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2010 AMC 12B Problem 13

Below is the professionally curated solution for Problem 13 of the 2010 AMC 12B, 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.

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Concepts:trigonometryspecial right trianglebounding to limit cases

Difficulty rating: 1560

13.

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

2\sqrt{2}

3\sqrt{3}

22

222\sqrt{2}

232\sqrt{3}

Solution:

A cosine plus a sine equals 22 only when each equals 1.1. So cos⁡(2A−B)=1\cos(2A-B)=1 and sin⁡(A+B)=1,\sin(A+B)=1, giving 2A−B=0∘2A-B=0^\circ and A+B=90∘.A+B=90^\circ.

Solving, A=30∘A=30^\circ and B=60∘,B=60^\circ, so △ABC\triangle ABC is a 30-60-9030\text{-}60\text{-}90 right triangle with the right angle at C.C.

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

Thus, the correct answer is C.

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