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Question Number 101851 by bemath last updated on 05/Jul/20

Answered by bobhans last updated on 05/Jul/20

OA = OB = radius ∠CAO = ∠OBA = β →∠AOP = 2β PA is tangent →∠OAP = 90^o then ∠APO = 90^o −2β ∠CPO = 45^o −β since PC bisect the angle APB hence ∠OMP = 135^o −β ∠ACM + ∠CMA + ∠MAC = 180^o ∠ACM = 180^o −135^o = 45^o answer (B) → (Bob− )

OA=OB=radius∠CAO=∠OBA=β→∠AOP=2βPAistangent→∠OAP=90othen∠APO=90o−2β∠CPO=45o−βsincePCbisecttheangleAPBhence∠OMP=135o−β∠ACM+∠CMA+∠MAC=180o∠ACM=180o−135o=45oanswer(B)→(Bob−)

Commented by bemath last updated on 05/Jul/20

coll....♡♥~♥~

Commented by bobhans last updated on 05/Jul/20

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