Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles? $\textbf{(A)}\ 3.5 \qquad$ $\textbf{(B)}\ 4 \qquad$ $\textbf{(C)}\ 4.5 \qquad$ $\textbf{(D)}\ 5 \qquad$ $\textbf{(E)}\ 5.5$
Solution
Let $P$ be the point on the ground directly below the airplane, and let $h$ be its altitude.
Since $P$ is due north of Alice with elevation angle $30^\circ$, the horizontal distance $AP=h/\tan30^\circ=h\sqrt3$.
Since $P$ is due west of Bob with elevation angle $60^\circ$, the horizontal distance $BP=h/\tan60^\circ=h/\sqrt3$.
Since $AP$ points north and $BP$ points west, $AP\perp BP$, so $AB^2=AP^2+BP^2$.
$100=3h^2+\dfrac{h^2}3=\dfrac{10h^2}3$, so $h^2=30$, $h=\sqrt{30}\approx 5.48$.
Alice and Bob live $20$ miles apart. Alice looks due north and sees an airplane; Bob looks due west and sees the same airplane. The angle of elevation is $30^\circ$ from Alice and $45^\circ$ from Bob. What is the airplane's altitude, in miles?
