2012 AMC 12A Problem 10

Question

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?

\dfrac{3}{10}

\dfrac{1}{3}

\dfrac{9}{20}

\dfrac{2}{3}

\dfrac{9}{10}

A median splits a triangle into two pieces of equal area

One half-triangle has sides

5

and

9

with included angle

\theta

and area

15

The median divides the triangle into two triangles of equal area

15.

One of them has the two sides of length

5

(half the base) and

9

(the median) meeting at angle

\theta.

Its area is

\tfrac12 \cdot 5 \cdot 9 \sin\theta = 15,

so

\sin\theta = \dfrac{2 \cdot 15}{5 \cdot 9} = \dfrac{2}{3}.

Thus, the correct answer is D.

Problem 10 in other years