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

(Created page with "==Problem 5== Triangle <math>ABC</math> has a right angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is <math>\textbf{(A)}\ \frac{3}{5}...")
 
(Solution)
Line 9: Line 9:
 
C=(0,0);
 
C=(0,0);
 
pair A,B,C;
 
pair A,B,C;
draw A--B--C--A
+
draw A--B--C--A;
 
</asy>
 
</asy>
 +
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1983|num-b=4|num-a=6}}
 
{{AHSME box|year=1983|num-b=4|num-a=6}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:25, 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

A=(0,1.7);
B=(2,0);
C=(0,0);
pair A,B,C;
draw A--B--C--A;
 (Error making remote request. Unknown error_msg)

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


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