The angle which is twice its supplement is:

120^{o}

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90^{o}

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60^{o}

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30^{o}

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Solution

The correct option is A $120^{o}$
Let supplement be $x$
$∴ A n g l e = 2 x$
$x + 2 x = 180$
$3 x = 180$
$x = 60^{o}$
Now , Supplement = $2 x = 2 \times 60 = 120^{o}$

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Similar questions

30^{o}

\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

30^{o}

∠ B

120^{o}

4 ({sin}^{4} 30^{o} + {cos}^{4} 60^{o}) - 3 ({cos}^{4} 45^{o} - {sin}^{2} 90^{o})

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