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Difference between revisions of "2012 AMC 12A Problems/Problem 10"

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<math> \textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10} </math>
 
<math> \textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10} </math>
  
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== Solution ==
 
  
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===Solution 1===
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==Solution 1==
  
 
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<center><asy>
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<math>AB</math> is the side of length <math>10</math>, and <math>CD</math> is the median of length <math>9</math>. The altitude of <math>C</math> to <math>AB</math> is <math>6</math> because the 0.5(altitude)(base)=Area of the triangle. <math>\theta</math> is <math>\angle CDE</math>. To find <math>\sin{\theta}</math>, just use opposite over hypotenuse with the right triangle <math>\triangle DCE</math>. This is equal to <math>\frac69=\boxed{\textbf{(D)}\ \frac23}</math>.
 
<math>AB</math> is the side of length <math>10</math>, and <math>CD</math> is the median of length <math>9</math>. The altitude of <math>C</math> to <math>AB</math> is <math>6</math> because the 0.5(altitude)(base)=Area of the triangle. <math>\theta</math> is <math>\angle CDE</math>. To find <math>\sin{\theta}</math>, just use opposite over hypotenuse with the right triangle <math>\triangle DCE</math>. This is equal to <math>\frac69=\boxed{\textbf{(D)}\ \frac23}</math>.
  
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===Solution 2===
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==Solution 2==
  
 
It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of <math>\triangle BCD = 15</math> in the above figure. Expressing the area in terms of <math>\sin{\theta}</math>, <math>\frac{1}{2} \cdot 5 \cdot 9 \cdot \sin{\theta} = 15</math>. Solving for <math>\sin{\theta}</math> gives <math>\sin{\theta} = \frac{2}{3}</math>. <math>\boxed{D}</math>.
 
It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of <math>\triangle BCD = 15</math> in the above figure. Expressing the area in terms of <math>\sin{\theta}</math>, <math>\frac{1}{2} \cdot 5 \cdot 9 \cdot \sin{\theta} = 15</math>. Solving for <math>\sin{\theta}</math> gives <math>\sin{\theta} = \frac{2}{3}</math>. <math>\boxed{D}</math>.
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==Solution 3==
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The area of a triangle with sides <math>a, b</math> and angle between them <math> \theta </math> is <math>\frac{1}{2} ab \sin{\theta}. </math> Therefore, <math> 30 = \frac{1}{2} (9 \cdot 5) \sin{\theta} + \frac{1}{2} (9 \cdot 5) \sin{(180^{\circ} - \theta)}, </math> as two angles along the same line must be supplementary. This simplifies to <math> \sin{\theta} + \sin{(180^{\circ} - \theta)} = \frac{4}{3} = \sin{\theta} + \sin{180^{\circ}}\cos{\theta} - \cos{180^{\circ}}\sin{\theta}. </math>
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<math> 2 \sin{\theta} = \frac{4}{3} \to \sin{\theta} = \frac{2}{3}. </math> <math> \boxed{D} </math>
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 00:59, 16 February 2021

Problem

A triangle has area $30$, one side of length $10$, and the median to that side of length $9$. Let $\theta$ be the acute angle formed by that side and the median. What is $\sin{\theta}$?

$\textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10}$


Solution 1

[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0); draw(A--B--C--cycle); draw(D--C); draw(E--C); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,SW); label("$E$",E,S); label("$9$",(D--C),NW); label("$6$",(E--C)); label("$10$",(A--B),SE); [/asy]

$AB$ is the side of length $10$, and $CD$ is the median of length $9$. The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$. To find $\sin{\theta}$, just use opposite over hypotenuse with the right triangle $\triangle DCE$. This is equal to $\frac69=\boxed{\textbf{(D)}\ \frac23}$.

Solution 2

It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of $\triangle BCD = 15$ in the above figure. Expressing the area in terms of $\sin{\theta}$, $\frac{1}{2} \cdot 5 \cdot 9 \cdot \sin{\theta} = 15$. Solving for $\sin{\theta}$ gives $\sin{\theta} = \frac{2}{3}$. $\boxed{D}$.

Solution 3

The area of a triangle with sides $a, b$ and angle between them $\theta$ is $\frac{1}{2} ab \sin{\theta}.$ Therefore, $30 = \frac{1}{2} (9 \cdot 5) \sin{\theta} + \frac{1}{2} (9 \cdot 5) \sin{(180^{\circ} - \theta)},$ as two angles along the same line must be supplementary. This simplifies to $\sin{\theta} + \sin{(180^{\circ} - \theta)} = \frac{4}{3} = \sin{\theta} + \sin{180^{\circ}}\cos{\theta} - \cos{180^{\circ}}\sin{\theta}.$

$2 \sin{\theta} = \frac{4}{3} \to \sin{\theta} = \frac{2}{3}.$ $\boxed{D}$

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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

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