If $\sin A = \frac{3}{4},$ calculate $\cos A$ and $\tan A$.

(N/A) Let $\triangle ABC$ be a right-angled triangle,right-angled at point $B$.
Given that,$\sin A = \frac{3}{4}$.
Since $\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC}$,we have $\frac{BC}{AC} = \frac{3}{4}$.
Let $BC = 3k$ and $AC = 4k$,where $k$ is a positive constant.
Applying the Pythagoras theorem in $\triangle ABC$:
$AC^2 = AB^2 + BC^2$
$(4k)^2 = AB^2 + (3k)^2$
$16k^2 = AB^2 + 9k^2$
$AB^2 = 7k^2$
$AB = \sqrt{7}k$
Now,$\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4}$.
And,$\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}}$.