Triangle Calculation Question In a triangle $ABC$, it is known that $ AC = m$, $AB = 3m$, $BC = Rm$. 
Find for which values of $R $ the triangle is: 
(i) An acute angle triangle. 
(ii) A right angle triangle. 
(iii) An obtuse angle triangle. 
I have tried several approaches including the law of cosine and the Pythagorean theorem, but it didn't help. 
 A: 
Hint: first to find when the triangle will be right triangle.($E,F$ is trivial)
Then check $A,B,C$ ,which kind of triangle it will form. the rest is easy.
Edit: following is full answer:

$B$ will on the circle green with radius $3m$, $AC=m<AB-AC=2m<BC<4m=AB+AC$
So if $\triangle ABC$ is right triangle, then $AC$ will be the leg .  to be the right triangle,  the only possible position for B is on $D$ or $E$, $\implies R=\sqrt{10} $ or $ R=\sqrt{8}$ for (ii)
if $B$  on between $DE$ , note the $BH$ is altitude, so $\angle BAC$ and $\angle BCA$ will be  acute angle, $AC$ is the shortest side, so $\angle ABC$ must be acute angle . so in this case , $\triangle ABC$ must be (i).
If$B$ is on $EG$ or $DF$, note $H_1$ or $H_2$ will be outside of $AC$, so $ $\angle BCA$ or $\angle BAC$ must be obtuse angle , that is case (iii)
the last step is to find $R$'s range.
let$AH=x \implies  BC^2= BH^2+HC^2=AB^2-AH^2+(|AH-AC|)^2=Ab^2+AC^2-2AH*AC=10m^2-2mx \implies R^2=10-\dfrac{2x}{m}$
so $R$ is mono decreasing fonction of $x$
$ \sqrt{8} <R <\sqrt{10}$ for case(i)
$R >\sqrt{10}$ or $R<\sqrt{8}$ for case(iii)
A: You can use the law of cosines 9 times, since we are unsure which angle will be the right or obtuse in those cases.  For the sake of simplicity, lets let $AB=c, AC=b, BC=a$.  Choose angle $A$ to be our first case; by the law of Cosines, 
$$a^2=b^2+c^2-2bc\cos{(A)}$$
The three cases are: if $A\lt{90}$ degrees, if $A=90$ degrees, and if $A\gt{90}$ degrees.  
If $A\lt{90}$, then $\cos{(A)}$ is positive which means
$$(Rm)^2=m^2+(3m)^2-2m(3m)\cos{(A)}$$
$$R^2m^2=10m^2-6m^2\cos{(A)}$$
Thus
$R=\sqrt{10-6\cos{A}}.$
If $A=90$ degrees, then $\cos{(A)}=0$ which means 
$$R^2m^2=m^2+9m^2$$
Thus
$R=\sqrt{10}$.
Finally if $A\gt{90}$ degrees, then $\cos{(A)}$ is negative which
$$R^2m^2=m^2+9m^2+2m(3m)\cos{(A)}$$
$$R^2m^2=10m^2+6m^2\cos{(A)}$$
Thus $R=\sqrt{10+6\cos{(A)}}$
Repeat this argument with $b^2=a^2+c^2-2ac\cos{(B)}$ and $c^2=a^2+b^2-2ab\cos{(C)}$.
EDIT:  with the other two cases, you will end up with quadratics in $R$, since the law of cosines with produce an $R^2$ term and an $R$ term.  For example, 
$$m^2=9m^2+R^2m^2-18R\cos{(B)}$$
Now you must solve
$$R^2-18\cos{(B)}R+8=0$$
When $B=90$ degrees it just becomes solving $R^2+8=0$ which produces complex solutions, not what we are after...
