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

(Solution)
(Solution)
Line 5: Line 5:
 
==Solution==
 
==Solution==
 
<asy>
 
<asy>
A=(0,1.7);
+
pair A,B,C;
B=(2,0);
+
A = (0,0);
C=(0,0);
+
B = (3,0);
pair (A,B,C);
+
C = (0,1);
draw (A - - B - - C - - A);
+
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>
 
</asy>
  

Revision as of 20:28, 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; 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]

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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