In ∆ACL, AC = CL
So, ∆ACL is an isosceles triangle.
In isosceles triangle, the angles which are opposite to equal sides are also equal.
So, ∠A = ∠L = x° ∠C = 56° (Vertically opposite angle)
In ∆ACL, we know ∠A + ∠C + ∠L = 180°
⇒ x + 56° + x = 180°
⇒ 2x + 56° – 56° = 180° – 56°
⇒ 2x =124°
⇒ 2x/2 = 124°/2
∴ x = 62°
Exterior angle at A = ∠C + ∠L
⇒ y = 56 + 62
⇒ y = 118°