If `costheta+((1)/(sqrt3))sintheta=(2)/(sqrt3)`, then find `theta` in circular measure.A. `pi^(c)/(10)`B. `pi^(c)/(9)`C. `pi^(c)/(6)`D. `pi^(c)/(3)`

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

1.	If `costheta+((1)/(sqrt3))sintheta=(2)/(sqrt3)`, then find `theta` in circular measure.A. `pi^(c)/(10)`B. `pi^(c)/(9)`C. `pi^(c)/(6)`D. `pi^(c)/(3)`
Answer» Correct Answer - C 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)`