$I n f i g u r e i f ∠ A O B = 90^{o} a n d ∠ A B C = 30^{o} t h e n ∠ C A O i s e q u a l t o$

30^{o}

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60^{o}

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90^{o}

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45^{o}

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Solution

The correct option is D $60^{o}$
$A B i s t h e c h o r d o f a c i r c l e w i t h c e n t r e O . A B s u b t e n d s ∠ A O B = 90^{o} t o t h e c e n t r e a n d ∠ A C B t o t h e c i r c u m f e r e n c e . ∠ A B C = 30^{o} . T o f i n d o u t - ∠ C A O = ? S o l u t i o n - A B s u b t e n d s ∠ A O B = 90^{o} t o t h e c e n t r e a n d ∠ A C B t o t h e c i r c u m f e r e n c e . ∴ ∠ A C B = \frac{1}{2} ∠ A O B = \frac{1}{2} \times 90^{o} = 45^{o} . (T h e a n g l e, s u b t e n d e d b y a c h o r d t o t h e c i r c u m f e r e n c e o f a c i r c l e, i s h a l f o f t h a t s u b t e n d e d t o t h e c e n t r e) . N o w i n Δ A C B w e h a v e ∠ A C B + ∠ A B C = 45^{o} + 30^{o} = 75^{o} . ∴ ∠ B A C = 180^{o} - (∠ A C B + ∠ A B C) = 180^{o} - 75^{o} = 105^{o} . (a n g l e s u m p r o p e r t y o f t r i a n g l e s) N o w i n Δ A O B w e h a v e A O = B O (r a d i i o f t h e s a m e c i r c l e) . ∴ ∠ O A B = ∠ O B A i . e ∠ O A B + ∠ O B A = 2 ∠ O A B . S o ∠ A O B + ∠ O A B + ∠ O B A = ∠ A O B + 2 ∠ O A B = 180^{o} . (a n g l e s u m p r o p e r t y o f t r i a n g l e s) . i . e 90^{o} + 2 ∠ O A B = 180^{o} ⟹ ∠ O A B = 45^{o} . ∴ ∠ C A O = ∠ B A C - ∠ O A B = 105^{o} - 45^{o} = 60^{o} . A n s - O p t i o n B .$

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