In a non right triangle P Q R , the median from R meets the sides P Q at S , the

In a non right triangle $PQR$, the median from $R$ meets the sides $PQ$ at $S$, the perpendicular from $P$ meets sides $QR$ at $E$ and $RS$ and $PE$ intersect at $O$. $p=\sqrt{3},q=1$ and circumradius of $PQR$ is $1$.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Antonio Dickerson
Note that $PE$ is a median and hence $PE=3OE$. So $\mathrm{△}PQR=3×\mathrm{△}OQR$.
The question states that $\mathrm{△}PQR$ is not a right angled triangle otherwise $\mathrm{\angle }R={90}^{0}$ is another obvious configuration for the given sides and circumradius.
As $p=\sqrt{3}>q=1,\mathrm{\angle }Q\phantom{\rule{thinmathspace}{0ex}}$ is acute. That leads to only other possible configuration of $\mathrm{\angle }P\phantom{\rule{thinmathspace}{0ex}}$ being obtuse.
If $C$ is the circumcenter, you can see why $\mathrm{\angle }P={120}^{0}\phantom{\rule{thinmathspace}{0ex}}PQ=PR$ and $E$ is the midpoint of $QR$.