Question

In a right triangle ABC, right-angled at A, if $\sin B = \dfrac{1}{4}$, then the value of $\sec B$ is

• A. 4
• B. $\dfrac{\sqrt{15}}{4}$
• C. $\sqrt{15}$
• D. $\dfrac{4}{\sqrt{15}}$

Hint:

Use right triangle identity: $\sec = \dfrac{\text{hypotenuse}}{\text{adjacent}}$.

Answer 1

The Correct Option is B

If $\sin B = \dfrac{1}{4} = \dfrac{\text{opposite}}{\text{hypotenuse}}$ → opposite = 1, hypotenuse = 4.
Using Pythagoras: adjacent = $\sqrt{4^2 - 1^2} = \sqrt{15}$
Then $\sec B = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{4}{\sqrt{15}}$