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Angle Subtended by an Arc of a Circle on the Circle and at the Center

In the given ...

Question

In the given figure, AC the diameter of the circle with centre O. CD and BE are parallel. Angle $∠$ AOB = $80^{o}$ and $∠$ ACE = $10^{o}$ . 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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