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?
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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.
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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 :)
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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.
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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.
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$\DeclareMathOperator{\d}{d}$ And now for a completely different approach altogether. Not necessarily the recommended one, but it has some interesting ideas in it.
First, those numbers really are a smidgeon unpleasant. Please allow me the luxury of working in half-centimetres rather than centimetres, so I get some nice wholesome integers to work with. That way $EF = 11$ half-centimetres and $DE - DF = 3$ half-centimetres. (Really I'm just scaling the lengths and keeping the angles the same. Think similarity.) Now let's consider $E$ fixed at $(0,0)$ and $F$ fixed at $(11,0)$. The locus of all points $D$ such that $\angle DEF = 75^{\circ}$ is just a ray through the origin; its Cartesian equation is $y = (\tan 75^{\circ}) x$. We shall find the locus of points that satisfy $DE - DF = 3$, and see if the two loci intersect. If they do, then any point of intersection satisfies both conditions and we can mark point $D$ of the triangle. If they do not, then there is no way for $D$ to satisfy both conditions simultaneously, and the triangle is impossible to construct.
What might the locus $\d(D,E) - \d(D,F) = 3$ look like? Well, Wikipedia has a nice article on the Cartesian oval, which is the plane curve traced out by all points $S$ with the same linear combination of distances from foci $P$ and $Q$:
$$\d(P,S) + m \d(Q,S) = a$$
So we're in business. Moreover, there's a list of special cases including where $m=-1$, which gives us a hyperbola. You might already have known it would be a hyperbola if you'd learned about conic sections before. Let's find its equation, by letting $E=(x,y)$ and using the Pythagorean distance formula to find $DE$ and $EF$:
$$\sqrt{x^2 + y^2} - \sqrt{(x-11)^2 + y^2} = 3$$
What an almighty mess! To render it more user-friendly, we can follow the steps laid out on this Maths SE thread about Cartesian ovals. First, move the square roots to opposite sides, and square.
$$x^2 + y^2 = \left(3 + \sqrt{(x-11)^2 + y^2} \right)^2 = x^2 + y^2 - 22x + 130 + 6\sqrt{(x-11)^2 + y^2}$$
Then simplify (e.g. we can cancel the quadratic terms and divide by two), isolate the square root on one side, then square again. Et voilà, no more radicals to be seen.
$$ \left(11x - 65\right)^2= \left(3\sqrt{(x-11)^2 + y^2}\right)^2 = 9 \left( (x-11)^2 + y^2 \right)$$
Skipping merrily through some algebra, we obtain something we can actually graph:
$$ 9y^2 = 112x^2 - 1232x + 3136 $$
The blue line is the $75^{\circ}$ ray (the angle does not look that size, but that is only because of the disparity in vertical and horizontal scale) and the red curve is the hyperbola. It seems they do intersect after all, near $(2.5,9.3)$, yet this is clearly closer to $E$ than $F$! What has happened?
Unfortunately, squaring the equation makes it "forget" whether we originally had $DE - DF = 3$ or, instead, $DF - DE = 3$. The hyperbola has two branches: the left branch corresponds to $D$ being three units closer to $E$ than $F$ (which is possible, hence the point of intersection) while the right branch corresponds to $D$ being three units closer to $F$ than $E$ (in which case it is apparently impossible to satisfy $\angle DEF = 75^{\circ}$, because the graph suggests the ray misses this branch of the hyperbola entirely). You might have noticed the branches have $x$-intercepts at $(4,0)$ and $(7,0)$ — the former being seven units from $F$ and only four from $E$ (three units closer!), the latter vice versa.
If we extend the ray down into the third quadrant, where $x$ and $y$ are both negative, we see it will intersect the (left branch of) the hyperbola again, about $(-94.7, -353.6)$. This isn't a solution to the original problem: not only would $D$ be three units closer to $E$ than $F$ whereas we seek the reverse, but $\angle DEF = 105^{\circ}$ (the supplement of $75^{\circ}$) because the ray is in the wrong quadrant. So why should we care about this point? Well, the existence of two distinct points of intersection on the left branch of the hyperbola proves that there can be no point of intersection on the right branch. A line can only intersect a hyperbola at most twice — a general result true for all conic sections, except for the degenerate ones. Solving their equations simultaneously would give rise to a quadratic, which can have at most two solutions. This is sufficient to show that the desired construction is impossible.
