1. O is the center of the circle and two tangents are drawn from a point P to this circle at points A and B. If ∠AOP = 50°, then what is the value (in degrees) of ∠APB?
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By: anil on 05 May 2019 01.46 am
Given : $$angle$$ AOP = $$50^circ$$ To find : $$angle$$ APB = $$2 heta$$ = ? Solution : $$angle$$ APO = $$frac{1}{2} imes$$ $$angle$$ APB => $$angle$$ APO = $$frac{1}{2} imes 2 heta= heta$$ Also, the radius of a circle intersects the tangent at the circumference of circle at $$90^circ$$ => $$angle$$ OAP = $$90^circ$$ In $$ riangle$$ AOP => $$angle$$ AOP + $$angle$$ APO + $$angle$$ OAP = $$180^circ$$ => $$ heta + 50^circ+90^circ=180^circ$$ => $$ heta=180^circ-140^circ$$ => $$ heta=40^circ$$ $$ herefore$$ $$angle$$ APB = $$2 heta=2 imes40=80^circ$$ => Ans - (B)
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