how to find values of x for which a triangle exists Given a $\triangle PQR , QR=3 , PR= x , PQ=2x$  and $\angle PQR= \theta$  calculate the values of $x$ for which the triangle exists.  I don't even know where to start.
 A: By the triangle inequality, the sum of any possible combination of two sides must be strictly greater than or equal to the third side (for a non degenerate triangle).
So:
$2x + 3 > x \implies x > -3$ (holds trivially, ignore).
$x+3 > 2x \implies x < 3$
$x+2x > 3 \implies x > 1$
So $1 < x < 3$ is the required range.
A: Hint: triangle inequality (what is the relation between the sum of lengths of any two sides and the length of the remaining side?)
A: The answer given by @Deepak is undoubtly the shortest possible and the one which is certainly awaited at your level.
Nevertheless, here is an answer that gives another proof for the result: 
$$1<x<3$$
in the framework of a "global understanding" (see figure below). 
It involves a classical result : the locus of points $P$ such that the ratio of their distances to two fixed points is constant (here $\dfrac{PQ}{PR}=2=\dfrac{2x}{x}$) is a circle (https://www.cut-the-knot.org/Curriculum/Geometry/LocusCircle.shtml)
(see commentary below) 
(more generally, constant 2 can be any positive constant).
On the figure, you see a triangle $PQR$ where $RQ$ is fixed and $P$ can be any point on the circle.
Now take the extreme points :


*

*$P$ is in $P_1$ iff $x=1$,

*$P$ is in $P_2$ iff $x=3$,
all values of $x$ in the range $[1,3]$ being allowed, with the restriction that these extreme points give flat triangles...
Commentary: this circle has its center on line $RQ$, with diameter $P_1P_2$ where $P_1$ and $P_2$ are the two extreme points on line $RQ$ mentionned above.

