In `Delta ABC, angle B=(pi)/(4),angle C=(pi)/(6)`. D is a point on BC which divides it in the ratio `1 : 3, angle DAB=beta`, then

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 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 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 Delta ABC angle A = (pi)/(6) & b:c= 2: sqrt(3) "then " angle B 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))

In a Delta ABC, angle B=100^(@), angle C=70^(@), AE _|_ BC & AD is angle bisector . Find angle EAD .