Why is not possible to draw this triangle? Why is it not possible to draw triangle $DEF$ with $EF=5.5cm$,$\angle E=75^0$ and $DE-DF=1.5cm$?(I used this method for consruction-http://gradestack.com/CBSE-Class-9th-Complete/Construction/Construction-of-a/14905-2953-4044-study-wtw)
I could see it follows all triangles inequalities I could make out.So,why is this triangle not possible to draw? 
 A: There are plenty of triangles that can be labeled $\triangle DEF$
with $EF=5.5$ and $DE−DF=1.5$.
They all satisfy the triangle inequalities
$DE + EF \leq DF$, $EF + DF\leq DE$, and $FD + DE \leq EF$.
But every one of those triangles has
less than a $75$-degree angle $\angle DEF$.
The construction to which you referred shows you why no such triangle
can exist. In order for $DE−DF=1.5$, there must be a point $P$ on the line
between $D$ and $E$ such that $DP = DF$ and $EP=1.5$.
The construction therefore tells you to construct the ray $ED$
such that $\angle DEF$ is $75$ degrees,
and construct the desired point $P$ on that ray at distance $1.5$ from $E$.
Do that. Now consider the right triangle that you get by
dropping a perpendicular from $F$ to the line $DE$.
Let the intersection of that perpendicular with $DE$ be the point $Q$
so $\angle EQF$ is the right angle in $\triangle EQF$.
Apply the definition of the cosine to find the length $EQ$:
this says 
$$EQ = 5.5 \cos(75^\circ) = 5.5 \times \frac14(\sqrt6 - \sqrt2)
\approx 1.4235.$$
Notice that this is less than $1.5$, which implies that $Q$ is between $E$
and $P$ which implies that $\triangle EPF$ is acute, which
implies that $\angle FPD$ is obtuse.
But you need $\triangle PDF$ to be an isoceles triangle
(such that $DP = DF$). It cannot have any obtuse angles.
So not only does this construction fail, we know that no construction
of such a triangle can possibly succeed.
More generally, 
the condition for being able to construct a triangle $\triangle ABC$
wtih $BC= a$, $AB-AC = d$, and $\angle ABC = \theta$
is that $AB-AC < a \cos \theta$, because when you put a point $X$ at
distance $d$ from $B$ in the direction of $A$, the internal angle at $X$
in the triangle $\triangle BXC$ must be obtuse
so that the external angle at $X$ can be one of the
base angles of an isoceles triangle.
A: Hint The Law of Cosines gives that
$$|DE|^2 + |EF|^2 - 2 |DE| |EF| \cos E = |DF|^2 .$$
We know that $|EF| = \frac{11}{2}$, $|DE| = |DF| + \frac{3}{2}$, and substituting these values and simplifying gives a linear equation in $|DF|$.

The resulting equation is $$(3 - 11 \cos 75^{\circ}) |DF| = \tfrac{33}{2} \cos 75^{\circ} - \tfrac{65}{2} .$$ The coefficient on the l.h.s. is positive but the constant on the r.h.s. is negative, so the only solution is negative, but $|DF|$ is a length and so cannot be negative.

A: You don't even need to solve this if you are looking for just an answer or an intuitive way. If triangle inequality is satisfied you can always draw a triangle unless there is any additional constraint on the lengths. In this case $DE - DF = 1.5\mathrm{cm}$ is required which poses a problem as proven by few people above. In this case the value cannot cross an upper limit which is less than 1.5cm :)
A: Let $DF=x$.  Then $DE=x+1.5$.  By the cosine rule,
$$x^2=(x+1.5)^2+(5.5)^2-2(5.5)(x+1.5)\cos75^\circ\ .$$
If you expand, the $x^2$ drops out which makes it very easy to find $x$.  However the value of $x$ is negative, which does not make sense for this problem.  So there is no solution.
A: It may help to understand this question if you visualize it. 
Click on the below image to see a larger version of it. 
The thick black line is $EF$,
the blue line is the ray along which point $D$ must lie
(because of the $75^\circ$ angle at $\angle E$),
and the orange curve is the locus of points along which $D$ must lie
to satisfy the $DE-DF=1.5$ constraint. 
This is not a proof (the other answers provide that),
but you can see that the two requirements have no intersection.
                
