Question on: JAMB Mathematics - 2002

In Δ PQS, ∠PSQ = X(base ∠s of isoc Δ PQS)

In Δ QRS, ∠RQS = ∠PSQ + X(Extr ∠ = sum of two intr. opp ∠s)

∴ ∠RQS = X + X

= 2X

Also ∠QRS = 2X(base ∠s of isoc Δ QRS in Δ PRS,

72 = ∠RPS + ∠PRS(Extr, ∠ = sum of two intr. opp ∠s)

∴ 72 = x + 2x

72 = 3x

x = ⁷²/₃

x = 24^o