Question

In a triangle abc angle c=60 degree and angle a=75 degree if d is a point on ac such that tbe area of the trianglr

Answer 1

Here is the full question:

In a triangle ABC 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 √3  times the area of the triangle BCD find angle ABD.

Answer:

First of all, angle B = 45 degree  by sine law

 BA/ BC  = sin 60/ sin 75

It will give us

BA/ BC = \sqrt6 / (\sqrt3  +  1)

The Area  = (1/2) a* b* sin\theta

Let \theta =  Angle ABD

Now

ar(tr  BAD) =  \sqrt3 ar (tr  BCD)

BA sin \theta / BC  = \sqrt3 sin(45 –\theta)

We will put the value of BA/ BC in the above equation

 Then solve and you will find that

\theta = 30 degree  = angle ABD