A geometry problem by A Former Brilliant Member

Geometry Level 2

A circle with center at point $O$ is shown above. Find the measure of $\angle DOC$ in degrees.

note: $AOD$ is a straight line

The answer is 40.

1 solution

A Former Brilliant Member
Jun 13, 2017

Since $OA=OB$ , $\triangle AOB$ is isosceles. Therefore $\angle OBA=30^\circ$ .

Let $\angle DOC= \theta$ and $\angle AOC= \phi$ , then by Thales Theorem

$\phi = 2(30+40)=2(70)=140$

It follows that,

$\theta = 180 - \phi = 180 - 140 =$ $\boxed{40^\circ}$

We don't have to use the Thales theorem:

$OA=OB=OC$ , becuase they are radiuses. Since $\angle OAB=30°, \angle ABO=30°$ and $\angle AOB=180°-2*30°=120°$ and $\angle OBC=40°=\angle OCB$ . From that $\angle BOD=60°$ and $\angle DOC=\angle BOC-\angle BOD=100°-60°=\boxed{40°}$ .

Áron Bán-Szabó - 3 years, 11 months ago

A geometry problem by A Former Brilliant Member

The answer is 40.

1 solution

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