Maths / Circles / Cyclic Quadrilateral

QUESTION

 In the figure, $\angle ABC=70^o\;and \; \angle BCD=160^o.$Measure of $\angle BAD+\anglr ADC$ is equal to

 OPTIONS A. $130^o$ B. $230^o$ C. 180 D. $90^o$
Right Option : A

EXPLANATION
Explain TypeExplanation Content
Text
 In the figure, $\angle ABC=70^o\;and \; \angle BCD=160^o.$$\dpi{120} \large \angle ABC + \angle ADC= 180^0$$\dpi{120} \large \angle ADC= 180^0 - \angle ABC = 180^0-70^0= 110^0$$\large \angle BAD + \angle BCD= 180^0$$\large \angle BAD= 180^0 - \angle BCD = 180^0-160^0= 20^0$$\angle BAD+\anglr ADC= 20^0 + 110^0=130^0$

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