Question

If secA=13/5 and A is acute, find sinA.

• A. 5/13
• B. 12/13
• C. 13/12
• D. 13/5

Answer 1

Answer:

B. 12/13

Given:

$\sec A = \frac{13}{5}$
$\Rightarrow \cos A = \frac{5}{13}$

Now use identity:

$\sin^2 A + \cos^2 A = 1$

Substitute:

$\sin^2 A = 1 - \left(\frac{5}{13}\right)^2$
$\sin^2A = 1 - \frac{25}{169} = \frac{144}{169}$

Since (A) is acute, (\sin A > 0):

$\sin A = \frac{12}{13}$

Therefore, sin A = 12/13.