2016 AMC 12B Problems/Problem 13
Contents
Problem
Alice and Bob live 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 from Alice's position and from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
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, . . So,
From Bob's point of view, . . So,
We know that + =
Solving the equation (by plugging in x and y), we get z= = about 5.5.
So, answer is
solution by sudeepnarala
Solution 2 (Use One Variable)
We have two 30-60-90 triangles and that are perpendicular and share leg (the altitude of the plane ). The shared leg is the shortest leg of one triangle and the longest leg of the other. and are Bob and Alice respectively.
Find and in terms of . Use Pythagorean Theorem on triangle to produce
EXAMPLE: Let angle be 30 degrees and angle be 60 degrees. , and by special triangle rules. By Pythagorean Theorem, , so .
(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 and , respectively. From the Pythagorean Theorem, . As both are triangles, the altitude of the airplane can be expressed as or . Solving the equation , we get . Plugging this into the equation , we get , or ( cannot be negative), so the altitude is , which is closest to
Solution 4: Diagram
~credit to mathmaster2012 for original diagram
~BakedPotato66 added/contributed some elements
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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