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

Revision as of 19:28, 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$",A,SW); label("$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]$

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.

Page

Toolbox

Search

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

Revision as of 19:28, 18 May 2016

Problem 5

Solution

See Also

@@ Line 10: / Line 10: @@
 A = (0,1);
 draw(A--B--C--A);
-draw(rightanglemark(B,A,C,8));
+draw(rightanglemark(B,C,A,8));
 label("$A$",A,SW);
 label("$B$",B,SE);