Answer to Question #88297 in Mechanics | Relativity for Falcon

Question

Prove that in ∆=r^2 cotalpha/2 cot beta/2 cot gamma/2

Answer 1

Consider a triangle ABC that has an incircle of radius r with center O. The incircle is tangent to AB in point P. Since the radius is perpendicular to AB, APO is right angle. Thus r is an altitude of of BOA.

Hence its area is

$$\Delta_{BOA}=\frac{1}{2}rc.$$

By analogy, the triangles CAO and CBO have respective areas:

$$\Delta_{CAO}=\frac{1}{2}rb, \space\space\space \Delta_{CBO}=\frac{1}{2}ra,$$

and the total area is

$$\Delta=\frac{1}{2}r(a+b+c).$$

Now look a bit closer at small triangles like APO or COD or GOB. They are right-angled triangles and AP in APO for instance is:

$$AP=r\text{cot}(\alpha/2).$$

The area of AOP:

$$\Delta_{APO}=\frac{1}{2}AP\cdot r=\frac{1}{2}\cdot r\text{cot}(\alpha/2)\cdot r=\frac{1}{2}r^2\text{cot}(\alpha/2).$$

We have 6 small right-angled triangles in total and 3 of them are similar, for instance, APO and ADO. By analogy repeat the calculations for the triangles POB (=BOG) or DOC (=GOC).

The area of quadrilateral APOD is:

$$A_{APOD}=r^2\text{cot}(\alpha/2).$$

The total area of the triangle ABC is a sum of areas of 3 quadrilaterals (or 3 pairs of small triangles):

$$\Delta=r^2 [\text{cot}(\alpha/2)+\text{cot}(\beta/2)+\text{cot}(\gamma/2)].$$