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Chord Properties of Circles

In the given ...

Question

In the given figure, $Δ A B C$ is inscribed in a circle. The bisector of $∠ B A C$ meets BC at D and the circle at E. If EC is joined then $∠ E C D = 30^{o}$ . Find $∠ B A C$ .

30^{o}

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40^{o}

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50^{o}

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60^{o}

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Solution

The correct option is C $60^{o}$
$G i v e n - Δ A B C i s i n s c r i b e d i n a c i r c l e . T h e a n g u l a r b i s e c t o r o f ∠ B A C m e e t s B C a t D a n d t h e c i r c l e a t E . E C i s j o i n e d . ∠ E C D = 30^{o} . T o f i n d o u t - ∠ B A C = ? S o l u t i o n - W e j o i n E B . T h e c h o r d E B h a s s u b t e n d e d ∠ E C D & ∠ E A B t o t h e c i r c u m f e r e n c e o f t h e g i v e n c i r c l e a t C & A r e s p e c t i v e l y . S o ∠ E C D = ∠ E A B = 30^{o} . s i n c e t h e a n g l e s, s u b t e n d e d b y a c h o r d o f a c i r c l e t o d i f f e r e n t p o i n t s o f t h e c i r c u m f e r e c e o f t h e s a m e c i r c l e, a r e e q u a l . B u t A D i s t h e a n g u l a r b i s e c t o r o f ∠ B A C . ∴ ∠ E A B = ∠ E A C = 30^{o} . ∴ ∠ E A B + ∠ E A C = ∠ B A C . ⟹ ∠ B A C = 30^{o} + 30^{o} = 60^{o} . A n s - O p t i o n C .$

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