<s>[INST] 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. 
 [/INST] 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...
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