Question

Two boats are equidistant from a lighthouse. The boats are 30 miles apart. The angle formed between the two boats, with the lighthouse as the vertex, measures 40°. Approximately how far is each boat from the lighthouse?
A.20 mi.
B.21 mi.
C.44 mi.
D.46 mi.

Answer 1

Answer:

Each boat are 43.86 miles from the lighthouse.

Step-by-step explanation:

From the diagram, at point A the lighthouse is placed and at B, C two boats are placed.

As the two boats are equidistant from a lighthouse, so AB=AC.

Hence, ΔABC is an isosceles triangle.

The angle formed between the two boats, with the lighthouse as the vertex, measures 40°. So m∠A=40°.

The altitude to the base of an isosceles triangle bisects the vertex angle.

Hence, $m\angle BAE=m\angle CAE=20^{\circ}$

The altitude to the base of an isosceles triangle bisects the base.

Hence, $BE=CE=15$

In ΔABE,

$\sin 20=\dfrac{BE}{AB}=\dfrac{15}{AB}$

$\Rightarrow AB=\dfrac{15}{\sin 20}=43.86$

As AB=AC, so AC=43.86

Answer 2

Using the Law of Sines  (sina/A=sinb/B=sinc/C for any triangle)

(sin40)/30=(sin(180-40)/2)/d

d=(30sin70)/sin40

d≈43.857mi

d≈44 mi to the nearest whole mile...