Question:
If \(\cos A = \dfrac{12}{13}\), where \(A\) is an acute angle, find \(\sin A\).
यदि \(\cos A = \dfrac{12}{13}\) है, जहाँ \(A\) एक न्यून कोण है, तो \(\sin A\) का मान ज्ञात कीजिए।
Options:
- \(\dfrac{5}{13}\)
- \(\dfrac{12}{13}\)
- \(\dfrac{13}{5}\)
- \(\dfrac{7}{13}\)
- \(\dfrac{5}{13}\)
- \(\dfrac{12}{13}\)
- \(\dfrac{13}{5}\)
- \(\dfrac{7}{13}\)
Correct Answer: a)
Explanation:
Since \(A\) is acute, \(\sin A = \sqrt{1 – \cos^2 A} = \sqrt{1 – \left(\dfrac{12}{13}\right)^2} = \sqrt{1 – \dfrac{144}{169}} = \sqrt{\dfrac{25}{169}} = \dfrac{5}{13}\).
\(\sin A = \sqrt{1 – \cos^2 A} = \sqrt{1 – \dfrac{144}{169}} = \sqrt{\dfrac{25}{169}} = \dfrac{5}{13}\)। अतः विकल्प (a) सही है।