Difference between revisions of "1983 AHSME Problems/Problem 5"

(Solution)
(Solution)
Line 11: Line 11:
 
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,SW);
+
label("$A$",A,W);
 
label("$B$",B,SE);
 
label("$B$",B,SE);
label("$C$",C,N);
+
label("$C$",C,SW);
label("$3$",B/2,S);
 
label("$1$",C/2,W);
 
label("$\sqrt{10}$",(C+B)/2,NE);
 
 
</asy>
 
</asy>
  

Revision as of 20:29, 18 May 2016

Problem 5

Triangle $ABC$ has a right angle at $C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is

$\textbf{(A)}\ \frac{3}{5}\qquad \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad \text{(D)}\ \sqrt{3}\qquad \text{(E)}\ 2$

Solution

[asy] pair A,B,C; C = (0,0); B = (3,0); A = (0,1); draw(A--B--C--A); draw(rightanglemark(B,C,A,8)); label("$A$",A,W); label("$B$",B,SE); label("$C$",C,SW); [/asy]

See Also

1983 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png