Question

a boat is heading towards a lighthouse, where alexander is watching from a vertical distance of 102 feet above the water. alexander measures an angle of depression to the boat at point a to be 21°. at some later time, alexander takes another measurement and finds the angle of depression to the boat (now at point b) to be 71°. find the distance from point a to point b. round your answer to the nearest tenth of a foot if necessary.

Explanation:

Step1: Find the distance from \(A\) to \(L\)

The angle of depression from the lighthouse to point \(A\) is \(21^{\circ}\). Using the tangent function in the right - triangle formed by the lighthouse height (\(h = 102\) feet) and the distance from \(A\) to \(L\) (\(x_{A}\)).
We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\tan(21^{\circ})=\frac{102}{x_{A}}\), so \(x_{A}=\frac{102}{\tan(21^{\circ})}\).
Since \(\tan(21^{\circ})\approx0.384\), then \(x_{A}=\frac{102}{0.384}\approx265.6\) feet.

Step2: Find the distance from \(B\) to \(L\)

The angle of depression from the lighthouse to point \(B\) is \(71^{\circ}\). Using the tangent function in the right - triangle formed by the lighthouse height (\(h = 102\) feet) and the distance from \(B\) to \(L\) (\(x_{B}\)).
We know that \(\tan(71^{\circ})=\frac{102}{x_{B}}\), so \(x_{B}=\frac{102}{\tan(71^{\circ})}\).
Since \(\tan(71^{\circ})\approx2.904\), then \(x_{B}=\frac{102}{2.904}\approx35.1\) feet.

Step3: Calculate the distance from \(A\) to \(B\)

The distance from \(A\) to \(B\) (\(d\)) is \(d=x_{A}-x_{B}\).
Substitute the values of \(x_{A}\) and \(x_{B}\): \(d = 265.6-35.1=230.5\) feet.

Answer:

The distance from point \(A\) to point \(B\) is \(230.5\) feet.