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 $\left\lfloor\frac{a}{30}\right\rfloor$.
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This looks like a homework problem. What have you tried? – Dashi Mar 11 '20 at 19:57
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I know that side is inversely proportional to altitude and proportionality constant is twice the triangle area. – user483801 Mar 11 '20 at 20:01
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Is $a$ the length of the side opposite angle A or the area? – marty cohen Mar 11 '20 at 20:02
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yes. It is side length opposite to vertex A. – user483801 Mar 11 '20 at 20:03
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Have you drawn a picture? I found two right triangles with the hypotenuse each side of the triangle. You know one leg of each. Making variables for the pieces of each side and writing the Pythagorean equations looks like it will work. – Ross Millikan Mar 12 '20 at 00:36
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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:
$$ \begin{cases} \angle\alpha=\arccos\left(\frac{27049}{70980}\right)\\ \\ \angle\beta=\arccos\left(\frac{39199}{76440}\right)\\ \\ \angle\gamma=\arccos\left(\frac{49001}{81900}\right) \end{cases}\tag3 $$
Jan Eerland
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