1954 AMC 12 Problem 31

Question

In triangle

ABC,

\overline{AB}=\overline{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:

110

35

140

55

70

The base angles of triangle

ABC

are both

70^\circ

If the equal angles are

x,

then the other part of angle

C

is

70^\circ-x

The base angles of the isosceles triangle are

\angle ABC=\angle BCA=70^\circ.

Put

\angle OBC=\angle OCA=x.

Then

\angle OCB=70^\circ-x.

In triangle

BOC,

\begin{aligned} \angle BOC &=180^\circ-x\\ &\quad{}-(70^\circ-x)\\ &=110^\circ. \end{aligned}

Thus, the correct answer is A.

Problem 31 in other years