In Delta ABC,m angle A = 90^{circ} and m angl | vedclass

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(B) In $\Delta ABC$,we are given $m \angle A = 90^{\circ}$ and $m \angle B = 30^{\circ}$.Since the sum of angles in a triangle is $180^{\circ}$,$m \an

Vedclass · Accepted Answer

$AC = \frac{1}{2} BC$

Vedclass · Answer

(B) In $\Delta ABC$,we are given $m \angle A = 90^{\circ}$ and $m \angle B = 30^{\circ}$.Since the sum of angles in a triangle is $180^{\circ}$,$m \angle C = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.In a right-angled triangle,the side opposite to the $30^{\circ}$ angle is half the length of the hypotenuse.The side opposite to $\angle B$ $(30^{\circ})$ is $AC$,and the hypotenuse is $BC$.Therefore,$AC = \frac{1}{2} BC$.

In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $m \angle B = 30^{\circ}$. Then,which of the following holds true?

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