# What's the measure of the $\angle PBN$ in the question below?

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...

$$D$$ is circumcenter

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

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

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