In a triangle `A B C ,/_c=60^0a n d/_A=75^0` . If `D` is a point on `A C` such that the area of the ` B C D ,t h e/_A B D`

Explanation

Let `h` be the length of perpendicular from B on AC

given that `(Delta BAD)/(Delta BCD) = sqrt3`
`rArr ((1)/(2) h. AD)/((1)/(2) h. DC) = (AD)/(CD) = sqrt3`...(1)
In `Delta BAD`, taking `angle ABD = alpha`, we have
`(AD)/(sin alpha) = (BD)/(sin 75^(@))`...(2)
And in `Delta BCD`, we have `(CD)/(sin (45^(@) - alpha)) = (BD)/(sin 60^(2))`...(3)
`:.` from (2) and (3), we get
`(AD sin (45^(@) - alpha))/(CD sin alpha) = (sin 60^(@))/(sin 75^(@))`
`rArr sqrt3 sin 75^(@) sin (45^(@) - alpha) = sin 60^(@) . sin alpha`
`rArr sqrt3 ((sqrt3 + 1)/(2 sqrt2)) ((cos alpha - sin alpha)/(sqrt2)) = (sqrt3)/(2) sin alpha`
`rArr (sqrt3 + 1) cos alpha = (3 + sqrt3) sin alpha`
`rArr tan alpha = 1//sqrt3`
`rArr alpha = pi//6 = 30^(@)`
Hence `angle ABD = 30^(@)`