Question

question
a boat is heading towards a lighthouse, whose beacon - light is 113 feet above the water. the boats crew measures the angle of elevation to the beacon, 15°. what is the ships horizontal distance from the lighthouse (and the shore)? round your answer to the nearest tenth of a foot if necessary.

Explanation:

Step1: Set up the tangent ratio

We know that in a right - triangle (where the height of the lighthouse is the opposite side and the horizontal distance is the adjacent side with respect to the angle of elevation), the tangent of an angle of elevation \(\theta\) is given by \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Let \(h = 113\) feet (height of the lighthouse, opposite side) and \(x\) be the horizontal distance (adjacent side). Given \(\theta = 15^{\circ}\), so \(\tan(15^{\circ})=\frac{113}{x}\).

Step2: Solve for \(x\)

We know that \(\tan(15^{\circ})=\tan(45^{\circ}- 30^{\circ})=\frac{\tan45^{\circ}-\tan30^{\circ}}{1 + \tan45^{\circ}\tan30^{\circ}}=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=\frac{3-\sqrt{3}}{3 + \sqrt{3}}=\frac{(3 - \sqrt{3})(3-\sqrt{3})}{(3+\sqrt{3})(3 - \sqrt{3})}=\frac{9-6\sqrt{3}+3}{9 - 3}=\frac{12-6\sqrt{3}}{6}=2-\sqrt{3}\approx0.2679\).
From \(\tan(15^{\circ})=\frac{113}{x}\), we can solve for \(x\) by \(x=\frac{113}{\tan(15^{\circ})}\). Substituting \(\tan(15^{\circ})\approx0.2679\), we get \(x=\frac{113}{0.2679}\approx421.8\).

Answer:

\(421.8\) feet