If you want to have a go at procuring the numerical results for these points of intersection for yourself, it may help you to know that $\tan 75^{\circ} = 2+\sqrt{3}$, so the line has equation $y=(2+\sqrt{3})x$. But rather than do that, I shall investigate for which angles $\theta = \angle DEF$ the construction is possible. We can do this algebraically by checking whether the line $y=mx$ intersects the right branch, where $m=\tan \theta$. It's clear from the graph that any ray through the origin will intersect the left branch, so for any desired $\angle DEF$ we can construct a triangle where $D$ is three units closer to $E$ than $F$. The question is whether the equations will yield a second solution, and if so, on which branch? Substituting $y=mx$ into the equation of the hyperbola gives us
$$ 9m^2x^2 = 112x^2 - 1232x + 3136$$
This is quadratic with coefficients $a=112-9m^2$, $b=-1232$ and $c=3136$. Note that $b$ is fixed and negative, $c$ is fixed and positive, whereas by suitable choice of $m$ we can attain any value $a \le 112$, positive or negative. Imagine running through the angles $\theta = 0^{\circ}$ to $90^{\circ}$: this means running through gradients $m = \tan \theta$ from zero to positive infinity, and hence $a$ runs through all values from $112$ to negative infinity. Beware this means passing through $a=0$. This creates difficulties with the standard quadratic formula,
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
When $a$ is near zero, both numerator (when the "plus" is taken at the $\pm$) and denominator are near zero; when $a=0$ we obtain a fraction of the form $\frac{0}{0}$. This makes careful analysis of the roots quite difficult. It would be much more convenient if $a$ only appeared in the equation once, and hence affected only one of the numerator and denominator. Fortunately, there is an alternative quadratic formula sometimes used (e.g. in Muller's method) for its numerical properties:
$$x=\frac{-2c}{b\pm\sqrt{b^2-4ac}}$$
Let's first consider taking the "minus" at the $\pm$. Remembering that $c>0$ and $b<0$, we see the denominator is negative (it is the sum of two negative terms), as is the numerator, so this root is always positive. When $\theta = 0^{\circ}$ and $a=112$, we find the root is
$$x = \frac{-2(3136)}{-1232 - \sqrt{(-1232)^2-4(112)(3136)}} = 4$$
As we swing our line up through larger $\theta$, the falling value of $a$ makes the denominator even more negative. The $x$ coordinate of our point of intersection remains positive, but smoothly decreases so the point moves to the left. As our line nears the vertical, and $\theta$ approaches a right angle, then $a$ and hence the entire denominator approach negative infinity, $x$ approaches zero, and our point of intersection nears the $y$-axis. Looking at the graph, it's clear we've traced a path along the left branch of the hyperbola. For each angle, we've found a point three units closer to $E$ than $F$.
Now switch our attention to the root which takes the "plus" at the $\pm$. Starting at $\theta = 0^{\circ}$, we can again substitute $a=112$ to find $x=7$. With an angle of zero, we must be at $(7,0)$, so you can see we are initially on the right branch of the hyperbola, three units closer to $F$ than $E$. As we pivot our line up through greater $\theta$, we must pay careful consideration to signs. Since $c>0$ and $b<0$, the numerator is clearly negative, but the denominator is the sum of positive and negative terms. Its sign will be positive if and only if $\sqrt{b^2-4ac}$ is greater than the magnitude of $b$.
To start with, this is clearly not the case. But as $a$ declines, the value of $\sqrt{b^2-4ac}$ rises, the denominator becomes less negative (i.e. closer to zero), and the $x$ increases so our point of intersection moves to the right. When $a$ is only just above zero, the $b + \sqrt{b^2-4ac}$ is only just below zero, and $x$ will move very far to the right indeed: in fact, as $a \to 0^+$ we have $x \to +\infty$. But when we reach the critical angle at which $a$ is exactly zero, then $x = \frac{-2c}{0}$ yields no root. What's going on here is that, with $a=0$, our "quadratic" equation is really linear, and only has one root. It's not here, and that's because we found it already: it's sitting happily on the left branch of the hyperbola, where we chose "minus" at the $\pm$ and the special case $a=0$ caused no particular problems. (If we had used the standard quadratic formula, putting $a=0$ gives indeterminate forms $0/0$ in both $\pm$ cases, obfuscating which root has really been "lost". Using the alternative formula avoids this: its drawback is that analogous trouble would break out if $c=0$, but that's not a problem for us here.)
A negligible increase in $\theta$ will tip $a$ into negative territory, and $\sqrt{b^2-4ac}>\mid b \mid$ (albeit only just). A tiny positive denominator ensures that $x$ is large and negative so we must have switched over to the third quadrant (for $0^{\circ}<\theta<90^{\circ}$ then $m=\tan \theta >0$ so $x<0$ implies $y<0$) and the left branch of the hyperbola. As $\theta$ approaches a right angle, $a$ monotonically approaches negative infinity and the denominator monotonically approaches positive infinity: this means $x$ approaches zero from the left, and our point of intersection is smoothly moving right and nearing the $y$-axis. There is no way for it to return to the right branch of the hyperbola after its jump to the left. For angles above the critical angle, the construction is impossible.
