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

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(A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.Given $m \angle B = 90^{\circ}$ and $m \angle A = 60^{\circ}$,then $m \angle C = 180^{\circ} -

Vedclass · Accepted Answer

$5$

Vedclass · Answer

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

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $m \angle A = 60^{\circ}$. If $AC = 10$,then $AB = \dots$

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