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In the given figure PA and PB are tangents to a circle with centre O. if `angleAPB= (2x +3)^@` and `angleAOB= (3x +7)^@`. then find the value of x. |
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Answer» As tangents from an external point are perpendicular to radius of the circle, `:. /_OAP = /_OBP = 90^@` Now, in quadrilateral `OAPB`, `/_OAP+/_APB+/_OBP+/_BOA = 360^@` `/_90^@+2x+3+90^@+3x+7 = 360^@` `5x+10 = 180^@` `x = 170/5` `=>x = 34` |
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