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

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

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) In $\Delta ABC$,we are given $m \angle B = 90^{\circ}$.Since the sum of angles in a triangle is $180^{\circ}$,we have $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Substituting $m \angle B = 90^{\circ}$,we get $m \angle A + m \angle C = 90^{\circ}$.Given the ratio $m \angle A : m \angle C = 1 : 2$,let $m \angle A = x$ and $m \angle C = 2x$.Thus,$x + 2x = 90^{\circ}$,which implies $3x = 90^{\circ}$,so $x = 30^{\circ}$.Therefore,$m \angle A = 30^{\circ}$ and $m \angle C = 60^{\circ}$.In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to $30^{\circ}$ is half the hypotenuse.Here,$BC$ is opposite to $\angle A = 30^{\circ}$,so $BC = \frac{1}{2} AC$.Given $BC = 4$,we have $4 = \frac{1}{2} AC$,which implies $AC = 8$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $m \angle A : m \angle C = 1 : 2$. If $BC = 4$,find $AC$.

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