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Byju's Answer

Standard XII

Mathematics

Circular Measurement of Angle

In the given ...

Question

In the given figure, $O$ is the centre of the circumcircle of $△ A B C$ . Tangents at $A$ and $B$ intersect at $T$ . If $∠ A T B = 80^{o}$ and $∠ A O C = 130^{o}$ . Then $∠ C A B$ is

115^{o}

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65^{o}

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50^{o}

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100^{o}

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Solution

The correct option is B $65^{o}$
$G i v e n - O i s t h e c e n t r e o f a c i r c l e t o w h i c h a p a i r o f t a n g e n t s A T & B T f r o m a p o i n t T t o u c h t h e c i r c l e a t A & B r e s p e c t i v e l y . ∠ A T B = 80^{o} . T o f i n d o u t - ∠ C A B = ? S o l u t i o n - W e j o i n O B . ∠ O A T = 90^{o} = ∠ O B T s i n c e t h e a n g l e, b e t w e e n a t a n g e n t t o a c i r c l e a n d t h e r a d i u s o f t h e s a m e c i r c l e p a s s i n g t h r o u g h t h e p o i n t o f c o n t a c t, i s 90^{o} . ∴ B y a n g l e s u m p r o p e r t y o f q u a d r i l a t e r a l s, w e g e t ∠ O A T + ∠ A T B + ∠ O B T + ∠ A O B = 360^{o} . ⟹ 90^{o} + 80^{o} + 90^{o} + ∠ A O B = 360^{o} . ⟹ ∠ A O B = 100^{o} . N o w t h e c h o r d A B s u b t e n d s ∠ A O B t o t h e c e n t r e a n d ∠ A C B t o t h e c i r c u m f e r e n c e o f t h e g i v e n c i r c l e a t C . ∴ ∠ A C B = \frac{1}{2} \times ∠ A O B = \frac{1}{2} \times 100^{o} = 50^{o} . S i n c e t h e a n g l e, s u b t e n d e d b y a c h o r d o f a c i r c l e a t t h e c e n t r e, i s d o u b l e t h e a n g l e s u b t e n d e d b y t h e s a m e c h o r d a t t h e c i r c u m f e r e n c e o f t h e c i r c l e . A g a i n t h e c h o r d A C s u b t e n d s ∠ A O C t o t h e c e n t r e a n d ∠ A B C t o t h e c i r c u m f e r e n c e o f t h e g i v e n c i r c l e a t B . ∴ ∠ A B C = \frac{1}{2} \times ∠ A O C = \frac{1}{2} \times 130^{o} = 65^{o} . S i n c e t h e a n g l e, s u b t e n d e d b y a c h o r d o f a c i r c l e a t t h e c e n t r e, i s d o u b l e t h e a n g l e s u b t e n d e d b y t h e s a m e c h o r d a t t h e c i r c u m f e r e n c e o f t h e c i r c l e . S o, i n Δ A B C, b y a n g l e s u m p r o p e r t y o f t r i a n g l e s, w e g e t ∠ A B C + ∠ A C B + ∠ C A B = 180^{o} ⟹ ∠ C A B = 180^{o} - (∠ A B C + ∠ A C B) ⟹ ∠ C A B = 180^{o} - (65^{o} + 50^{o}) = 65^{o} . A n s - O p t i o n B .$

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In the given figure, O is the centre of the circumcircle of △ABC. Tangents at A and B intersect at T. If ∠ATB=80o and ∠AOC=130o. Then ∠CAB is

In the given figure, $O$ is the centre of the circumcircle of $△ A B C$ . Tangents at $A$ and $B$ intersect at $T$ . If $∠ A T B = 80^{o}$ and $∠ A O C = 130^{o}$ . Then $∠ C A B$ is