Difference between revisions of "1983 AHSME Problems/Problem 5"
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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; | ||
Line 16: | Line 17: | ||
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 has a right angle at . If , then is
Solution
Since is opposite side over the hypotenuse of a right triangle, we can label the diagram as shown. By the Pythagorean Theorem, we have:
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.