# 1983 AHSME Problems/Problem 5

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## Problem 5

Triangle $ABC$ has a right angle at $C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is $\textbf{(A)}\ \frac{3}{5}\qquad \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad \textbf{(D)}\ \frac{\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{5}{3}$

## Solution

Since $\sin$ can be seen as "opposite over hypotenuse" in a right triangle, we can label the diagram as shown. $[asy] pair A,B,C; C = (0,0); B = (2,0); A = (0,1.7); draw(A--B--C--A); draw(rightanglemark(B,C,A,8)); label("A",A,W); label("B",B,SE); label("C",C,SW); label("2x",(B+C)/2,S); label("3x",(A+B)/2,NE); label("y",(A+C)/2,W); [/asy]$ By the Pythagorean Theorem, we have: $$y^2+(2x)^2=(3x)^2$$ $$y=\sqrt{9x^2-4x^2}$$ $$y=\sqrt{5x^2}$$ $$y=x\sqrt{5}$$ so $\tan{B} = \frac{x \sqrt{5}}{2x} = \frac{\sqrt{5}}{2}$, which is choice $\boxed{\textbf{(D)}}$.

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