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Question

In the given figure, AC the diameter of the circle with centre O. CD and BE are parallel. Angle AOB = 80o and ACE = 10o. Calculate :

(i) Angle BEC,

(ii) Angle BCD,

(iii) Angle CED.

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Solution

(i) \angle BOC = 180^o-80^o=100^o (Since AC is a straight line)

\angle BOC = 2 \angle BEC (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\Rightarrow \angle BEC = \frac{100}{2} = 50^o

(ii) DC \parallel EB

\angle DCE = \angle BEC = 50^o

\angle AOB = 80^o (given)

\Rightarrow \angle ACB = \frac{1}{2} \angle AOB = 40^o (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

\angle BCD = \angle ACB + \angle ACE + \angle DCE = 40^o+10^o+50^o = 100^o

(iii) \angle BED = 180^o-\angle BCD = 180^o-100^o = 80^o (cyclic quadrilateral)

\angle CED + 50^o = 80^o \Rightarrow \angle CED = 30^o


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