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Difference between revisions of "2016 AMC 12B Problems/Problem 13"

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solution by sudeepnarala
 
solution by sudeepnarala
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==Solution 2==
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Non-trig solution by e_power_pi_times_i
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Set the disance from Alice's and Bob's position to the point directly below the airplane to be <math>x</math> and <math>y</math>, respectively. From the Pythagorean Theorem, <math>x^2 + y^2 = 100</math>. As both are <math>30-60-90</math> triangles, the altitude of the airplane can be expressed as <math>\dfrac{x\sqrt{3}}{3}</math> or <math>y\sqrt{3}</math>. Solving the equation <math>\dfrac{x\sqrt{3}}{3} = y\sqrt{3}</math>, we get <math>x = 3y</math>. Plugging this into the equation <math>x^2 + y^2 = 100</math>, we get <math>10y^2 = 100</math>, or <math>y = \sqrt{10}</math> (<math>y</math> cannot be negative), so the altitude is <math>\sqrt{3*10} = \sqrt{30}</math>, which is closest to $\boxed{\textbf{B)}\ 5.5}.
  
 
==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 12:56, 28 February 2016

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

Non-trig solution by e_power_pi_times_i


Set the disance 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{B)}\ 5.5}.

See Also

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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