The Integral value of $\lfloor\frac{a}{30}\rfloor$

Question

The altitudes from the angular points $A$, $B$ and $C$ on the opposite sides $BC$, $CA$ and $AB$ of $ABC$ are $210$, $195$ and $182$, respectively. Then what is the the value of $\left\lfloor\frac{a}{30}\right\rfloor$.

Answer 1

For a $\triangle\text{ABC}$:

$$ \begin{cases} \left|\text{A}\right|^2=\left|\text{B}\right|^2+\left|\text{C}\right|^2-2\left|\text{B}\right|\left|\text{C}\right|\cos\angle\alpha\\ \\ \left|\text{B}\right|^2=\left|\text{A}\right|^2+\left|\text{C}\right|^2-2\left|\text{A}\right|\left|\text{C}\right|\cos\angle\beta\\ \\ \left|\text{C}\right|^2=\left|\text{A}\right|^2+\left|\text{B}\right|^2-2\left|\text{A}\right|\left|\text{B}\right|\cos\angle\gamma \end{cases}\tag1 $$

By setting $\left|\text{A}\right|=210$, $\left|\text{B}\right|=195$ and $\left|\text{C}\right|=182$, we get:

$$ \begin{cases} 210^2=195^2+182^2-2\cdot195\cdot182\cdot\cos\angle\alpha\\ \\ 195^2=210^2+182^2-2\cdot210\cdot182\cdot\cos\angle\beta\\ \\ 182^2=210^2+195^2-2\cdot210\cdot195\cdot\cos\angle\gamma \end{cases}\tag2 $$

So:

$$ \begin{cases} \angle\alpha=\arccos\left(\frac{27049}{70980}\right)\\ \\ \angle\beta=\arccos\left(\frac{39199}{76440}\right)\\ \\ \angle\gamma=\arccos\left(\frac{49001}{81900}\right) \end{cases}\tag3 $$