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If \(\sin\theta = \frac{3}{5}, 0° < \theta < 90&de...

If \sin\theta = \frac{3}{5}, 0° < \theta < 90°, evaluate \cos(180 - \theta).

A.

\frac{4}{5}

B.

\frac{3}{5}

C.

\frac{-3}{5}

D.

\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}