If `costheta+((1)/(sqrt3))sintheta=(2)/(sqrt3)`, then find `theta` in circular measure.

Options

A. `pi^(c)/(10)`
B. `pi^(c)/(9)`
C. `pi^(c)/(6)`
D. `pi^(c)/(3)`

Answer

C

Explanation

Given
`costheta+((1)/(sqrt3))sintheta=(2)/(sqrt3)`
`rArr((sqrt3)/(2))costheta+((1)/(2))sintheta=1`
`rArrsin60^(@)costheta+sinthetacos60^(@)=1`
`sin(60^(@)+theta)=sin90^(@)`
`60^(@)+theta=90^(@)`
`rArrtheta=30^(@)=(pi^(c))/(6)`