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

2010 AMC 12B Problem 13

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13.

\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}

Both a cosine and a sine are at most

1,

so each must equal

1

Solve

2A-B=0^\circ

and

A+B=90^\circ

to identify the triangle

A cosine plus a sine equals

2

only when each equals

1.

\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,

\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.