In the figure above PQR is a straight line segm... - JAMB Mathematics 2001 Question

In Δ PQT,
∠PTQ = 25^o(base ∠s of isosceles Δ)
In Δ QSR,
∠RQS = ∠QPT + ∠QTP
(Extr = sum of interior opposite ∠s)
∠RQS = 25 + 25
= 50^o
Also in Δ QSR,
75 + ∠RQS + ∠QSR = 180^o
(sum of ∠s of Δ)
∴75 + 50 + ∠QSR = 180
125 + ∠QSR = 180
∠QSR = 180 - 125
∠QSR = 55^o
But ∠QSR and ∠RST are the same
∠RST = 55^o