If three sides of an acute triangle is $3$ cm, $4$ cm, and $x$ cm. What are the possible values of $x$? Problem: If three sides of an acute triangle is 3 cm, 4 cm, and $x$ cm. What are the possible values of $x$?
2 of the choices: (A) $1<x<5$, and (B) $0<x<7$
Solution: According to triangle theorems, 2 sides of a triangle is greater than the 3rd side and their difference is less than the 3rd side. 
From one of the choices, (ranges $0<x<7$) when $x = 6$:


*

*$3 + 4 > 6$

*$6 + 4 > 3$

*$3 + 6 > 4$
From the above, it means $6$ is a possible value for $x$, and since choice A doesn't have $6$ in it, I chose B as the correct answer. However that isn't the case, the correct answer according to the book is A, which I don't know why. So how was it $1<x<5$? Any help would be appreciated. 
 A: Let the angles opposite the sides of lengths $x,3,4$ be $\theta,\omega, \phi$ respectively. Then by the Cosine Rule we obtain the relations: $$x^2=3^2+4^2-2\cdot3\cdot4\cos \theta,$$ $$3^2=x^2+4^2-2\cdot x\cdot4\cos \omega$$ and $$4^2=x^2+3^2-2\cdot x\cdot3\cos \phi$$ for $0<\cos \theta, \cos \omega,\cos \phi<1$.
From the first relation we have that $x^2=25-24\cos\theta$, which tells us that the extremes of $x^2$ are $25-24\times 1=1$ and $25-24\times 0=25$, so that $1<x^2<25$, or $1<x<5$.
From the second relation we deduce that $x=4\cos\omega\pm\sqrt{16\cos^2\omega-7}$, from which we learn that $x=\sqrt{-7}$ when $\cos\omega=0$ and $x=1,7$ when $\cos\omega=1$. Any way we look at these, they tell us nothing interesting.
From the third relation we have that $x=3\cos\phi\pm\sqrt{9\cos^2\phi+7}$, which tells us that $x=\sqrt7$ when $\cos\phi=0$ and $x=-1,7$ when $\cos\phi=1$. We have the possible interesting relations $\sqrt7<x<7$.
Finally, combining the inequalities $1<x<5$ and $\sqrt 7<x<7$, we have the tighter bounds $\sqrt 7<x<5$.
PS. I have tacitly assumed throughout that the sinusoidal functions $x(\theta),x(\omega),x(\phi)$ which we analysed are either strictly increasing or strictly decreasing (indeed they're all strictly decreasing) wherever they are defined in the intervals $0<\theta,\omega,\phi<π/2$ respectively. This can however be easily justified if needed.
A: If $a,b,c$ are all lengths of sides of a (non-degenerate) triangle, the triangle inequality implies the following conditions:
$$1.~~~~~a+b>c$$
$$2.~~~~~a+c>b$$
$$3.~~~~~b+c>a$$
Further, if $a,b,c$ are all lengths of sides of an acute triangle, we also learn the following conditions:
$$4.~~~~~a^2+b^2>c^2$$
$$5.~~~~~a^2+c^2>b^2$$
$$6.~~~~~b^2+c^2>a^2$$

Letting $a=3,b=4,c=x$, plugging these values into inequalities 1,2,4 and 5 respectively we transform the above into the following:
$$3+4>x$$
$$3+x>4$$
$$3^2+4^2>x^2$$
$$3^2+x^2>4^2$$
Simplifying arithmetic and moving $x$ to a single side this simplifies further as:
$$7>x$$
$$x>1$$
$$25>x^2$$
$$x^2>7$$
Remembering that $x$ should be positive we can take square roots of each side of the third and fourth inequalities here and recognizing these conditions as being more restrictive than the first two inequalities we learn that
$$5>x$$
$$x>\sqrt{7}\approx 2.64$$
We have therefore the following as the maximal restrictions we can place on the value of $x$ being:
$$\sqrt{7}<x<5$$
Indeed, one can make an acute triangle with sidelengths $3,4,x$ for every value of $x$ from the set $\{x~:~\sqrt{7}<x<5\}$ and no value of $x$ outside of that set makes a valid acute triangle with sidelengths $3,4,x$ either because it would be a right triangle, an obtuse triangle, or not a triangle at all.  For example a 2-3-4 triangle is obtuse.
A: If there are only two choices given for the solution it might be easier to prove one of the solutions to be incorrect rather than the other one to be correct. x must be greater than 4 - 3 = 1, therefore the second solution is wrong and the first solution must be the correct one by default!
