Question 73

Let O be the centre of a circle and AC be its diameter. BD is a chord intersecting AC at E. Point A is joined to B and D. If $$\angle$$BOC = $$50^\circ$$ and $$\angle$$AOD = $$110^\circ$$, then $$\angle$$BEC = ?

Solution

$$\angle AOD=110$$

$$\angle ABD=\frac{1}{2}\angle AOD=55$$

AC is Diameter

$$\angle ABC=90$$ (angle on semicircle)

$$\angle CBE=90-55=35$$

In triangle BOC

OB=OC = Radius

$$\angle OBC=\angle OCB$$

$$\angle OBC+\angle OCB+\angle BOC = 180$$

$$2\angle OCB+50=180$$

$$\angle OCB=65$$

In triangle BEC

$$\angle CBE+\angle BEC+\angle ECB=180$$

$$\angle BEC+65+35=180$$

$$\angle BEC=80$$

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