Question

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

Answer 1

SInce,

`\begin{gathered} \sin B=(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=}(opposie)/(adjacent)=\frac{3}{\sqrt[]{55}}`