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4 years ago

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Two boats are equidistant from a lighthouse. The boats are 30 miles apart. The angle formed between the two boats, with the ligh

thouse 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.

2 answers:

ryzh [129]4 years ago

6 0

Answer:

<em>Each boat are </em><em>43.86</em><em> miles from the lighthouse.</em>

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

anastassius [24]4 years ago

3 0

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...

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$\huge\text{Hey there!}$

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