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Angles in Same Segment of a Circle

In the given ...

Question

In the given figure, AB is parallel to DC. $B C E = 80^{\circ} a n d ∠ B A C = 25^{\circ} .$

Find :

(i) $∠ C A D$ (ii) $C B D$ (iii) $A D C$

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Solution

Given $\angle = 80^o$ and $\angle BAC = 25^o$ . Also $AB \parallel DC$

Therefore $\angle CAB = \angle ACD$ (alternate angles)

Hence $\angle ACD = 25^o$

Therefore $\angle ACB = 180^o - 80^o - 25^o = 75^o$

and $\angle DCB = 25^o + 75^o = 100^o$

$\angle DAB = 180^o - 100^o = 80^o$

Therefore $\angle DAC = 80^o - 25^o = 55^o$

In $\triangle ABC$ :

$\angle ABC = 180^o - 25^o - 75^o = 80^o$

Therefore $\angle ADC = 180^o = 80^o = 100^o$

$\angle CBD = \angle DAC = 55^o$ (angles in the same segment)

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