<s>[INST] equations of sides of triangles; find largest angle If the sides of a triangle are $2x+3$, $x^2 + 3x + 3$, and $x^2 + 2x$, find the greatest interior angle of a triangle.
The answer is $120$ degrees.
I was hoping to find a formula to relate all the three sides, then use cosine law to find an angle. If I use the sine law, I don't have an angle.... If I use the cosine law, I also don't have an angle (and you also have to square terms; takes more time?). Pythagorean doesn't assume a right triangle. 
Do I assume a value for $x$? How so? I am totally lost. Any hint?
 [/INST] Use cosine rule. It's really quite trivial when you keep the algebra simple.
The longest side is $x^2 + 3x + 3$, so the greatest angle lies opposite this. To show that that's the longest side, you simply need to sketch the curves for $x>0$, which is the region that matters, since for $x \leq 0$, the side of length $x^2 + 2x$ becomes zero or negative, which is impossible.
Apply the cosine rule:
$(x^2 + 3x + 3)^2 = (2x+3)^2 + (x^2 + 2x)^2 - 2(2x+3)(x^2+2x)\cos\theta$
When simplifying, avoid opening up the brackets immediately. Use identities like $a^2 - b^2 = (a+b)(a-b)$ to keep your life simple.
Working in this careful fashion, we get:
$(x+3)(2x^2 + 5x + 3) - (2x+3)^2= -2(2x+3)(x^2 + 2x)\cos \theta$
$(x+3)(2x +3)(x+1) - (2x+3)^2= -2(2x+3)(x^2 + 2x)\cos \theta$
It is permitted to cancel $2x+3$ from both sides as this cannot be zero (since $x>0$).
So:
$(x+3)(x+1) - 2x - 3 = -2(x^2 + 2x)\cos\theta$
$x^2 + 4x + 3 - 2x - 3 = -2(x^2 + 2x)\cos\theta$
$(x^2 + 2x) = -2(x^2 + 2x)\cos\theta$
giving $\cos \theta = -\frac 12$ and $\theta = 120^{\circ}$
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