$□ A B C D$ is cyclic. If $m ∠ B = 60^{o}$ then, $m ∠ D =$ ?

120^{o}

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100^{o}

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30^{o}

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90^{o}

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Solution

The correct option is B $120^{o}$
In a cyclic quadrilateral, opposite angles are supplementary
$∴ m ∠ B + m ∠ D = 180^{o}$
$∴ m ∠ D = 180^{o} - 60^{o} = 120^{o}$
Hence, the answer is $120^{o}$

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Similar questions

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

∠ D =

∠ B = 60^{o}

∠ D =

\frac{{tan}^{2} 60^{o} + 4 {sin}^{2} 45^{o} + {sin}^{2} 90^{o}}{3 {sec}^{2} 30^{o} + {csc}^{2} 60^{o} - {cot}^{2} 30^{o}}

\frac{{tan}^{2} 60^{o} + 4 {cos}^{2} 45^{o} + 3 {sec}^{2} 30^{o} + 5 {cos}^{2} 90^{o}}{c o sec 30^{o} + sec 60^{o} - {cot}^{2} 30^{o}}

A B C D

∠ B A D = 120^{o}

m ∠ B C D

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