In Delta ABC,m angle C = 90^{circ} and tan A | vedclass

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(C) Given: $	an A = \frac{1}{\sqrt{3}}$.In a right-angled triangle,$	an A = \frac{	ext{Opposite side}}{	ext{Adjacent side}} = \frac{BC}{AC}$.There

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$\frac{1}{2}$

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(C) Given: $	an A = \frac{1}{\sqrt{3}}$.In a right-angled triangle,$	an A = \frac{	ext{Opposite side}}{	ext{Adjacent side}} = \frac{BC}{AC}$.Therefore,$\frac{BC}{AC} = \frac{1}{\sqrt{3}}$.Let $BC = k$ and $AC = \sqrt{3}k$,where $k > 0$.Using the Pythagorean theorem: $AB^2 = BC^2 + AC^2$.$AB^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2$.Thus,$AB = 2k$.Now,$\sin A = \frac{	ext{Opposite side}}{	ext{Hypotenuse}} = \frac{BC}{AB} = \frac{k}{2k} = \frac{1}{2}$.Alternatively,since $	an A = \frac{1}{\sqrt{3}}$,we know that $A = 30^{\circ}$.Therefore,$\sin A = \sin 30^{\circ} = \frac{1}{2}$.

In $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\tan A = \frac{1}{\sqrt{3}}$,then $\sin A = \ldots$

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