Question

The $△ABC$ has a right angle at C. If $\sin A=\frac{2}{3}$ then $\tan B$ is

• A. $\frac{3}{5}$
• B. $\frac{\sqrt{5}}{3}$
• C. $\frac{2}{\sqrt{5}}$
• D. $\frac{\sqrt{5}}{2}$

Answer 1

The correct option is D $\frac{\sqrt{5}}{2}$

$\Rightarrow \sin A=\frac{2}{3}=\frac{BC}{AB}$

In right angled $△ABC,$

$\Rightarrow A{C}^2+B{C}^2=A{B}^2$

$\Rightarrow A{C}^2=A{B}^2-B{C}^2$

$=9-4$

$\Rightarrow AC=\sqrt{5}$

$\Rightarrow \tan B=\frac{AC}{BC}=\frac{\sqrt{5}}{2}$

Hence, the answer is $\frac{\sqrt{5}}{2}.$