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Circle and Line in a Plane

In the given ...

Question

In the given figure, O is the centre of the circle $∠$ OAB = 30 $^{o}$ and $∠$ OCB = 40 $^{o}$ . Then $∠$ AOC is

70^{o}

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20^{o}

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60^{o}

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120^{o}

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Solution

The correct option is A $70^{o}$
$G i v e n - O i s t h e c e n t r e o f a c i r c l e . O n i t s c i r c u m f e r e n c e t h r e e p o i n t s A, B & C a r e s u c h t h a t ∠ O A B = 30^{o} & ∠ O C A = 40^{o} . T o f i n d o u t - ∠ A O C = ? S o l u t i o n - W e j o i n A C . I n Δ O A C w e h a v e O A = O C (r a d i i o f t h e s a m e c i r c l e) . ∴ Δ O A C i s i s o s c e l e s w i t h A C a s b a s e . ∴ ∠ O A C = ∠ O C A = x (s a y) . ∴ ∠ A O C = 180^{o} - (x + x) = 180^{o} - 2 x . . . . . . . . . (i) . N o w t h e c h o r d A C s u b t e n d s ∠ A O C t o t h e c e n t r e O a n d ∠ A B C t o t h e c i r c u m f e r e n c e a t B . ∴ ∠ A O C = 2 ∠ A B C ⟹ ∠ A B C = \frac{1}{2} ∠ A O C = \frac{1}{2} \times (180^{o} - 2 x) = 90^{o} - x (f r o m i) . . . . . . . (i i) . A g a i n i n Δ A B C w e h a v e ∠ B A C = ∠ O A B + ∠ O A C = 30^{o} + x a n d ∠ B C A = ∠ O C A + ∠ O C B = 40^{o} + x . ∴ ∠ A B C = 180^{o} - (∠ B A C + ∠ B C A) = 180^{o} - (30^{o} + x + 40^{o} + x) = 110^{o} - 2 x . . . . . . (i i i) . (b y a n g l e s u m p r o p e r t y o f t r i a n g l e s) F r o m (i i) & (i i i) 110^{o} - 2 x = 90^{o} - x ⟹ x = 20^{o} . ∴ ∠ A O C = 90^{o} - x = 90^{o} - 20^{o} = 70^{o} (f r o m i i) A n s - O p t i o n A .$

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Similar questions

∠ O A B = 30^{\circ}

∠ O C B = 40^{\circ}

∠ A O C

∠ O A B

∠ O C B

30^{\circ}

40^{\circ}

∠ A O C

In the given figure O is the centre of circle. If ∠OAB = 30^o and ∠OCB = 40^o, then measure of ∠AOC is :

∠ O A B = 30^{\circ}

∠ O C B = 55^{\circ}

∠ B O C

∠ A O C

∠ O A B = 20 °

∠ O C B = 50 °

∠ A O C

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