2012 AMC 12A Problem 10

Below is the professionally curated solution for Problem 10 of the 2012 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12A solutions, or check the answer key.

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Concepts:triangle area median (geometry)trigonometry

Difficulty rating: 1610

10.

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 $\sin\theta?$

$\dfrac{3}{10}$

$\dfrac{1}{3}$

$\dfrac{9}{20}$

$\dfrac{2}{3}$

$\dfrac{9}{10}$

Solution:

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.

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