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CBSE Class 10 Answered

Plz solve the question. 
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Asked by adilaman.java | 08 Feb, 2019, 06:03: PM
answered-by-expert Expert Answer
begin mathsize 12px style angle end styleAFC + begin mathsize 12px style angle end styleAEC = begin mathsize 12px style angle end styleBFD+begin mathsize 12px style angle end styleAEC = 360 - begin mathsize 12px style angle end styleEBF - begin mathsize 12px style angle end styleEDF ...................(1)
in eqn. (1) we used the relation begin mathsize 12px style angle end styleAFC = begin mathsize 12px style angle end styleBFD (vertically opposite angles)
begin mathsize 12px style angle end styleEBF = 180-begin mathsize 12px style angle end styleABC ....................(2)
begin mathsize 12px style angle end styleEDF = 180-begin mathsize 12px style angle end styleADC ...................(3)
using (2) and (3) in eqn.(1), we have begin mathsize 12px style angle end styleAFC+begin mathsize 12px style angle end styleAEC = 360-(180-begin mathsize 12px style angle end styleABC)-(180-begin mathsize 12px style angle end styleADC) = begin mathsize 12px style angle end styleABC + begin mathsize 12px style angle end styleADC ....................(4)
 
we have begin mathsize 12px style angle end styleABC = (1/2)begin mathsize 12px style angle end styleAOC    and  begin mathsize 12px style angle end styleADC = (1/2)begin mathsize 12px style angle end styleAOC  because angle subtended by a chord at center is twice the angle subtended at
circumference by the same chord . using these relations in eqn.(4), we have  begin mathsize 12px style angle end styleAFC+begin mathsize 12px style angle end styleAEC = begin mathsize 12px style angle end styleAOC
Answered by Thiyagarajan K | 08 Feb, 2019, 08:48: PM

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