The two tangents from an external point P to a circle with centre at O are PA and PB. If ∠APB=70°, then the value of ∠ AOB is:(2 mark)

130°

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120°

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100°

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110°

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Solution

The correct option is D 110°
∠ OAP = 90°

∠ APO = 35°

(As, ∠ APB = 70°)

∠ AOP = 180° – (90° + 35°) (1 mark)

= 180° – 125° = 55°

∠ AOB = 2 x 55° = 110°(1 mark)

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Similar questions

If PA and PB are two tangents to a circle with centre O such that $∠$ AOB = $110^{o}$ then $∠$ APB is equal to

(a) $55^{o}$ (b) $60^{o}$ (c) $70^{o}$ (d) $90^{o}$

< A P B = 45^{0}

< A O B + < A P B

P

P A

P B

O

∠ P A B = 50^{\circ},

∠ A O B

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