The correct option is D 25^o nbsp;Given PA & PB are two tangents to a circle with centre O at A & B r ...

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Quadrilateral Formed by Centre, Any Two Points on the Circle and Point of Intersection of Tangents

∠ OAB is

Question

$∠ O A B$ is

50^{o}

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75^{o}

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110^{o}

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25^{o}

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Solution

The correct option is D $25^{o}$
$G i v e n - P A & P B a r e t w o t a n g e n t s t o a c i r c l e w i t h c e n t r e O a t A & B r e s p e c t i v e l y . ∠ A P B = 50^{o} . T o f i n d o u t - ∠ O A B = ? . S o l u t i o n - w e j o i n O A & O B . T h e n O A & O B a r e r a d i i t h r o u g h t h e p o i n t s o f c o n t a c t A & B w i t h t h e t a n g e n t t o t h e g i v e n c i r c l e . ∴ O A ⊥ P A & O B ⊥ P B . i . e ∠ O A P = 90^{o} = ∠ O B P . ∴ ∠ O A P + ∠ O B P = 90^{o} + 90^{o} = 180^{o} . ∴ ∠ A O B + ∠ A P B = 360^{o} - (∠ O A P + ∠ O B P) = 360^{o} - 180^{o} = 180^{o} . ∴ ∠ A O B = 180^{o} - ∠ A P B = 180^{o} - 50^{o} = 130^{o} . N o w i n Δ O A B O A = O B (r a d i i o f t h e s a m e c i r c l e) . S o ∠ O A B = ∠ O B A i . e ∠ O A B + ∠ O B A = 2 ∠ O A B . ∴ ∠ A O B + 2 ∠ O A B = 180^{o} (a n g l e s u m p r o p e r t y o f t r i a n g l e s .) ⟹ ∠ O A B = \frac{1}{2} \times (180^{o} - ∠ A O B) = \frac{1}{2} \times (180^{o} - 130^{o}) = 25^{o} . A n s - O p t i o n A .$

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Similar questions

∠ C A B = 75^{o}

∠ C B A = 50^{o}

∠ D A B + ∠ A B D

110^{o}

50^{o}

40^{o}

△ A B C, ∠ A = 25^{o}, ∠ B = 50^{o}, ∠ C = 105^{o}

A B C D

∠ C A D = 25^{o}

∠ A B C = 50^{o}

∠ A C B = 35^{o}

∠ D A B

P Q R S

P Q = 3.8

Q R = 6.8

∠ P = 100^{o}, ∠ R = 110^{o}

∠ S = 75^{o}

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