Question

2. If sin B = 3/8, what is tan B?

Answer 1

SInce,

$\begin{gathered} \sin B=\frac{opposite}{hypotenuse} \\ \text{opposite}=3 \\ \text{hypotenuse}=8 \end{gathered}$

Therefore using Pythagoras theorem, we have,

$x=\sqrt[]{8^2-3^2}=\sqrt[]{64-9}=\sqrt[]{55}$

Therefore, tan B can be calculated as,

$\tan \text{ B=}\frac{opposie}{adjacent}=\frac{3}{\sqrt[]{55}}$