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?
$\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5$
The attached image shows my solution: $A$ and $B$ each denote where Alice and Bob are, and from their angles of elevation, I was able to figure out that the plane would be where the right angle of a $30-60-90$° triangle would be. Letting $a$ be the altitude, I set up $10\cdot a = 5\cdot 5\sqrt{3}$ and solved by estimating $\sqrt{3} \approx 1.7$ ($\sqrt{3}>1.7$). Solving for $a$ gave me $4.25$, but since $\sqrt{3}$ was rounded down in my estimation (but not too much), I assumed the answer would be (C). However, it turns out that I am wrong. But why doesn't my method work? 
Best Answer
I don't understand your diagram. You don't seem to have used the parts which specify that Alice was looking due North and Bob due West. You have to think three-dimensionally.
Let the altitude be $h$.
Designate the point on the ground directly (vertically) under the plane $P$.
The distance from Alice to point $P$ is $\frac h{\tan 30^{\circ}} = \sqrt 3 h$. Furthermore $P$ lies due North of Alice.
The distance of Bob to point $P$ is $\frac h{\tan 60^{\circ}} = \frac h{\sqrt 3}$. Furthermore $P$ lies due West of Bob.
You can now construct a right angled triangle on the ground (assumed flat) with sides $h\sqrt 3, \frac h{\sqrt 3}, 10$ where $10 \ \mathrm {miles}$ is the hypotenuse.
By Pythagoras, $h^2(3 + \frac 13) = 10^2 \implies h = \sqrt{30} \approx 5.5 \ \mathrm {miles}$, and the closest is option $E$.