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 = ?

$\frac{5}{13}$ $\frac{2}{13}$ $\frac{7}{13}$ $\frac{9}{13}$

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

$\sin A=\frac{5}{13},\ A+B=90^\circ \cos B=\sin A=\frac{5}{13}$

Get latest updates, jobs, results & quizzes directly on your device.

Scroll to Top