Let’s write the measurement of the angles of ΔEHG in the figure beside.

Given: AB||CD, FE as the transversal line, ∠AFH = 110°, ∠HGE = 60°

To find the angles of ΔEHG, i.e., ∠GHE and ∠GEH

Now in the given figure, PQ is parallel to TS, so

∠AFH = ∠FHG as they form is alternate interior angles.

So ∠FHG = 110°….(i)

Now FHE is a straight line, so

∠FHE = 180°

⇒∠FHG + ∠GHE = 180°

Now substituting the value from equation (i), we get

⇒110° + ∠GHE = 180°

⇒∠GHE = 180° - 110°

⇒∠GHE = 70°……..(ii)

Now consider ΔEHG,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠GHE + ∠GEH + ∠HGE = 180°

Substituting the values from given criteria and equation(ii), we get

70° + ∠GEH + 60° = 180°

⇒ ∠GEH = 180° - 60° - 70°

⇒ ∠GEH = 50°

So the measurement of the unknown angle of the given triangle EHG are 50° and 70°.