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Difference between revisions of "1983 AHSME Problems/Problem 5"

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Triangle <math>ABC</math> has a right angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is
 
Triangle <math>ABC</math> has a right angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is
  
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<math>\textbf{(A)}\ \frac{3}{5}\qquad \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad \text{(D)}\ \sqrt{3}\qquad \text{(E)}\ 2</math>
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<math>\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}</math>
 
==Solution==
 
==Solution==
 
Since <math>\sin</math> is opposite side over the hypotenuse of a right triangle, we can label the diagram as shown.
 
Since <math>\sin</math> is opposite side over the hypotenuse of a right triangle, we can label the diagram as shown.

Revision as of 17:20, 26 January 2019

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 $\sin$ is opposite side over the hypotenuse of 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: \[y^2+(2x)^2=(3x)^2\] \[y=\sqrt{9x^2-4x^2}\] \[y=\sqrt{5x^2}\] \[y=x\sqrt{5}\]

See Also

1983 AHSME (Problems • Answer Key • Resources)
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


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