The correct option is C 110^o nbsp;Given PA & PB are two tangents to a circle with centre O at A & B ...

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Tangent

∠ AMB is

Question

$∠ A M B$ is

220^{o}

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80^{o}

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110^{o}

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170^{o}

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Solution

The correct option is C $110^{o}$
$G i v e n - P A & P B a r e t w o t a n g e n t s t o a c i r c l e w i t h c e n t r e O a t A & B r e s p e c t i v e l y . ∠ A P B = 40^{o} . M i s a p o i n t o n t h e c i r c u m f e r e n c e o f t h e c i r c l e . T o f i n d o u t - ∠ A M B = ? . S o l u t i o n - w e j o i n O A & O B . T h e n O A & O B a r e r a d i i t h r o u g h t h e p o i n t s o f c o n t a c t A & B w i t h t h e t a n g e n t t o t h e g i v e n c i r c l e . ∴ O A ⊥ P A & O B ⊥ P B . i . e ∠ O A P = 90^{o} = ∠ O B P . ∴ ∠ O A P + ∠ O B P = 90^{o} + 90^{o} = 180^{o} . ∴ ∠ A O B + ∠ A P B = 360^{o} - (∠ O A P + ∠ O B P) = 360^{o} - 180^{o} = 180^{o} . ∴ ∠ A O B = 180^{o} - ∠ A P B = 180^{o} - 40^{o} = 140^{o} . ∴ r e f l e x ∠ A O B = 360^{o} - 140^{o} = 220^{o} . N o w r e f l e x ∠ A O B i s t h e a n g l e a t t h e c e n t r e a n d ∠ A M B i s t h e a n g l e a t t h e c i r c u m f e r e n c e s u b t e n d e d b y t h e s a m e a r c A M B . ∴ ∠ A M B = \frac{1}{2} \times r e f l e x ∠ A O B = \frac{1}{2} \times 220^{o} = 110^{o} s i n c e t h e a n g l e, s u b t e n d e d b y a n a r c o f a c i r c l e a t t h e c e n t r e, i s d o u b l e t h e a n g l e s u b t e n d e d b y t h e s a m e a r c a t t h e c i r c u m f e r e n c e o f t h e c i r c l e . A n s - O p t i o n C .$

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