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CEMC Fermat, 2024

Geometry Difficulty 1.3 Junior Find the answer Canada

In the diagram, a semi-circle has centre OO and diameter ACAC. Also, BB is a point on the circumference such
that ∠BAC=25°\angle BAC = 25\degree.

Figure 0

The measure of ∠BOC\angle BOC is

Pick one

Solution

Since OO is the centre of the
circle, then OA=OB=OCOA = OB = OC.

This means that △AOB\triangle AOB and
△COB\triangle COB are both isosceles
with ∠ABO=∠BAO=∠BAC=25°\angle ABO = \angle BAO = \angle BAC = 25\degree.

Thus, ∠AOB=180°−∠ABO−∠BAO=130°\angle AOB = 180\degree - \angle ABO - \angle BAO = 130\degree.

Since ∠AOC\angle AOC is a straight
angle, then ∠BOC=180°−∠AOB=180°−130°=50°\angle BOC = 180\degree - \angle AOB = 180\degree - 130\degree = 50\degree.

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