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 \ | + | <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 = ( | + | B = (2,0); |
| − | A = (0,1); | + | 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)); | ||
| − | label("$A$",A, | + | label("$A$",A,W); |
label("$B$",B,SE); | label("$B$",B,SE); | ||
| − | label("$C$",C, | + | label("$C$",C,SW); |
| − | label("$ | + | label("$2x$",(B+C)/2,S); |
| − | label("$ | + | label("$3x$",(A+B)/2,NE); |
| − | label("$ | + | label("$y$",(A+C)/2,W); |
</asy> | </asy> | ||
| + | By the Pythagorean Theorem, we have: | ||
| + | <cmath>y^2+(2x)^2=(3x)^2</cmath> | ||
| + | <cmath>y=\sqrt{9x^2-4x^2}</cmath> | ||
| + | <cmath>y=\sqrt{5x^2}</cmath> | ||
| + | <cmath>y=x\sqrt{5}</cmath> | ||
| + | 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 23:42, 19 February 2019
Problem 5
Triangle 
has a right angle at 
. If 
, then 
is
Solution
Since 
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]](http://latex.artofproblemsolving.com/1/7/a/17a88c19f38f78c73773241b6eb0ed13fa71886f.png)
By the Pythagorean Theorem, we have:
![\[y^2+(2x)^2=(3x)^2\]](http://latex.artofproblemsolving.com/6/8/6/686ffa510d576731905d732f69ba2b00e324bffd.png)
![\[y=\sqrt{9x^2-4x^2}\]](http://latex.artofproblemsolving.com/0/c/0/0c069169c6cc689aab68ba5af18ac088b8625ced.png)
![\[y=\sqrt{5x^2}\]](http://latex.artofproblemsolving.com/3/b/6/3b64384db2e84286e4b644dbbc9980910d12a6d1.png)
![\[y=x\sqrt{5}\]](http://latex.artofproblemsolving.com/8/c/a/8cada19913fbbaf0d998ba591decab85cfbd8154.png)
so 
, which is choice 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
