In a `!ABC` , there is a point `D` on the side BC such that `(BD)/(DC)` =`1/3`.If `angleB=pi/3,angleC=pi/4` and `sinangle(CAD)=lamdasinangleBAD` then `lamda` is equal to

Options

A. `1/(sqrt6)`
B. `sqrt6`
C. `1/(sqrt3)`
D. `sqrt3`

Explanation

Let `angleCAD=theta_(1)` and `angleBAD=theta_(2)`

Using sine rule in `Delta' S` ADB and CAD, we get
`(BD)/(sintheta_(2))=(AD)/(sinB)` and `(CD)/(sintheta_(1))=(AD)/(sinC)`
`rArrBD=(ADsintheta_(2))/(sinpi/3)` and, `CD=(ADsintheta_(1))/(sinpi/4)`
`rArr(BD)/(CD)=(sintheta_(2))/(sintheta_(1))xx(sinpi/4)/(sinpi/3)`
`rArr1/3=(sintheta_(2))/(sintheta_(1))xxsqrt(2/3)rArr(sintheta_(1))/(sintheta_(2))=sqrt6rArrlamda=sqrt6`