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.Q3. PQ and PR are tangents to a circle with centre Alf ZQPA=27° then find ZQAR |
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Answer» hey MATE here is your answer... Step-by-step explanation: ΔAQP+ΔARP AP=APcommon ∠Q=∠R[90 ∘ ] AQ=AR[RADII] So1∠QPA=∠RPA=27 ∘
∠P+∠Q+∠A+∠R=360 ∘
54+90+∠A+90=360 ∘
∠A=360 ∘ −180−54 ∠A=180−54 ∠A=126 ∘
∠QAR=126 ∘ ANS.
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