1992 AHSME Problems/Problem 25

Problem

In $\triangle{ABC}$ , $\angle ABC=120^\circ,AB=3$ and $BC=4$ . If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$ , then $CD=$

$\text{(A) } 3\quad \text{(B) } \frac{8}{\sqrt{3}}\quad \text{(C) } 5\quad \text{(D) } \frac{11}{2}\quad \text{(E) } \frac{10}{\sqrt{3}}$

Solution

$\fbox{E}$

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1992 AHSME Problems/Problem 25

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