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 $\angle BOC = 180 - (x + 70 - x) = \boxed{\textbf{(A) } 110^{\circ}}$.

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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