InterviewSolution
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If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of `80^@`, then `/_P O A` is equal to |
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Answer» We can draw the figure as per the given details. Please refer to the video for the diagram. We know tangent at any point `P` is always perpendicular to line joining the center `O` of that circle. So, `/_OAP = /_OBP =90^@` Now, in `Delta OAP` and `Delta OBP`, `OA = OB,OP= OP, /_OAP = /_OBP =90^@` So, `Delta OAP ~ Delta OBP`. It means, `/_APO = /_BPO` We are given, `/_APB = 80^@` So, `/_APO = /_BPO = 80/2 = 40^@` In `Delta AOP`, `/_AOP+/_APO+/_OAP = 180^@` `/_AOP = 180 -40-90 = 50^@` |
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