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

Given:
In right-angled triangle ABC, right-angled at A.
\(\sin B = \dfrac{1}{4}\)

Step 1: Use definition of sine
\[ \sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{4} \]
So, opposite side to angle \(B = 1\), hypotenuse = 4
Step 2: Use Pythagoras theorem to find adjacent side
Let adjacent side = \(x\)
\[ x^2 + 1^2 = 4^2 \Rightarrow x^2 + 1 = 16 \Rightarrow x^2 = 15 \Rightarrow x = \sqrt{15} \]
Step 3: Use definition of secant
\[ \sec B = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{4}{\sqrt{15}} = \frac{\sqrt{15}}{4} \text{ (rationalizing the denominator)} \]

Final Answer:
\[ \sec B = \frac{\sqrt{15}}{4} \]