1983 AHSME Problems/Problem 5

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 $\boxed{\textbf{(D)}}$.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png