Difference between revisions of "2016 AMC 12B Problems/Problem 13"
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We know that <Math>x^2</Math> + <Math>y^2</Math> = <Math>10^2</Math> | We know that <Math>x^2</Math> + <Math>y^2</Math> = <Math>10^2</Math> | ||
| − | + | Solving the equation (by plugging in x and y), we get <Math>sqrt(30)</Math> = about 5.5. | |
| + | |||
| + | So, answer is <Math>E) 5.5</Math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=12|num-a=14}} | {{AMC12 box|year=2016|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:52, 21 February 2016
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
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/
. So, y=z*.
From Bob's point of view, tan(θ)=z/x. tan(60)=sin(60)/cos(60)=. So, x=z/.
We know that 
+ 
= 
Solving the equation (by plugging in x and y), we get 
= about 5.5.
So, answer is 
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
