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

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 $\sin{\theta} + \sin{(180^{\circ} - \theta)} = \frac{4}{3} = \sin{\theta} + \sin{180^{\circ}}\cos{\theta} - \cos{180^{\circ}}\sin{\theta}.$

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

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

Page

Toolbox

Search