Finding sides of triangle

Question

Given : $$\triangle ABC$$ $$M \in AB,N \in BC ,P \in AC$$ are the points at which the incircle crosses the triangle $$MN=3\sqrt{10}$$ $$NP=2\sqrt{20}$$ $$PM=10$$ I have to find the sides of the triangle. I have no idea how to solve this exercise, I can't use the formula $$r = \frac{a+b-c}{2}$$ since the triangle isn't right(doesn't have right angle), nor I can prove it. The answer from my textbook is : $$AB = 32$$ $$BC = 16$$ $$CA=24$$

Answer 1

Since $\angle{MNB}=\angle{NMB}=\angle{MPN}$, we have $$BM=BN=\frac{MN/2}{\cos{\angle{MPN}}}$$ You can find $\cos\angle{MPN}$ by the law of cosines. Similarly do for $AM=AP,CP=CN$.

Answer 2

One can even do WITHOUT law of cosines, us the law of sines. Here $r$ is the radius of the in-circle. Check that $\angle MNI= \angle B/2$

$\frac{MN}{sin(180^0-B)}=\frac{r}{sin\frac{B}{2}}$

So it stands that $\frac{MN}{cos(B/2)}=\frac{PM}{cos(A/2)}=\frac{PN}{cos(C/2)}=\frac{PN}{sin((A+B)/2)}=2r$

Now solving is easy..3 equations 2 unknowns..