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

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

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$5$

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(D) In $\Delta ABC$,we are given $m \angle B = 90^{\circ}$ and $m \angle C = 30^{\circ}$.Since the sum of angles in a triangle is $180^{\circ}$,$m \angle A = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}$.In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to $30^{\circ}$ is half the hypotenuse.Here,$AB$ is opposite to $30^{\circ}$ (at $C$),so $AB = \frac{1}{2} AC$.Given $AB = 5$,we have $5 = \frac{1}{2} AC$,which implies $AC = 10$.In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Therefore,$BD = \frac{1}{2} AC = \frac{1}{2} 	imes 10 = 5$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is a median. If $m \angle C = 30^{\circ}$ and $AB = 5$,find $BD$.

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