# inatriangleABC, angle C =60 degree  and angle A =75 degree. if D is a point on AC such that the area of the triangle BAD is $\sqrt{}$3  timess the area of the triangle BCD find angle ABD

Arun
25757 Points
5 years ago
first of all, angle B = 45 degree
by sine law

BA/ BC  = sin 60/ sin 75
which gives
BA/ BC = $\sqrt$6 / ($\sqrt$3  +  1)

we know that area  = (1/2) a* b* sin$\theta$
let $\theta$ =  angle ABD
now
ar(tr  BAD) =  $\sqrt$3 ar (tr  BCD)
BA sin $\theta$ / BC  = $\sqrt$3 sin(45 –$\theta$)
put the value of BA/ BC in the above equation

then solve and you will find that
$\theta$ = 30 degree  = angle ABD