Question

a boat is heading towards a lighthouse, whose beacon - light is 126 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: Recall tangent formula

We know that in a right - triangle, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Here, the height of the lighthouse above water is the opposite side ($h = 126$ feet) and the horizontal distance of the boat from the lighthouse is the adjacent side ($x$), and the angle of elevation $\theta=15^{\circ}$. So, $\tan\theta=\tan(15^{\circ})=\frac{126}{x}$.

Step2: Calculate $\tan(15^{\circ})$

We know that $\tan(15^{\circ})=\tan(45^{\circ}- 30^{\circ})=\frac{\tan45^{\circ}-\tan30^{\circ}}{1 + \tan45^{\circ}\tan30^{\circ}}=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=\frac{3 - \sqrt{3}}{3+\sqrt{3}}$. Rationalize the denominator: $\frac{(3 - \sqrt{3})(3-\sqrt{3})}{(3+\sqrt{3})(3 - \sqrt{3})}=\frac{9-6\sqrt{3}+3}{9 - 3}=\frac{12-6\sqrt{3}}{6}=2-\sqrt{3}\approx2 - 1.732 = 0.268$.

Step3: Solve for $x$

Since $\tan(15^{\circ})=\frac{126}{x}$, then $x=\frac{126}{\tan(15^{\circ})}$. Substitute $\tan(15^{\circ})\approx0.268$ into the formula, we get $x=\frac{126}{0.268}\approx470.1$.

Answer:

$470.1$ feet