Given : $$cos x=sin y$$
=> $$sin(90^\circ-x)=siny$$
=> $$(90^\circ-x)=y$$
=> $$x+y=90^\circ$$ ---------(i)
Also, $$cot(x-40^\circ)=tan(50^\circ-y)$$
=> $$cot(x-40^\circ)=tan[90^\circ-(40^\circ+y)]$$
Using, $$tan(90^\circ-\theta)=cot \theta$$
=> $$cot(x-40^\circ)=cot(40^\circ+y)$$
=> $$x-40^\circ=40^\circ+y$$
=> $$x-y=80^\circ$$ -----------(ii)
Adding equations (i) and (ii), => $$2x=80^\circ+90^\circ=170^\circ$$
=> $$x=\frac{170^\circ}{2}=85^\circ$$
and $$y=90^\circ-85^\circ=5^\circ$$
=> Ans - (C)
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