1986 AHSME Problems/Problem 3

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Problem

$\triangle ABC$ has a right angle at $C$ and $\angle A = 20^\circ$. If $BD$ ($D$ in $\overline{AC}$) is the bisector of $\angle ABC$, then $\angle BDC =$

$\textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ\qquad \textbf{(E)}\ 60^\circ$

Solution

Since $\angle C = 90^{\circ}$ and $\angle A = 20^{\circ}$, we have $\angle ABC = 70^{\circ}$. Thus $\angle DBC = 35^{\circ}$. It follows that $\angle BDC = 90^{\circ} - 35^{\circ} = 55^{\circ}$, which is $\boxed{D}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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