2016 AMC 12B Problems/Problem 13

Problem

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

This problem needs a solution. If you have a solution for it, please help us out by adding it. Let's set the altitude = z, distance from Alice to airplane's ground position (point right below airplane)=y and distance from Bob to airplane's ground position=x

From Alice's point of view, tan(θ)=z/y. tan(30)=sin(30)/cos(30)=1/ $sqrt(3)$ . So, y=z* $sqrt(3)$ .

From Bob's point of view, tan(θ)=z/x. tan(60)=sin(60)/cos(60)= $sqrt(3)$ . So, x=z/ $sqrt(3)$ .

We know that $x^2$ + $y^2$ = $10^2$

Solving the equation (by plugging in x and y), we get $sqrt(30)$ = about 5.5.

So, answer is $E) 5.5$

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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2016 AMC 12B Problems/Problem 13

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