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Question

In the given figure, AB is parallel to DC. BCE=80 and BAC=25.

Find :

(i) CAD (ii) CBD (iii) ADC

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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)

d26Hence \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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