Some Triangle

As $\angle A_1A_2A_3=50^{\circ}$ $\implies \angle A_2A_1H_1=40^{\circ}$ because $A_2A_1H_1$ is a right triangle and notice that the quadrilatral $A_1H_2HH_3$ is cyclic.

It is the property of a triangle that $H$ (orthocentre) of the triangle is the incentre of the pedal triangle (triangle formed by joining the feet of the perpendiculars).

Therefore $\angle H_3A_1H=H_3H_2H=40^{\circ}$ as the quadrilateral is cyclic and thus the angles in the same segment are equal.

Now just double the angle to get $\angle H_1H_2H_3=\boxed{80^{\circ}}$