$~ \Delta~ ABC~$ is a triangle with altitudes from $~B~$ and $~C~$ meeting $~AC,~ AB~$ at $~E~$ & $~F~$.

Question

Points $~O~$ and $~M~$ denote circumcenter of $~\Delta ~ABC~$ & midpoint of $~BC~$ respectively. If $~AE=3,~ AF=4,~ A=60^\circ,~$ then find $~OM~$.

The answer is $~\sqrt{\frac{13}{3}}~.$

(I only know upto high school level mathematics)

Answer 1

I assume that you've drawn a picture, otherwise it will be hard for you to follow the solution.

Remember that central angle $\angle BOC$ is twice the inscribed angle $\angle A $, so $\angle BOC = 120^\circ$ and $\angle MOC = \frac{1}{2}\angle BOC = 60^\circ$.

Since $AB = 6$, $AC = 8$ and $\angle A = 60^\circ$, from cosine formula it follows that $BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos 60^\circ = (2 \sqrt{13})^2$and $MC = \frac{1}{2} BC = \sqrt{13}$.

Now we have all required components of $\triangle OMC$.

$OM = \cot (\angle MOC)\cdot MC = \frac{1}{\sqrt{3}} \cdot \sqrt{13}$.