2012 AMC 12A Problems/Problem 10

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

$[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0); draw(A--B--C--cycle); draw(D--C); draw(E--C); pair[] ps={A,B,C,D,E}; dot(ps); label("$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 $\frac69=\boxed{\textbf{(D)}\ \frac23}$ .

2012 AMC 12A (Problems • Answer Key • Resources)
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2012 AMC 12A Problems/Problem 10

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