Given \(\sin \theta = \frac{3}{5}  \implies opp = 3, hyp = 5\)

Using Pythagoras' Theorem, we have \( adj = \sqrt{5^{2} - 3^{2}} = \sqrt{16}  = 4\)

\(\therefore \cos \theta = \frac{4}{5}, 0° < \theta < 90°\)

In the quadrant where \(180° - \theta\) lies is the 2nd quadrant and here, only \(\sin \theta = +ve\).

\(\therefore \cos (180 - \theta) = -ve = \frac{-4}{5}\)