Given sinθ=35 ⟹ opp=3,hyp=5\sin\theta = \frac{3}{5} \implies \text{opp} = 3, \text{hyp} = 5sinθ=53⟹opp=3,hyp=5
Using Pythagoras' Theorem, we have adj=52−32=16=4adj = \sqrt{5^{2} - 3^{2}} = \sqrt{16} = 4adj=52−32=16=4
∴cosθ=45,0∘<θ<90∘\therefore \cos\theta = \frac{4}{5}, 0^{\circ} < \theta < 90^{\circ}∴cosθ=54,0∘<θ<90∘
In the quadrant where 180∘−θ180^{\circ} - \theta180∘−θ lies is the 2nd quadrant and here, only sinθ=+ve\sin\theta = +\text{ve}sinθ=+ve.
∴cos(180−θ)=−ve=−45\therefore \cos(180 - \theta) = -\text{ve} = \frac{-4}{5}∴cos(180−θ)=−ve=5−4