If a chord AB of a circle subtends an angle `theta(nepi//3)` at a point C on the circumference such that the triangle ABC has maximum area , then

Options

A. `A=pi/3+(theta)/2,B=(2pi)/3-(3theta)/2`
B. `A=(pi)/4+(theta)/2,B=(3pi)/4-(3theta)/2`
C. `A=pi/6+theta,B=(5pi)/6+2theta)`
D. none of these

Explanation

We have,
`Delta`= Area of `Delta` ABC
`rArrDelta=1/2(ABxxACxxsinA)~
`rArrDelta=1/22Rsinthetaxx2RsinBxxsinA` `[becauseAB=2Rsintheta` and `AC=2RsinB]`
`rArrDelta=2R^(2)sinAsinBsintheta`
`rArrDelta=R^(2)sintheta[cos(A-B)-cos(A+B)}`
`rArrDelta=R^(2)sintheta[cos(pi-2B-theta)-cos(pi-theta)]` `[becauseA+B+C=pi]`
`rArrDelta=R^(2)sintheta[costheta-cos(2B+theta)}`

Clearly, `Delta` is maximum when
`cos(2B+theta)=-1rArr2B+theta=pirArrB=pi/2-(theta)/2`
`thereforeA+B+theta=piA=pi/2-(theta)/2`
Hence, area is maximum when `A=B=(pi)/2-(theta)/2`