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Answer» Correct Answer - C
We have `cos y=sqrt(3)/2`
`rArr y=pi/6`
Let `cot (x-y)=t`
`:. t^(2)-t-sqrt(3) t+sqrt(3)=0`
`rArr (t-1) (t-sqrt(3))=0`
`:. cot (x-y)=1 or cot (x-y)=sqrt(3)`
Given that `x in (0, pi)`
If `x-y=pi/6` then `x=pi/6+pi/6=pi/3 rArr (x, y) =(pi/3, pi/6)`
If `x-y=pi/4` then `x=pi/6+pi/4=(5pi)/12 rArr (x, y) = ((5pi)/12, pi/6)`
Possible ordered pairs (x, y) are `(pi/3, pi/6), ((5pi)/12, pi/6)`.


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