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:

$110$

$35$

$140$

$55$

$70$

Answer: A

Concepts:isosceles triangle angle chasing angle sum

Difficulty rating: 1780

Small Hint:

The base angles of triangle $ABC$ are both $70^\circ$

Big Hint:

If the equal angles are $x,$ then the other part of angle $C$ is $70^\circ-x$

Solution:

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.

1954 AMC 12 Problem 31

Answer: A

Difficulty rating: 1780

Small Hint:

Big Hint:

Solution:

Problem 31 in Other Years