# A ship is anchored off a long straight shoreline that runs north and south. From two observation points 15 mi apart on shore, the bearings of the ship are N 31 degrees E and S 53 degrees E. What is the shortest distance from the ship to the shore?

##### 1 Answer
Jul 13, 2018

color(magenta)(6.2 $\text{miles to the nearest 1 decimal place}$

#### Explanation:

$\therefore \text{Let the triangle with the ship and shorline be ABC and let the shortest distance be AD}$

$\therefore \angle A D C \mathmr{and} \angle A D B = {90}^{\circ}$

$\therefore I n \triangle A B D \angle \text{BAD} = {180}^{\circ} - \left({90}^{\circ} + {31}^{\circ}\right) = {59}^{\circ}$

$\therefore I n \triangle A C D \angle \text{CAD} = {180}^{\circ} - \left({90}^{\circ} + {53}^{\circ}\right) = {37}^{\circ}$

$\therefore \angle B A C = {59}^{\circ} + {37}^{\circ} = {96}^{\circ}$

Sine rule

$\therefore \frac{A B}{\sin} C = \frac{B C}{\sin} A$

$\therefore A B = \frac{B C \cdot \sin C}{\sin} A$

$\therefore A B = \frac{15 \cdot \sin 53}{\sin} {96}^{\circ}$

$\therefore A B = 12.046 m i$

$\therefore A D = \sin {31}^{\circ} \cdot 12.046$

:.color(magenta)(=6.204mi="shortest distance"

check:-

$\therefore \frac{A C}{\sin} B = \frac{A B}{\sin} C$

$\therefore \frac{A C}{\sin {31}^{\circ}} = \frac{12.046 \cdot \sin 31}{\sin} {53}^{\circ}$

$\therefore A C = 7.768 m i$

:.color(magenta)(AD=cos37^@*7.768=6.204mi