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makvit [3.9K]
3 years ago
9

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 capta

in 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?

Mathematics
1 answer:
grin007 [14]3 years ago
5 0

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

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