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A boat is 400 feet away from one dock and 500 feet away from another dock.

The angle between the two paths is 45°. What is the approximate distance
between the docks?
O
A. 695 feet
O
O
O
B. 540 feet
C. 166 feet
D. 357 feet

1 Answer

3 votes

The approximate distance between the docks is 356.59 feet approximately.

Solution:

Given, A boat is 400 feet away from one dock and 500 feet away from another dock.

The angle between the two paths is 45°.

Now, Let d be the distance between the docks, which forms a third leg of a triangle for which the angle opposite it has measure 45 degrees.

The other two legs have lengths 400 ft and 500 ft.

So, let us use the law of cosines:


c^(2)=a^(2)+b^(2)-2 a b * \cos (C)

where a, b, c are sides of triangle and C is angle opposite to side c.

Here in our problem, c = d, a = 400, b = 500 and C = 45


\begin{array}{l}{\text { Then, } a^(2)=400^(2)+500^(2)-2 * 400 * 500 * \cos (45)} \\\\ {d^(2)=160000+250000-2 * 200000 * (1)/(√(2))} \\\\ {d^(2)=410000-√(2) * 20000 }} \\\\ {d^(2)=410000-282842.712} \\\\ {d^(2)=127157.2875} \\\\ {d=√(127157.2875)} \\\\ {d=356.59}\end{array}

Hence, the distance between two docks is 356.59 feet approximately.

User Alberto Montalesi
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