Since \( a \) is in quadrant IV, the sine value will be negative. We use the Pythagorean identity:
\[
\sin^2 a + \cos^2 a = 1
\]
Substituting \( \cos a = \frac{5}{13} \):
\[
\sin^2 a + \left( \frac{5}{13} \right)^2 = 1
\]
\[
\sin^2 a + \frac{25}{169} =
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