If $a^2+b^2 = c^2 + \sqrt{3}$, find value of $\angle C$. 
In a triangle with sides $a$, $b$, $c$, consider this relationship: 
  $$a^2+b^2 = c^2 +\sqrt 3$$
  Now find value of $\measuredangle C$ .

My try: I tried to use Laws of Cosines and Law of Sinee, but I didn't get any result.
 A: All that can really be concluded about $\angle C$ is, as others have said, that $$\cos(C) = \frac{\sqrt 3}{2ab} \tag{1}$$ 
and, as others have not said, that $$\angle C \lt 90^\circ \tag{2}$$
We can conclude statement $2$ by seeing that the right hand side of equation $1$ is always positive. And the cosine function is only positive for angles in the intervals $0^\circ \le \angle C \lt 90^\circ$ and $270^\circ \lt \angle C \le 360^\circ$. As no angle in a triangle is greater than $180^\circ$, the conclusion follows.
A: Applying the comments mentioned, we have $ab \cos \measuredangle C = \dfrac {\sqrt 3}{2}$.
Let us consider the special case for a = b = 1 first. We build the red-dotted unit circle centered at C with radius = CB = CA = 1. Then $\measuredangle C = … = 30^0$.

If we don’t like the above isosceles triangle solution, we can build another one, which is A’B’C (the green triangle). See below.
We assume that $\alpha = A’C \lt B’C = \beta$. (To avoid confusion, I let the legs of angle C be $\alpha$ and $\beta$.)
We construct the black circle (centered at C with radius $ = \alpha$. Locate A’’ on CB. We further located B’, which is the inversion point of A’’ about the red-dotted circle such that $\alpha \times \beta = 1$. Or equivalently, $A’’B’ = \beta - \alpha$.
By varying A’, we can obtain infinitely many triangles meeting the requirement. Another infinite set of solution can be obtained if we let $\beta \lt \alpha$. However, I cannot exhaust other possibilities that $\angle C =$ an angle with other sizes. 
A: Note: In what follows I am interpreting $a^2+b^2 = c^2 +\sqrt 3$ as meaning that $c^2$ is larger than what would be obtained using the Pythagorean theorem by $\sqrt 3$. 
Since you must have $$a^2+b^2 = c^2 +\sqrt 3$$ the law of cosines states that in this case $$- 2\ a\ b\ cos (\measuredangle C) = \sqrt 3$$ Therefore the angle is determined by the formula $$\measuredangle C = \arccos(-\frac{\sqrt 3}{2\ a\ b})$$.
Note: if the interpretation should have been that $a^2 + b^2$ is smaller than $c^2$ (as obtained using the Pythagorean theorem) by $\sqrt 3$ then the above equation is the same but without the minus sign.
$\measuredangle C$ varies with the values of $a$ and $b$ therefore there are an infinite number of values for $\measuredangle C$.  For example, in the figure below (showing both possible interpretations)

when $a$ and $b$ are both 1, then $\measuredangle C$ is $150 ^{\circ}$.  When $a$ and $b$ are both $2$, then $\measuredangle C$ is $102.5^{\circ}$.  In both cases $a^2 + b^2 = c^2 + \sqrt 3$ as required.  Given any values of $a$ and $b$, the value of $\measuredangle C$ can be calculated to make $a^2 + b^2 = c^2 + \sqrt 3 $.
A: We have by simple triangle trigonometry 
$$ a\,b \cos C= \sqrt 3/2 $$
which is the scalar dot product of vector forces $ (A,B)$ with angle  $C$ included between them, This fixed/given amount of mechanical work can be performed in infinitely many ways with variabilities among the three available degrees of freedom, one for each parameter $a,b,C.$  It can have no unique solution to determine angle $C$ .. as required by the question. Often the underlying physics can be exploited to gain clarity.
