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)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bib7lik1ogfou53bh3cmvd12f2p59jdx5z.png)
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}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/idcqw3w83ioy3ckgcungf80647u5wzexit.png)
Hence, the distance between two docks is 356.59 feet approximately.