Answer:
Let us mark the points R and S on line p, T and U on line q, A and B on line l and C and D on line m.
Suppose the lines p and q intersect the line l at K and L respectively and line m at N and M respectively.
Since, AB is a straight line and ray KN stands on it, then
∠NKL + ∠NKA = 180∘ (Angles in a linear pair)
⇒110∘ + a = 180∘
⇒a = 180∘ − 110∘ = 70∘
Since, line p || line q and line l is a transversal intersecting them at K and L, then
∠TLB = ∠NKA (Alternate exterior angles)
⇒b = a
⇒b = 70∘
Since, line p || line q and line m is a transversal intersecting them at N and M, then
∠CNK = ∠NML (Corresponding angles)
⇒c = 115∘
Since, CD is a straight line and ray ML stands on it, then
∠NML + ∠LMD = 180∘ (Angles in a linear pair)
⇒115∘ + d = 180∘
⇒d = 180∘ − 115∘ = 65∘
