d_2 = 200 \cdot \tan(55^\circ 5') \approx 617.47 \, \text{ft} $$$$ \text{Distance traveled} = 617.47 \, \text{ft} - 57.29 \, \text{ft} \approx 560.18 \, \text{ft} $$Therefore, the boat traveled approximately $560.18$ feet.

Solve the Problem 15∘57′15 ^ { \circ } 57 ^ { \prime }15∘57′

Solve the Problem $15 ^ { \circ } 57 ^ { \prime }$