Exercise 362

A ship sails $10$ miles from $A$ to $B$, turns through an angle of exactly $45^\circ$, and sails another $30$ miles to $C$. What is $AC^2$?

Solution

If $\beta$ is the turn angle, $AC^2=AB^2+BC^2+2\cdot AB\cdot BC\cos\beta$ (since the interior angle at $B$ is $180^\circ-\beta$).
With $AB=10,BC=30,\beta=45^\circ$: $AC^2=100+900+600\cdot\frac{\sqrt2}2=1000+300\sqrt2$.
$AC^2=\boxed{1000+300\sqrt2}$.