In\(\Delta\) ABC, right-angled at C, sin A = $ \frac{5}{13}$, then COS B?
In\(\Delta\) ABC, right-angled at C, sin A = $ \frac{5}{13}$, then COS B?
\(\Delta\) ABC में, C समकोण है, यदि Sin A = $ \frac{5}{13} $ तो COS B = ?
Detailed Solution & Logic
Correct Answer:
\(\frac{5}{13}\)
In triangle ABC, the triangle is right angles at C. Therefore, the two acute angles A and B are complementary.
A + B = 90 Degree
$\sin A=\frac{5}{13},\ A+B=90^\circ \cos B=\sin A=\frac{5}{13}$
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