1954 AHSME Problems/Problem 31

Problem

In $\triangle ABC$ , $AB=AC$ , $\angle A=40^\circ$ . Point $O$ is within the triangle with $\angle OBC \cong \angle OCA$ . The number of degrees in $\angle BOC$ is:

$\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}$

Solution

Since $\triangle ABC$ is an isosceles triangle, $\angle ABC = \angle ACB = 70^{\circ}$ . Let $\angle OBC = \angle OCA = x$ . Since $\angle ACB = 70$ , $\angle OCB = 70 - x$ . The angle of $\triangle OBC$ add up to $180$ , so $\angle BOC = 180 - (x + 70 - x) = \boxed{\textbf{(A) } 110^{\circ}$ (Error compiling LaTeX. Unknown error_msg).

1954 AHSC (Problems • Answer Key • Resources)
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1954 AHSME Problems/Problem 31

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