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

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Question

(B) In $\Delta ABC$,the sum of angles is $180^{\circ}$.Since $m \angle B = 90^{\circ}$ and $m \angle C = 30^{\circ}$,we have $m \angle A = 180^{\circ}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) In $\Delta ABC$,the sum of angles is $180^{\circ}$.Since $m \angle B = 90^{\circ}$ and $m \angle C = 30^{\circ}$,we have $m \angle A = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.This is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to the $30^{\circ}$ angle is half the hypotenuse.The side opposite to $30^{\circ}$ is $AB$.The hypotenuse is $AC = 10$.Therefore,$AB = \frac{1}{2} 	imes AC = \frac{1}{2} 	imes 10 = 5$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $m \angle C = 30^{\circ}$. If $AC = 10$,then $AB = \ldots$

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