Angle bisector divides the triangle into two triangles. Find the area of one of them.

Question

In $\triangle ABC, AB = 12, AC = 10$.
$I$ is incenter $∠BIC = 105 ^{\circ}$.
Find area of $\triangle ABD$ where $AD$ is angle bisector.

I've drawn the following figure:

Now, $∠IBD + ∠ICB =75 ^{\circ} $
Hence $∠A=180-150=30^{\circ} $
By law of cosines ,
$BC^2=244-120\sqrt3$
Now , $x+y=12$
$x+z=10$
$y+z=\sqrt{244-120\sqrt3}$

Now I want the length of perpendicular from $A$ to $BC$ . The first problem is that I can't find it.
The second problem is that I think my algebraic approach is not at all elegant , so can anyone find a better way to solve this problem?

Answer 1

We know $\frac{\angle A}{2}=15^{\circ}$.

Now by summing the two areas $(ABD)$ and $(ACD)$ we get $(ABC)$, using the area formula $2\Delta=ac\sin B$, we can write it as: $$AD\cdot12\sin15^{\circ}+AD\cdot10\sin 15^{\circ}=120\sin 30^{\circ}$$ So, $$AD=\frac{30}{11\sin15^{\circ}}$$ Thus, $(ABD)=\frac12\cdot12\cdot AD \cdot\sin 15^{\circ}= 6\cdot\frac{30}{11\sin15^{\circ}}\cdot\sin 15^{\circ}=\frac{180}{11}$.

Answer 2

I don't know whether its invertendo - compenendo - invertendo or something else. These answers were intended for me so they are written in the language that I can understand. So if you didn't got something do ask.