1953 AHSME Problems/Problem 34

Problem 34

If one side of a triangle is $12$ inches and the opposite angle is $30^{\circ}$, then the diameter of the circumscribed circle is:

$\textbf{(A)}\ 18\text{ inches} \qquad \textbf{(B)}\ 30\text{ inches} \qquad \textbf{(C)}\ 24\text{ inches} \qquad \textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}$

Solution

By the Extended Law of Sines, the diameter, or twice the circumradius $R$, is given by \[2R=\frac{12\text{ inches}}{\sin30^\circ}=\boxed{\textbf{(C)}\ 24\text{ inches}}.\]

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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