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.
 A: Case 1:  Suppose that $a,b$ are the legs of the right triangle. Let then $c$ be the length of the hypotenuse, and $C = \frac{\pi}{2}, A, B$ the respective angles.
It follows by inspection that:
$$ 7 = \frac{c \; \cos A}{\cos A/2} \quad \text{and} \quad 4 = \frac{c \; \cos B\; }{\cos B/2}$$
Since $\cos \frac{x}{2}=\sqrt\frac{1 + \cos x}{2}$ and $B = \frac{\pi}{2} - A$, the above  gives:
$$7^2 = \frac{2 c^2 \cos^2 A}{1+ \cos A} \quad \text{and} \quad 4^2 = \frac{2 c^2 \sin^2 A}{1+ \sin A}$$
After dividing the two equalities:
$$\left(\frac{4}{7}\right)^2 = \tan^2 A \;\frac{1+ \cos A}{1+ \sin A}$$
In terms of $t=\tan A/2$ (with $\cos A = \frac{1 - t^2}{1+t^2}, \sin A = \frac{2t}{1+t^2}$) the equation becomes:
$$\require{cancel}\frac{2 \cdot \bcancel{8}}{49}=\frac{\bcancel{4} t^2}{(1-t^2)^2} \cdot \frac{\bcancel{2}}{\cancel{1+t^2}} \cdot \frac{\cancel{1+t^2}}{1+2 t+t^2}$$
$$2(1-t^2)^2(1+t)^2 - 49 t^2 = 0$$
$$\Big(\sqrt{2}(1-t^2)(1+t) - 7 t\Big)\Big(\sqrt{2}(1-t^2)(1+t) + 7 t\Big) = 0$$
$$\Big(\sqrt{2} + (\sqrt{2}-7) t- \sqrt{2} t^2-\sqrt{2} t^3\Big)\Big(\sqrt{2} + (\sqrt{2}+7) t- \sqrt{2} t^2-\sqrt{2} t^3\Big) = 0$$
Since $A/2 \in (0, \pi / 4)$ the roots of interest are $t = \tan A/2 \in (0,1)$ and it can be shown that the only such root belongs to the first cubic. The value can be calculated exactly by solving the cubic using Cardano's formula, for example, though the resulting expression is not pretty.
Numerically, the eligible root is $t \approx 0.23579$ corresponding to $A \approx 26.53447^{\circ}$ and the hypotenuse $c \approx 7.61533$ (which agrees with the Maple solver result posted by Han de Brujin in his answer).
Case 2:  Suppose that one of $a,b$ is the hypotenuse. Since the shortest bisector corresponds to the largest angle, the right angle bisector must be $4$ and side $b$ must be the hypotenuse. Then (by the bisector length formula for the former):
$$ 4^2 = \frac{2 a^2 c^2}{(a+c)^2} = \frac{2 \;b^2 \;\sin^2 A \;\cos^2 A}{(\sin A + \cos A)^2} \quad \text{and} \quad 7 = \frac{b \;\cos A}{\cos A/2}$$
$$ 4^2 = \frac{2 \;b^2 \;\sin^2 A \;\cos^2 A}{1 + 2 \;\sin A \;\cos A} \quad \text{and} \quad 7^2 = \frac{2 b^2 \;\cos^2 A}{1 + \cos A}$$
Similar to the previous case, dividing and rationalizing in $t = \tan A/2$ gives:
$$\left(\frac{7}{4}\right)^2 = \frac{1 + 2 \;\sin A \;\cos A}{(1 + \cos A) \sin^2 A} = \frac{(1+t^2)^2 + 4 t (1-t^2)}{(1+t^2)^2} \cdot \frac{(1+t^2)^3}{8 t^2}$$
$$\cdots$$
$$2 t^6-8 t^5+6 t^4-43 t^2+8 t+2 = 0$$
The sextic can be shown to have a unique real root in $(0,1)$ which is numerically $t \approx 0.32999$ and corresponds to $A \approx 36.52438^{\circ}$ and the hypotenuse $b \approx 8.27202$.
A: 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.
