Difference between revisions of "1988 AHSME Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | + | In the problem we are given that <math>\sin{(x)}=3\cos{(x)}</math>, and we want to find <math>\sin{(x)}\cos{(x)}</math>. We can divide both sides of the original equation by <math>\cos{(x)}</math> to get <cmath>\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.</cmath> | |
| − | + | We can now use right triangle trigonometry to finish the problem. | |
| + | <asy> | ||
| + | pair A,B,C; | ||
| + | A = (0,0); | ||
| + | B = (3,0); | ||
| + | C = (0,1); | ||
| + | draw(A--B--C--A); | ||
| + | draw(rightanglemark(B,A,C,8)); | ||
| + | label("$A$",A,SW); | ||
| + | label("$B$",B,SE); | ||
| + | label("$C$",C,N); | ||
| + | label("$3$",B/2,S); | ||
| + | label("$1$",C/2,W); | ||
| + | label("$\sqrt{10}$",(C+B)/2,NE); | ||
| + | </asy> | ||
| + | |||
| + | Since the problem asks us to find <math>\sin{(x)}\cos{(x)}</math>. | ||
| + | <cmath>\sin{(x)}\cos{(x)}=\frac{3}{\sqrt{10}}\frac{1}{\sqrt{10}}=\frac{3}{10}.</cmath> | ||
| + | So <math>\boxed{\textbf{(E)}\ \frac{3}{10}}</math> is our answer. | ||
== See also == | == See also == | ||
Revision as of 09:26, 2 January 2016
Problem
If 
then what is 
?
Solution
In the problem we are given that 
, and we want to find 
. We can divide both sides of the original equation by 
to get ![\[\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.\]](http://latex.artofproblemsolving.com/d/c/8/dc885d8a953b450b93c19a12d44ad499db3d198d.png)
We can now use right triangle trigonometry to finish the problem.
![[asy] pair A,B,C; A = (0,0); B = (3,0); C = (0,1); draw(A--B--C--A); draw(rightanglemark(B,A,C,8)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$3$",B/2,S); label("$1$",C/2,W); label("$\sqrt{10}$",(C+B)/2,NE); [/asy]](http://latex.artofproblemsolving.com/f/7/9/f79fa3de0726b6a6cc5f8ebee031fa281fe7fd25.png)
Since the problem asks us to find .
![\[\sin{(x)}\cos{(x)}=\frac{3}{\sqrt{10}}\frac{1}{\sqrt{10}}=\frac{3}{10}.\]](http://latex.artofproblemsolving.com/a/c/a/aca5db71ee5f8a0b1c82acac37c3401ae4f2d6dd.png)
So 
is our answer.
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
