Question

A ship is anchored off a long straight shoreline that runs north and south. From the ship, there are two observation points 18 miles apart. One is 54° north of East, and the other is 36° south of East. What is the shortest distance from the ship to the shore. Round to the nearest tenth of a mile.

Answer 1

First, draw a diagram of the situation to visualize the problem:

Notice that the triangle ASB is a right triangle because the angle ASB has a measure of 90º. The segment SP is perpendicular to AB, then the triangles BSP and ASP are also right triangles.

Notice that:

$\sin (\angle SAP)=\frac{SP}{SA}=\frac{x}{SA}$

On the other hand:

$\sin (\angle ABS)=\frac{SA}{AB}$

Isolate SA and replace AB=18 and ABS=54º:

$SA=AB\cdot\sin (\angle ABS)=18\cdot\sin (54º)$

From the first equation, isolate x and replace the value of SA and the measure of the angle SAP=36º:

$x=SA\cdot\sin (\angle SAP)=18\cdot\sin (54º)\sin (36º)$

Use a calculator to find the value of x:

$x=18\cdot\sin (54º)\cdot\sin (36º)=8.5595\ldots\approx8.6$

Therefore, to the nearest tenth of a mile, the distance from the ship to the shore is 8.6 miles.