Triangle ABC has a right angle at C. If sin A=23, then tan B is

1983 AMC 12 Problem 5

Attempt Problem 5 of the 1983 AMC 12 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 1983 AMC 12 solutions, or check the answer key.

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Triangle

ABC

has a right angle at

C.

\sin A=\frac23,

then

\tan B

\frac35

\frac{\sqrt5}{3}

\frac{2}{\sqrt5}

\frac{\sqrt5}{2}

\frac53

Angles

A

and

B

are complementary

Use

\cos A=\sqrt{1-\sin^2A}

and

\tan B=\cot A

Since

\sin A=\frac{2}{3},

\cos A=\sqrt{1-\frac{4}{9}}=\frac{\sqrt5}{3}.

Because

B=90^\circ-A,

we have

\tan B=\cot A

and

\frac{\cos A}{\sin A}=\frac{\sqrt5}{2}.

Therefore, the correct answer is D.