In the given figure, $A B C D$ is a cycle quadrilateral in which $∠ B A D = 120^{o}$ , then $∠ B C D$ is

240^{o}

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60^{o}

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120^{o}

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180^{o}

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Solution

The correct option is B $60^{o}$
$G i v e n - A B C D i s a c y c l i c q u a d r i l a t e r a l . ∠ B A D = 120^{o} . T o f i n d o u t - ∠ B C D = ? S o l u t i o n - W e k n o w t h a t t h e s u m o f t h e o p p o s i t e a n g l e s o f a c y c l i c q u a d r i l a t e r a l i s 180^{o} . ∴ ∠ B A D + ∠ B C D = 180^{o} ⟹ ∠ B C D = 180^{o} - ∠ B A D = 180^{o} - 120^{o} = 60^{o} . A n s - O p t i o n B .$

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Similar questions

A B C D

∠ B A D = 120^{o}

m ∠ B C D

∠ A = 120^{o}, ∠ B = 60^{o}, ∠ C = 100^{o}

∠ D =

∠ B A D = 120^{\circ}

∠ B C D

△ A B C

\frac{t a n A + t a n (A + 120^{o}) + t a n (A + 240^{o})}{t a n 3 A} = ?

A B C D

A B = B C = 3.0 c m, A D = C D = 5.0 c m

∠ A B C = 120^{o}

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