Question

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

Answer 1

Let $\Delta$ABC be a right-angled triangle, right-angled at B.
We know that, sin A = $\frac{BC}{AC}$ = $\frac{3}{4}$
Let BC be 3k and AC will be 4k where k is a positive real number.

By Pythagoras theorem; we get,
$A{C}^2=A{B}^2+B{C}^2$
$(4k{)}^2=A{B}^2+(3k{)}^2$
$16k^2-9k^2=A{B}^2$
$A{B}^2=7k^2$
AB = $\sqrt{7}k$
cos A = $\frac{AB}{AC}$ = $\frac{\sqrt{7}k}{4k}=\frac{\sqrt{7}}{4}$
tan A = $\frac{BC}{AB}$ = $\frac{3k}{\sqrt{7}k}=\frac{3}{\sqrt{7}}$