Maths Olympiad Prep

Track / Stage 4 / 221 of 340 #481 of 1964

Problem 481

AMC 12 late, AIME early
Geometry Difficulty 4.8 Find the answer

In the diagram, PQRP Q R is a straight line.

The value of xx is

Pick one

Official solution

Solution 1

Since PQS\angle P Q S is an exterior angle of QRS\triangle Q R S, then PQS=QRS+QSR\angle P Q S=\angle Q R S+\angle Q S R, so 136=x+64136^{\circ}=x^{\circ}+64^{\circ} or x=13664=72x=136-64=72.

## Solution 2

Since PQS=136\angle P Q S=136^{\circ}, then RQS=180PQS=180136=44\angle R Q S=180^{\circ}-\angle P Q S=180^{\circ}-136^{\circ}=44^{\circ}.

Since the sum of the angles in QRS\triangle Q R S is 180180^{\circ}, then 44+64+x=18044^{\circ}+64^{\circ}+x^{\circ}=180^{\circ} or x=1804464=72x=180-44-64=72.

ANSWER: (A)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.