In a triagnle ABC, `angle B=pi/3 " and " angle C = pi/4` let D divide BC internally in the ratio 1:3 .Then `(sin (angle BAD))/((Sin (angle CAD))` is equal to

In triangle ABC, angle B = pi//3. " and " angle C = pi//4 . Let D divide BC inernally in the ratio 1 : 3 . Then (sin angle BAD)/(sin angle CAD) equals

In a DeltaABC, angleB=(pi)/(3) and angleC=(pi)/(4) let D divide BC internally in the ratio 1:3 , then (sin(angleBAD))/(sin(angleCAD)) is equal to :

In a DeltaABC, angleB=(pi)/(3) and angleC=(pi)/(4) let D divide BC internally in the ratio 1:3 , then (sin(angleBAD))/(sin(angleCAD)) is equal to :

In a triangle ABC, angleB = pi/3 and angleC= pi/4 . Let D divide side BC internally in the ratio 1:3. Then sin(angleBAD)/sin(angleCAD) is

In triangle ABC,/_B=(pi)/(3), and /_C=(pi)/(4)* Let D divided BC internally in the ratio 1:3. Then (sin/_BAD)/(sin/_CAB) equals (a) (1)/(sqrt(6)) (b) (1)/(3)(c)(1)/(sqrt(3)) (d) sqrt((2)/(3))