If \( sinA \ = \ \frac{3}{4} \) , calculate cos A and tan A.

Given,
\( sinA \ = \ \frac{3}{4} \ = \ \frac{P}{H} \ = \ \frac{BC}{CA} \)

Let BC = 3k and AC = 4k.



Then, \( AB \ = \ \sqrt{AC^2 - BC^2} \ \)
\( = \ \sqrt{(4k)^2 - (3k)^2} \ \)
\( = \ \sqrt{16k^2 - 9k^2} \ \)
\( = \ \sqrt{7k^2} \ = \ \sqrt{7}k \)

\( \therefore cosA \ = \ \frac{b}{h} \ = \ \frac{AB}{AC} \ = \ \frac{ \sqrt{7}k}{4k} \ = \ \frac{ \sqrt{7}}{4} \)

\( tanA \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{3k}{ \sqrt{7}k} \ = \ \frac{3}{ \sqrt{7}} \)