\frac{4}{5}
\frac{3}{5}
\frac{-3}{5}
\frac{-4}{5}
Correct answer is D
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}
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