2010 AMC 12B Problem 13 — Solution | LIVE by Po-Shen Loh

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2010 AMC 12B Problem 13

Attempt Problem 13 of the 2010 AMC 12B 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 2010 AMC 12B solutions, or check the answer key.

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

13.

In $\triangle ABC,$ $\cos(2A-B)+\sin(A+B)=2$ and $AB=4.$ What is $BC?$

$\sqrt{2}$

$\sqrt{3}$

$2$

$2\sqrt{2}$

$2\sqrt{3}$

Answer: C

Concepts:trigonometry special right triangle bounding to limit cases

Difficulty rating: 1560

Solution:

A cosine plus a sine equals $2$ only when each equals $1.$ So $\cos(2A-B)=1$ and $\sin(A+B)=1,$ giving $2A-B=0^\circ$ and $A+B=90^\circ.$

Solving, $A=30^\circ$ and $B=60^\circ,$ so $\triangle ABC$ is a $30\text{-}60\text{-}90$ right triangle with the right angle at $C.$

With hypotenuse $AB=4,$ the side $BC$ opposite the $30^\circ$ angle is half the hypotenuse, so $BC=2.$

Thus, the correct answer is C.

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