Find the supplement of the following angle.
$40^{\circ}$

140

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40

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10

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None of these

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Solution

The correct option is D $140$
We know that the supplement angle
$= 180^{0} - θ$

So,
The supplement angle of $40^{0}$ will be
$= 180^{0} - 40^{0} = 140^{0}$

Hence, this is the answer.

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∠ B = 40 °

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= 20^{\circ} 30^{'} 40^{''}

Class interval	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$	$100 - 120$	$120 - 140$
Frequency	$6$	$8$	$10$	$12$	$6$	$5$	$3$

40

40

140

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