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

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, $\tan(\theta)=\frac{z}{y}$. $\tan{30}=\frac{\sin{30}}{\cos{30}}=\frac{1}{\sqrt{3}}$. So, $y=z*\sqrt{3}$

From Bob's point of view, $\tan(\theta)=\frac{z}{x}$. $\tan{60}=\frac{\sin{60}}{\cos{60}}=\sqrt{3}$. So, $x = \frac{z}{\sqrt{3}}$

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

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

So, answer is $E) 5.5$

solution by sudeepnarala

Solution 2 (Use One Variable)

We have two 30-60-90 triangles $ABC$ and $DBC$ that are perpendicular and share leg $BC$ (the altitude of the plane $p$). $AD=10$ The shared leg is the shortest leg of one triangle and the longest leg of the other. $A$ and $B$ are Bob and Alice respectively.


Find $AC$ and $DC$ in terms of $p$. Use Pythagorean Theorem on triangle $ADC$ to produce $p=\sqrt{30}\implies\boxed{\textbf{E)}\ 5.5}$

EXAMPLE: Let angle $A$ be 30 degrees and angle $B$ be 60 degrees. $AC=p\sqrt{3}$, and $DC=\frac{p}{\sqrt{3}}$ by special triangle rules. By Pythagorean Theorem, $AC^2+DC^2=AB^2$, so $(p\sqrt{3})^2+(\frac{p}{\sqrt{3}})^2=10^2$. $p=\sqrt{30}$

(Solution by BJHHar)

Solution 3

Non-trig solution by e_power_pi_times_i


Set the distance from Alice's and Bob's position to the point directly below the airplane to be $x$ and $y$, respectively. From the Pythagorean Theorem, $x^2 + y^2 = 100$. As both are $30-60-90$ triangles, the altitude of the airplane can be expressed as $\dfrac{x\sqrt{3}}{3}$ or $y\sqrt{3}$. Solving the equation $\dfrac{x\sqrt{3}}{3} = y\sqrt{3}$, we get $x = 3y$. Plugging this into the equation $x^2 + y^2 = 100$, we get $10y^2 = 100$, or $y = \sqrt{10}$ ($y$ cannot be negative), so the altitude is $\sqrt{3*10} = \sqrt{30}$, which is closest to $\boxed{\textbf{E)}\ 5.5}$

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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