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(B) Consider a right-angled triangle $ABC$ where $\angle B = 90^{\circ}$ and $\angle C = 60^{\circ}$.In this triangle,the side opposite to $\angle C$

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(B) Consider a right-angled triangle $ABC$ where $\angle B = 90^{\circ}$ and $\angle C = 60^{\circ}$.In this triangle,the side opposite to $\angle C$ is $AB$ and the hypotenuse is $AC$.Using the trigonometric ratio $\sin 	heta = \frac{	ext{opposite side}}{	ext{hypotenuse}}$,we have:$\sin 60^{\circ} = \frac{AB}{AC}$Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get:$\frac{\sqrt{3}}{2} = \frac{AB}{AC}$Therefore,$AB = \frac{\sqrt{3}}{2} AC$.Thus,the measure of the side opposite to the angle of $60^{\circ}$ is $\frac{\sqrt{3}}{2}$ times the measure of the hypotenuse.

In a right triangle,if the measure of one of the angles is $60^{\circ}$,then the measure of the side opposite to the angle with measure $60^{\circ}$ is $\ldots \ldots$ times the measure of the hypotenuse.

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