Question

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

• A. 4
• B. \(\frac{4}{\sqrt{15}}\)
• C. \(\sqrt{15}\)
• D. \(\frac{\sqrt{15}}{4}\)

Hint:

Use Pythagoras theorem to find the missing side, then apply trigonometric ratios.

Answer 1

The Correct Option is B

Given:
Right-angled triangle \(ABC\) with right angle at \(A\).
\[ \sin B = \frac{1}{4} \]

Step 1: Recall identity for \(\sin^2 B + \cos^2 B = 1\)
Calculate \(\cos B\):
\[ \cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \]

Step 2: Calculate \(\sec B = \frac{1}{\cos B}\)
\[ \sec B = \frac{1}{\frac{\sqrt{15}}{4}} = \frac{4}{\sqrt{15}} \]

Final Answer:
\[ \boxed{\frac{4}{\sqrt{15}}} \]