A ship sails $10$ miles in a straight line from $A$ to $B$, turns through an angle between $45^{\circ}$ and $60^{\circ}$, and then sails another $20$ miles to $C$. Let $AC$ be measured in miles. Which of the following intervals contains $AC^2$? $\textbf{(A)}\ [400,500] \qquad$ $ \textbf{(B)}\ [500,600] \qquad$ $ \textbf{(C)}\ [600,700] \qquad$ $ \textbf{(D)}\ [700,800] \qquad$ $ \textbf{(E)}\ [800,900]$
Solution
If the turn angle is $\beta$, the interior angle at $B$ is $180^\circ-\beta$, so $AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cos(180^\circ-\beta)=AB^2+BC^2+2\cdot AB\cdot BC\cos\beta$.
With $AB=10,BC=20$: $AC^2=500+400\cos\beta$. At $\beta=45^\circ$: $AC^2=500+200\sqrt2\approx782.8$. At $\beta=60^\circ$: $AC^2=700$.
So $AC^2\in[700,782.8]\subset[700,800]$, answer (D).
$\textbf{(A)}\ [400,500] \qquad$ $ \textbf{(B)}\ [500,600] \qquad$ $ \textbf{(C)}\ [600,700] \qquad$ $ \textbf{(D)}\ [700,800] \qquad$ $ \textbf{(E)}\ [800,900]$ 
