If sec A = 13/5 and A is acute, find sin A .

Question

If sec A = 13/5 and A is acute, find sin A .

1. A. 5/13
2. B. 12/13 (Correct answer)
3. C. 13/12
4. D. 13/5

Correct Answer

Option B — 12/13

Detailed Solution & Explanation

The correct answer is 12/13.

Key Points

• $\sec A = \dfrac{13}{5}$ means $\cos A = \dfrac{5}{13}$, since secant is the reciprocal of cosine.
• Reading this as a right triangle: adjacent 5, hypotenuse 13, so the opposite side is $\sqrt{169 - 25} = 12$.
• Therefore $\sin A = \dfrac{12}{13}$.

Additional Information

• (5, 12, 13) is a Pythagorean triple, so the third side comes out whole — recognising it removes the square-root step.
• A is acute, so every ratio is positive and no sign ambiguity arises.
• Alternatively use $\sin^2 A + \cos^2 A = 1$: $\sin^2 A = 1 - \tfrac{25}{169} = \tfrac{144}{169}$, giving $\sin A = \tfrac{12}{13}$.

प्रश्न (हिन्दी में)

यदि sec A = 13/5 और A एक न्यूनकोण है, तो sin A ज्ञात कीजिए।

1. A. 5/13
2. B. 12/13
3. C. 13/12
4. D. 13/5