Ratio of radius of two circles in a triangle 
Given a triangle with two circles and apex angle equals $\theta$.
Find the ratio of radius of the two circles in terms of $\theta$.
My approach: treat the circles as incircle and excircle by drawing a line parallel to base.
We know that  $$ \frac{r}{R}=s-\frac{a}{s}$$  where $s=$semiperimeter and $$\sin \frac\theta 2= \sqrt{(s-b)(s-c)}/bc$$
and proceeding for equilateral triangle and right angled triangle,
I get $$ \left(1+\sin \frac\theta 2\right)/\left(1-\sin \frac\theta 2\right).$$
I have proved it for the following angles: $\frac{\pi}{4}$ and $\frac{\pi}{3}$.
But I am yet to prove it for a scalene triangle.
Thanks.
 A: The question does not actually mention the base edge
You introduced the tangent where the two circles touch, which necessarily creates a small isosceles triangle.  You can then make the base tangent parallel to this, ensuring a similar large triangle which is then isosceles
So there is no scalene case that you need to consider

Looking back, that explanation was a little brief.  Consider this diagram where the red triangle and circles are given.  Your calculations only involved the blue lines and two isosceles triangles even though the original triangle was scalene.

A: Use the lower picture and note that the red lines are perpendicular to the outer edges.
Using this knowledge we can set up the following system.
$$\sin\dfrac{\theta}{2} = \dfrac{r}{r+s}$$
$$\sin\dfrac{\theta}{2} = \dfrac{R}{R+2r+s}$$
Equate both equations and solve for $s=s(r,R)$. Then take the first equation. Plug in $s=s(r,R)$ divide the denominator and numerator by $R$ and you should only get ratios $r/R$. Then solve for this ratio.

