Respuesta :
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:
[tex]c^{2}=a^{2}+b^{2}-2 a b \times \cos (C)[/tex]
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
[tex]\begin{array}{l}{\text { Then, } a^{2}=400^{2}+500^{2}-2 \times 400 \times 500 \times \cos (45)} \\\\ {d^{2}=160000+250000-2 \times 200000 \times \frac{1}{\sqrt{2}}} \\\\ {d^{2}=410000-\sqrt{2} \times 20000 }} \\\\ {d^{2}=410000-282842.712} \\\\ {d^{2}=127157.2875} \\\\ {d=\sqrt{127157.2875}} \\\\ {d=356.59}\end{array}[/tex]
Hence, the distance between two docks is 356.59 feet approximately.
