Question

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

• A. $\cos A=\frac{\sqrt{3}}{4}\tan A=\frac{2}{5}$
• B. $\cos A=\frac{\sqrt{7}}{4}\tan A=\frac{3}{\sqrt{7}}$
• C. $\cos A=\frac{\sqrt{3}}{2}\tan A=\frac{2}{\sqrt{5}}$
• D. $\cos A=\frac{\sqrt{5}}{7}\tan A=\frac{7}{\sqrt{3}}$

Answer 1

Answer: (B)

$\cos A=\frac{\sqrt{7}}{4}\tan A=\frac{3}{\sqrt{7}}$

Given, $\sin A=\frac{3}{4}$
$\Rightarrow \frac{BC}{DC}=\frac{3}{4}$
$\Rightarrow BC=3k$ and $AC=4k$
where $k$ is the constant of proportionality.
By Pythagoras theorem, we have
$A{B}^2=A{C}^2-B{C}^2=(4k{)}^2-(3k{)}^2=7k^2$
$\Rightarrow AB=\sqrt{7}k$
So, $\cos A=\frac{AB}{AC}=\frac{\sqrt{7}k}{4k}=\frac{\sqrt{7}}{4}$
And $\tan A=\frac{BC}{AB}=\frac{3k}{\sqrt{7}k}=\frac{3}{\sqrt{7}}$