Internal angle bisectors and right-angled triangle? Imagine a right angled triangle. Two of its internal bisectors are of length $7cm$ and $4cm$, respectively. The first internal angle bisector belongs to the angle of the side $a$ and the second one belongs to the angle of the side $b$ Calculate the length of the hypotenuse of the right angled triangle.

Here is a sketch of the triangle and its two internal bisectors $\overline{AP}=7$ , $\overline{BQ}=4$ :

We shall need a cosine law:
$$
\cos(A) = 2\cos^2(\frac{1}{2}A)-1 \quad ; \quad
\cos(B) = 2\cos^2(\frac{1}{2}B)-1
$$
Three highly non-linear equations with three unknowns can be derived then:
$$
2\cos^2(\frac{1}{2}A)-1 = 2\left(\frac{b}{7}\right)^2-1 = \frac{b}{c} = \cos(A) \\
2\cos^2(\frac{1}{2}B)-1 = 2\left(\frac{a}{4}\right)^2-1 = \frac{a}{c} = \cos(B) \\
c^2 = a^2 + b^2
$$
Since finding a solution by hand seems to be hopeless, I decided to feed this
into some computer algebra system (MAPLE) and solve anything numerically:

> eqns := {2*(b/7)^2-1=b/c,2*(a/4)^2-1=a/c,c^2=a^2+b^2};
> sols := {a,b,c}; fsolve(eqns,sols,0..infinity);
  { a = 3.402041954, b = 6.813171845, c = 7.615326654 }

Where $c$ is the length of the hypotenuse that is asked for.
