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Theorem of Cyclic Quadrilateral

In the given ...

Question

In the given figure, O is the centre of the circle. If $∠ A O C = 140^{o}$ ,AB=AC. Then $∠ O A C$ is

∠ O A C = 20^{o}

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∠ O A C = 40^{o}

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∠ O A C = 25^{o}

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∠ O A C = 30^{o}

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Solution

The correct option is C $∠ O A C = 20^{o}$
$G i v e n - O i s t h e c e n t r e o f a c i r c l e . A, B & C a r e p o i n t s o n t h e c i r c u m f e r e n c e o f t h e c i r c l e . ∠ A O C = 140^{o} .$

$T o f i n d o u t - ∠ O A C = ?$

$S o l u t i o n - O A & O C a r e r a d i i o f t h e s a m e c i r c l e .$

$∴ O A = O C .$

$i . e Δ O A C i s i s o s c e l e s .$

$∴ ∠ O A C = ∠ O C A$

$⟹ ∠ O A C + ∠ O C A = 2 ∠ O C A .$

$∴ B y a n g l e s u m p r o p e r t y o f t r i a n g l e s, w e h a v e$

$∠ A O C + 2 ∠ O C A = 180^{o}$

$⟹ 2 ∠ O C A = 180^{o} - ∠ A O C = 180^{o} . - 140^{o} . = 40^{o}$

$∠ O C A = \frac{40^{o}}{2} = 20^{o} .$

$A n s - O p t i o n A .$

Suggest Corrections

Similar questions

In the given figure, O is the centre of the circle and $∠ AOB = 70^{\circ}$ .

Calculate the values of (i) $∠ OCA (i i) ∠ OAC .$

O

∠ A O B = 140^{0}

∠ O A C = 50^{0}

∠ A C B

∠

^{o}

∠

∠ A O B = 140^{o}

∠ O A C = 50^{o}

(i) ∠ A C B,

(i i) ∠ O B C,

(i i i) ∠ O A B,

(i v) ∠ C B A .

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