1.

2 mi Pare ngents to the circle with centre O such that LAPB S0P ind theB.

Answer»

Since OA is perpendicular to PA and also, OB is perpendicular to PB

∠APB + ∠AOB = 180°

50°+ ∠AOB = 180°

∠AOB = 180° – 50° = 130°

In △AOB,

OA = OB = radii of same circle

∠OAB = ∠OBA = x ( say )

Again, ∠OAB + ∠OBA + ∠AOB = 180°

x +x + 130° = 180°

2x = 180° – 130° = 50°

X = 25°

Hence, ∠OAB =25°



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