in a circle PA and PB are two tangent to a circle with centre o. if angle APB = (2x+3)° and angle AOB = (3x+7)°, then find the value of x.
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tangent always perpendicular upon line radius.
now , quadrantal ABOP ,
angle AOB+angle APB+angle OAP+ angle OBP=360°.
(3x+7)°+(2x+3)°+90°+90°=360°
5x+10+180=360
5x=170
x=34°
now , quadrantal ABOP ,
angle AOB+angle APB+angle OAP+ angle OBP=360°.
(3x+7)°+(2x+3)°+90°+90°=360°
5x+10+180=360
5x=170
x=34°
kunjamrsk:
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thanks
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