# In a triangle ABC angle B=60 angle C=45 AND D divides BC internally in the ratio 1:3.?

## THEN (sin BAD)/(sin CAD) IS?

Sep 22, 2017 In the above figure

In $\Delta A B C$, $\angle A B C = {60}^{\circ} , \angle A C B = {45}^{\circ}$

$\frac{B D}{D C} = \frac{1}{3.} \ldots . . \left[1\right]$

Now applying sine rule for $\Delta A B D$ we can write

$\frac{\sin \angle B A D}{\sin \angle A D B} = \frac{B D}{A B} \ldots \ldots \left[2\right]$

And applying sine rule for $\Delta A D C$ we can write

$\frac{\sin \angle C A D}{\sin \angle A D C} = \frac{D C}{A C}$

$\implies \frac{\sin \angle C A D}{\sin \left(180 - \angle A D B\right)} = \frac{D C}{A C}$

$\implies \frac{\sin \angle C A D}{\sin \angle A D B} = \frac{D C}{A C} \ldots . . \left[3\right]$

Dividing  by  we get

$\frac{\sin \angle B A D}{\sin \angle C A D} = \frac{B D}{D C} \times \frac{A C}{A B} \ldots . . \left[4\right]$

Now applying sine rule for $\Delta A B C$ we can write

$\frac{A C}{A B} = \frac{\sin {60}^{\circ}}{\sin {45}^{\circ}} \ldots \ldots \left[5\right]$

Combining ,  and 

$\frac{\sin \angle B A D}{\sin \angle C A D} = \frac{1}{3} \times \frac{\sin {60}^{\circ}}{\sin {45}^{\circ}}$

$\implies \frac{\sin \angle B A D}{\sin \angle C A D} = \frac{1}{3} \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{1} = \frac{1}{\sqrt{6}}$