Question

A ship is sailing due north. At a certain​ point, the bearing of a lighthouse 3.7 km away is N37.6degreesE. Later​ on, the captain notices that the bearing of the lighthouse has become Upper S 34.5 degrees E.
How far did the ship travel between the two observations of the​ lighthouse?

Answer 1

Answer:

6.2 km

Step-by-step explanation:

Let x be the distance travel by ship between the two observations of the lighthouse.

AC=3.7 km

$\angle B=34.5^{\circ}$

$\angle A=37.6^{\circ}$

In triangle ABC,

$\angle A+\angle B+\angle C=180^{\circ}$

Triangle angle sum property

Substitute the values

$37.6+34.5+\angle C=180$

$\angle C=107.9^{\circ}$

sin law:$\frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C}$

Taking $\frac{b}{sin B}=\frac{c}{sin C}$

Substitute the values then we get

$\frac{3.7}{sin 34.5}=\frac{x}{sin 107.9}$

$x=\frac{3.7\times sin 107.9}{sin 34.5}$

$x=6.2 km$

Hence, the ship travel between the two observations of the light house=6. 2 km