In conclusion, for any desired angle $0^{\circ} < \angle DEF < 90^{\circ}$ we can construct a point $D$ such that $D$ lies three units nearer to $E$ than to $F$. But if we want $D$ three units closer to $F$ than to $E$, this can only be achieved below a critical angle for which $a=112-9(\tan \theta)^2=0$. This gives $\tan \theta = \frac{4\sqrt{7}}{3}$ and $\theta \approx 74.17^{\circ}$. If you are trying to see how this ties into other answers, it's interesting that this angle has a cosine of $\frac{3}{11}$. You might want to consider which constructions are possible for an obtuse $\angle DEF$: the hyperbolas are still valid, you just need to consider the intersections with the appropriate rays.
A slightly more general method: consider a hyperbola with foci at $(-c, 0)$ and $(c,0)$ hence centred at $(0,0)$. Consider the locus of a point whose distances from the foci differ by $2a$, then the Cartesian equation of the hyperbola is:
$$\sqrt{(x-c)^2 + y^2} - \sqrt{(x+c)^2 + y^2} = \pm 2a$$
Moving the square roots to opposite sides and squaring, we obtain:
$$(x-c)^2 + y^2 = 4a^2 + (x+c)^2 + y^2 \pm 4a \sqrt{(x+c)^2 + y^2}$$
Expanding the brackets, simplifying, and making the term with square root the subject,
$$\mp a \sqrt{(x+c)^2 + y^2} = cx + a^2$$
Squaring and simplifying, we obtain
$$a^2 y^2 + (a^2 - c^2) x^2 = a^4 - a^2 c^2$$
Which can be put into the more familiar form for a hyperbola,
$$\frac{x^2}{a^2} - \frac{y^2}{c^2 - a^2} = 1$$
Note that the asymptotes of the hyperbola are at $y = \pm \frac{b}{a} x$ where $b^2 = c^2 - a^2$ is the denominator of the $y^2$ fraction. If we want change the origin to be the focus on the left, we just translate right by $c$ so that the hyperbola has centre $(c,0)$:
$$\frac{(x-c)^2}{a^2} - \frac{y^2}{c^2 - a^2} = 1$$
The asymptotes still have slope $b/a$ but go through $(c,0)$, so have equation $y = \pm \frac{b}{a} (x - c)$. The equation of the hyperbola in your case can be rearranged into the desired form, or we can procure it by using $c=5.5$ and $a=1.5$ (since we demanded foci a distance $2c = EF = 11$ apart, and the distances from the foci had do differ by $2a = 3$) in the general formula given:
$$\frac{(x-5.5)^2}{2.25} - \frac{y^2}{28} = 1$$
The slope of the upwards asymptote is $\frac{b}{a} = \sqrt{\frac{28}{2.25}} = \frac{4\sqrt{7}}{3}$, which we previously identified as the critical slope. The key idea here is that the right branch of the hyperbola is entirely to the right of the asymptotes through $(c,0)=(5.5,0)$, although it does approach close to them. If the ray from the origin is as steep or steeper than the upwards asymptote, which will occur when the $\tan \angle DEF$ is more than or equal to the gradient of the asymptote, then the ray can never cross the asymptote and hence never intersects the hyperbola: the triangle can not be constructed. If it is less steep, then it is bound to cross the asymptote, and this means that it will intersect the hyperbola and the triangle can be constructed.
The final argument, that the ray crossing the asymptote guarantees it intersects the hyperbola, can be proven either using the algebraic root-finding approach outlined above, or an argument from analysis: to the right of the point of intersection, the ray lies below the asymptote — by an ever-increasing distance the further right you go — yet the hyperbola is a continuous curve that lies below the asymptote and must become arbitrarily close to it. The height of the ray above the hyperbola must be positive where it crosses the asymptote, but must eventually become negative, so by the Intermediate Value Theorem there must be a point where that height is zero.
More generally, the critical slope is $\frac{b}{a}$ and so the critical angle satisfies $\tan^2 \theta = \frac{b^2}{a^2}$. To put this in terms of the original $a$ and $c$, we can write $\tan^2 \theta = \frac{c^2 - a^2}{a^2} = \frac{c^2}{a^2} - 1$. Since $\sec^2 \theta = \tan^2 \theta + 1$ we find that $\sec^2 \theta = \frac{c^2}{a^2}$, and so
$$\cos \theta = \frac{a}{c} = \frac{2a}{2c} $$
So the cosine of the critical angle is simply the ratio between the difference in distances from $D$ to the foci, and the distance between the two foci. This is why, in our case, we simply obtained $\cos \theta = \frac{3}{11}$.
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.