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If `sintheta=-1/2` and `tantheta=1/sqrt(3)` then `theta` is equal to,

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If `sintheta=-1/2` and `tantheta=1/sqrt(3)` then `theta` is equal to,
A. `30^(@)`
B. `150^(@)`
C. `210^(@)`
D. none of these
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Let us first find out `theta` lying between 0 and `360^(@)`.
Since `sintheta=-1/2 rArr theta=210^(@)` or `330^(@)` and `tantheta=1/sqrt(3) rArr theta=30^(@)` or `210^(@)`
Hence, `theta=210^(@)` or `(7pi)/6` is the value satisfying both.
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