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For reference (exact copy of question): In the acute triangle $ABC$, the heights AH and BN are plotted. Extended H intersects the circumcircle at $P$. Calculate $\angle PBN$ if $\angle AON=30^\circ.$ $O$ is the orthocenter of triangle $ABC$. (answer $120^\circ$)

My progress: Here is the picture I made and the relationships found. I drew some auxiliary lines...

enter image description here

$D$ is circumcenter

$ABPC$ is cyclic $\implies \angle PAB = \angle PCB\\ \angle BDA = \angle NAP=60^o=\angle NBC$

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  • $\begingroup$ $\angle OAN=60^{\circ}.$ So $ABC$ cannot be equilateral. $\endgroup$
    – Sathvik
    Oct 6, 2021 at 14:34
  • $\begingroup$ @SathvikAcharya..thanks for alert... $\endgroup$
    – peta arantes
    Oct 6, 2021 at 15:06

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$$\angle HBO=90^{\circ}-\angle BOH=90^{\circ}-\angle AON=60^{\circ}$$ Since $ABPC$ is cyclic, $$\angle PBC=\angle PAC=90^{\circ}-\angle AON=60^{\circ}$$ Therefore, $$\angle PBN=\angle PBC+\angle CBN=\boxed{120^{\circ}}$$

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