<s>[INST] 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.
 [/INST] 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 $.
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