Question

In a `triangle ABC`, `angleB=pi/3`, `angleC=pi/4` and D divides `BC` internally in the ratio `1 : 3` Then `(angleBAD)/(angleCAD)=` is equal to (a) `1/sqrt6` (b) `1/3` (c) `1/sqrt3` (d) `sqrt(2/3)`
A. `(1)/(sqrt(6))`
B. `(1)/(3)`
C. `(1)/(sqrt(3))`
D. `sqrt((2)/(3))`

Answer 1

Correct Answer - A
In `DeltaABD`, applying since rule, we get

`(AD)/(sin pi//3)=(x)/(sin alpha)`
`rArr AD=(3)/(sqrt(2))x sin beta " "....(i)`
and in `DeltaACD`, applying since rule, we get
`(AD)/(sin//pi)=(3x)/(sin beta)`
`rArrAD=(3)/(sqrt(2))x sin beta" "....(ii)` From Eqs. (i) and (ii) , `(sqrt(3)x)/(2 sin alpha)=(3x)/(sqrt(2)sin beta)`
`rArr=(sin alpha)/(sin beta)=(1)/(sqrt(6))`