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

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Set the distance 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 <math>\boxed{\textbf{E)}\ 5.5}</math>
 
Set the distance 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 <math>\boxed{\textbf{E)}\ 5.5}</math>
  
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== Solution 4: Diagram ==
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==Solution 4 (Formal)==
 +
 
 +
Let Alice be at point <math>A</math>, Bob be at point <math>B</math>. Let the plane be at point <math>P</math> and <math>X</math> be the projection of <math>P</math> onto the ground (the plane with contains <math>A</math> and <math>B</math>). Let the height of the plane, or <math>PX</math>, be <math>h</math>. So, because of the <math>30-60-90</math> triangles,
 +
<cmath>AX = \dfrac{h}{\sqrt{3}}, BX = h\sqrt{3}</cmath>
 +
By [[Pythagorean Theorem]] on <math>\triangle ABX</math>,
 +
<cmath>\dfrac{h^2}{3} + 3h^2 = 100 \implies \dfrac{10h^2}{3} = 100 \implies h = \sqrt{30},</cmath>
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which is clossest to <math>\boxed{\textbf{(E)}\ 5.5}.</math>
 +
 
 +
~sml1809
 +
 
 +
==Solution 5: Diagram ==
 
<asy>
 
<asy>
 
import graph; usepackage("amsmath"); size(7.2cm);
 
import graph; usepackage("amsmath"); size(7.2cm);
Line 80: Line 90:
  
 
~credit to mathmaster2012 for original diagram
 
~credit to mathmaster2012 for original diagram
 +
 
~BakedPotato66 added/contributed some elements
 
~BakedPotato66 added/contributed some elements
 +
 +
==Video Solution by CanadaMath (Problem 11-20)==
 +
Fast Forward to 9:16 for problem 13
 +
https://www.youtube.com/watch?v=4osvFatUv1o
 +
 +
~THEMATHCANADIAN
  
 
==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}}

Latest revision as of 00:07, 10 November 2024

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 $\boxed{\textbf{E)}\ 5.5}$

Solution 4 (Formal)

Let Alice be at point $A$, Bob be at point $B$. Let the plane be at point $P$ and $X$ be the projection of $P$ onto the ground (the plane with contains $A$ and $B$). Let the height of the plane, or $PX$, be $h$. So, because of the $30-60-90$ triangles, \[AX = \dfrac{h}{\sqrt{3}}, BX = h\sqrt{3}\] By Pythagorean Theorem on $\triangle ABX$, \[\dfrac{h^2}{3} + 3h^2 = 100 \implies \dfrac{10h^2}{3} = 100 \implies h = \sqrt{30},\] which is clossest to $\boxed{\textbf{(E)}\ 5.5}.$

~sml1809

Solution 5: Diagram

[asy] import graph; usepackage("amsmath"); size(7.2cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 10.1, ymin = -4.44, ymax = 6.3; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((-0.12,5.52)--(-1.34,-0.82)--(4.78,-0.88)--cycle); draw(arc((-1.34,-0.82),0.6,41.93351449887411,79.10776933246521)--(-1.34,-0.82)--cycle, qqwuqq); draw(arc((4.78,-0.88),0.6,127.43857157233303,167.225815720283)--(4.78,-0.88)--cycle, qqwuqq); /* draw figures */ draw((-0.12,5.52)--(-1.34,-0.82)); draw((-1.34,-0.82)--(4.78,-0.88)); draw((4.78,-0.88)--(-0.12,5.52)); draw((-0.12,5.52)--(-0.16,0.24)); draw((-0.16,0.24)--(4.78,-0.88)); draw((-1.34,-0.82)--(-0.16,0.24)); label("$60^{\circ}$",(3.82,-0.06),SE*labelscalefactor); label("$30^\circ$",(-1.12,0.3),SE*labelscalefactor); draw((-0.15694767653526337,0.6429066973452366)--(0.22,0.58)); draw((0.22,0.58)--(0.21801530426333052,0.1542961253492044)); draw((-0.15694767653526337,0.6429066973452365)--(-0.44,0.4)); draw((-0.44,0.4)--(-0.42432750397456265,0.0025532591414946237)); draw((-0.42432750397456265,0.0025532591414946237)--(-0.16,0.24)); draw((-0.16,0.24)--(-0.42432750397456265,0.0025532591414946237)); draw((-0.42432750397456265,0.0025532591414946237)--(-0.02,-0.12)); draw((-0.02,-0.12)--(0.21801530426333054,0.15429612534920437)); /* dots and labels */ dot((-0.12,5.52),dotstyle); label("$\text{Plane}$", (-0.52,5.74), NE * labelscalefactor); dot((-1.34,-0.82),dotstyle); label("$\text{Alice}$", (-1.8,-1.2), NE * labelscalefactor); dot((4.78,-0.88),dotstyle); label("$\text{Bob}$", (4.86,-0.68), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ label("$10$", (-1.8,-1.2)--(4.86,-0.68), S); [/asy]

~credit to mathmaster2012 for original diagram

~BakedPotato66 added/contributed some elements

Video Solution by CanadaMath (Problem 11-20)

Fast Forward to 9:16 for problem 13 https://www.youtube.com/watch?v=4osvFatUv1o

~THEMATHCANADIAN

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. AMC logo.png

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