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
  
<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> can be seen as "opposite over hypotenuse" in a right triangle, we can label the diagram as shown.
 
<asy>
 
<asy>
 
pair A,B,C;
 
pair A,B,C;
 
C = (0,0);
 
C = (0,0);
B = (3,0);
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B = (2,0);
A = (0,1);
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A = (0,1.7);
 
draw(A--B--C--A);
 
draw(A--B--C--A);
 
draw(rightanglemark(B,C,A,8));
 
draw(rightanglemark(B,C,A,8));
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label("$B$",B,SE);
 
label("$B$",B,SE);
 
label("$C$",C,SW);
 
label("$C$",C,SW);
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label("$2x$",(B+C)/2,S);
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label("$3x$",(A+B)/2,NE);
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label("$y$",(A+C)/2,W);
 
</asy>
 
</asy>
 +
By the Pythagorean Theorem, we have:
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<cmath>y^2+(2x)^2=(3x)^2</cmath>
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<cmath>y=\sqrt{9x^2-4x^2}</cmath>
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<cmath>y=\sqrt{5x^2}</cmath>
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<cmath>y=x\sqrt{5}</cmath>
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so <math>\tan{B} = \frac{x \sqrt{5}}{2x} = \frac{\sqrt{5}}{2}</math>, which is choice <math>\boxed{\textbf{(D)}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 00:42, 20 February 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 $\sin$ 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
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All AHSME Problems and Solutions


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