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