Difference between revisions of "1986 AHSME Problems/Problem 3"

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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
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Revision as of 02:29, 24 October 2014

Problem

$\triangle ABC$ is 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

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AHSME Problems and Solutions

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