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

(Solution)
(Solution)
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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>
 
<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>
 
==Solution==
 
==Solution==
 +
Since <math>\sin</math> is opposite side over the hypotenuse of a right triangle, we can label the diagram as shown.
 
<asy>
 
<asy>
 
pair A,B,C;
 
pair A,B,C;
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label("$2x$",(B+C)/2,S);
 
label("$2x$",(B+C)/2,S);
 
label("$3x$",(A+B)/2,NE);
 
label("$3x$",(A+B)/2,NE);
 +
label("y",(A+C)/2,W);
 
</asy>
 
</asy>
 +
By the Pythagorean Theorem, we have:
 +
<cmath>y^2+(2)x^2=(3x)^2</cmath>
 +
<cmath>y=\sqrt{9x^2-4x^2}</cmath>
 +
<cmath>y=\sqrt{5x^2}</cmath>
 +
<cmath>y=x\sqrt{5}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 19:37, 18 May 2016

Problem 5

Triangle $ABC$ has a right angle at $C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is

$\textbf{(A)}\ \frac{3}{5}\qquad \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad \text{(D)}\ \sqrt{3}\qquad \text{(E)}\ 2$

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: \[y^2+(2)x^2=(3x)^2\] \[y=\sqrt{9x^2-4x^2}\] \[y=\sqrt{5x^2}\] \[y=x\sqrt{5}\]

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


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