2012 AMC 12A Problem 10

Attempt Problem 10 of the 2012 AMC 12A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

A triangle has area 30,30, one side of length 10,10, and the median to that side of length 9.9. Let θ\theta be the acute angle formed by that side and the median. What is sinθ?\sin\theta?

310\dfrac{3}{10}

13\dfrac{1}{3}

920\dfrac{9}{20}

23\dfrac{2}{3}

910\dfrac{9}{10}

Answer: D
Concepts:triangle areamedian (geometry)trigonometry
Difficulty rating: 1610
Solution:

The median divides the triangle into two triangles of equal area 15.15. One of them has the two sides of length 55 (half the base) and 99 (the median) meeting at angle θ.\theta.

Its area is 1259sinθ=15,\tfrac12 \cdot 5 \cdot 9 \sin\theta = 15, so sinθ=21559=23.\sin\theta = \dfrac{2 \cdot 15}{5 \cdot 9} = \dfrac{2}{3}.

Thus, the correct answer is D.

← Problem 9#9
Full Exam

Problem 10 in Other Years