<s>[INST] Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$ 

Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.


My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{aligned}
& \cos \left(50^{\circ}\right)<\cos \left(45^{\circ}\right) \\
\Rightarrow \quad & 2 \cos \left(50^{\circ}\right)<\sqrt{2} \\
\Rightarrow \quad & x<\sqrt{3}+\sqrt{2}<3.14
\end{aligned}$$
Now I am struggling hard to prove that $x>3$
 [/INST] It is known that $\cos(50^o)=\frac{\phi}2$. So, we need to find the floor of $\phi+\sqrt{3}$. $\sqrt{3}$ is less than $2$ since it is less than $\sqrt{4}$. $\phi$ is less than $2$ since $(1+\sqrt{5})/2<(1+\sqrt{9})/2=2$. We could do a similar proof that $\sqrt{3},\phi>1$. So, the answer must be between $\lfloor4\rfloor$ and $\lfloor2\rfloor$. The answer is thus $3$.
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