Question

# ΔABC with all acute angles is given. The point whose coordinates are ($$\cos B\;-\;\sin A, \sin B\;-\;\cos A$$) can be in which quadrant.

Solution:

$$\angle A\;+\;\angle B\;+\;\angle C = \pi$$
$$\Rightarrow \angle A\;+\;\angle B = \pi\; – \;\angle C$$

And from question $$\angle A\; + \; \angle B > \frac{\pi}{2}$$
$$\Rightarrow \angle A > \frac{\pi}{2} – \angle B$$

Taking $$\cos$$ on both sides
$$\cos (\frac{\pi}{2} – B) > \cos A$$
$$\Rightarrow \sin B – \cos A > 0$$

Now taking $$\sin$$ on both sides
$$\sin A > \sin(\frac{\pi}{2} – B)$$
$$\Rightarrow \sin A > \cos B$$
$$\Rightarrow \cos B – \sin A < 0$$

Therefore Point($$\cos B\;-\;\sin A, \sin B\;-\;\cos A$$) lies in 2nd quadrant.