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2016 AMC 12B Problem 13

Problem 13 of 25IntermediateGeometry

Alice and Bob live 1010 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∘30^\circ from Alice’s position and 60∘60^\circ from Bob’s position. Which of the following is closest to the airplane’s altitude, in miles?

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Solution

Let the airplane be at C,C, directly above point DD on the ground at altitude h.h. Triangles ACDACD and BCDBCD are 3030-6060-9090 right triangles, so AD=3 hAD=\sqrt3\,h and BD=h3.BD=\dfrac{h}{\sqrt3}. Since Alice looks north and Bob looks west, ∠ADB=90∘,\angle ADB=90^\circ, so AD2+BD2=AB2=100.AD^2+BD^2=AB^2=100. Then 3h2+h23=10h23=100,3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100, giving h=30≈5.48,h=\sqrt{30}\approx5.48, closest to 5.5.5.5. Thus, the correct answer is E.
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Tagged: special right triangle · Pythagorean Theorem · 3D geometry

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