Q.
In a $ \triangle ABC$, $\angle B = \frac{\pi}{3} \, and \, \angle C = \frac{\pi}{4}. $ Let D divides BC
internally in the ratio 1: 3, then $ \frac{ sin \, \angle BAD }{ sin \, \angle CAD} $ is equal to
IIT JEEIIT JEE 1995
Solution:
In $\triangle$ ABD, applying sine rule, we get
= $ \frac{AD}{ sin \, \pi / 3} = \frac{ x}{sin \, \alpha} $
$\Rightarrow AD = \frac{\sqrt 3 }{2} x \, sin \, \alpha$ $$ ...(i)
and in $\triangle$ ACD, applying sine rule, we get
$ \frac{AD}{ sin \, \pi / 4} = \frac{ 3x}{sin \, \beta}$
$\Rightarrow AD = \frac{ 3}{ \sqrt 2} x \, sin \, \beta$ $$ ...(i)
From Eqs. (i) and (ii).
$ \frac{ \sqrt 3 \, x}{ 2 \, sin \, \alpha} = \frac{ 3x}{ \sqrt 2 \, sin \, \beta} $
$\Rightarrow \frac{ sin \, \alpha }{ sin \, \beta} = \frac{1}{ \sqrt 6} $
