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In a "DeltaA B C"," "/"C"=60" "&\ /A=75....

In a `"DeltaA B C"," "/_"C"=60" "&\ /_A=75.` If `D` is a point on `A C` such that area of the `"DeltaB A D"` is `sqrt(3)` times the area of the `"DeltaB C D",` then the `/_"A B D"=` `60^0` (b) `30^0` (c) `90^0` (d) none of these

Text Solution

Verified by Experts

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^(@)`
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