1988 AHSME Problems/Problem 13: Difference between revisions

Revision as of 18:02, 2 January 2016

Problem

If $\sin(x) =3 \cos(x)$ then what is $\sin(x) \cdot \cos(x)$ ?

$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{5}\qquad \textbf{(C)}\ \frac{2}{9}\qquad \textbf{(D)}\ \frac{1}{4}\qquad \textbf{(E)}\ \frac{3}{10}$

Solution

In the problem we are given that $\sin{(x)}=3\cos{(x)}$ , and we want to find $\sin{(x)}\cos{(x)}$ . We can divide both sides of the original equation by $\cos{(x)}$ to get $\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.$ 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 $\sin{(x)}\cos{(x)}$ . $\sin{(x)}\cos{(x)}=\left(\frac{3}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right)=\frac{3}{10}.$ So $\boxed{\textbf{(E)}\ \frac{3}{10}}$ is our answer.

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 28: / Line 28: @@
 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>
+<cmath>\sin{(x)}\cos{(x)}=\left(\frac{3}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right)=\frac{3}{10}.</cmath>
 So <math>\boxed{\textbf{(E)}\ \frac{3}{10}}</math> is our answer.

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1988 AHSME Problems/Problem 13: Difference between revisions

Revision as of 18:02, 2 January 2016

Problem

Solution

See also