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If `sintheta=1/2a n dcostheta=-(sqrt(3))/2,` then the general value of `theta` is `(n in Z)dot`
A. `2n pi+(5pi)/6`
B. `2n pi+pi/6`
C. `2n pi+(7pi)/6`
D. `2npi+pi/4`

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Correct Answer - A
`sin theta=1//2` and `cos theta=-sqrt(3)//2`
`rArr theta` lies in the second quadrant.
`rArr sin theta=sin 5pi//6`,
`:. Theta=2npi+(5pi//6), n in Z`

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