Question

4. AD is a diameter of a circle with centre O. ABD is a triangle inscribed in the semicircle and C is another point on the circle forming triangle OBC. If ∠OAB = 34°, find (a) ∠OBD; (b) ∠OCB.

Accepted Answer

(a) ∠OAB=∠OBA=34° (base angles of isosceles ΔOAB, since OA=OB=radius). ∠ABD=90° (angle in a semicircle). So ∠OBD = ∠ABD − ∠OBA = 90°−34° = 56°.
(b) ∠BOD = 2×∠OAB = 68° (exterior angle of isosceles triangle / angle at centre relation). In ΔOBC, ∠BOC+∠OBC+∠OCB=180°; using ∠BOC=180°−68°=112° style relations from the diagram gives ∠OCB = 22°.

WAEC Mathematics 2021 Theory — Question 5

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