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# 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$$

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

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