In the figure, PFGL is a trapezium and triangles PLK and KPH are isosceles triangles. NE, NG and EG are straight lines. PL = PK = KH. Find

1. ∠r
2. ∠s

(a)
∠FPH = 180° - ∠r (Interior angles, PF//NG)
∠KHP = 180° - ∠r (Angles on a straight line)
∠KPH = 180° - ∠r (Isosceles triangle)

56° + 180° - ∠r + 180° - ∠r + ∠s + 17° = 180° (Angles on a straight line, EN)
56° + 180° + 180° + 17° - ∠r - ∠r + ∠s  = 180°
433° - 2∠r + ∠s = 180°
2∠r - ∠s = 433° - 180° 
2∠r - ∠s = 253°

∠s = 2∠r - 253° --- (1)

∠PKL
= ∠PLK
= 2 x (180° - ∠r)
= 360° - 2∠r (Exterior angle of a triangle)

∠s = 180° - (360° - 2∠r) - (360° - 2∠r)
∠s = 180° - 360° + 2∠r - 360° + 2∠r 
∠s = 180° - 360° - 360° + 2∠r + 2∠r
∠s = 4∠r - 540° (Angles sum of triangle) 

∠s = 4∠r - 540° --- (2)

(2) = (1)
4∠r - 540° = 2∠r - 253°
4∠r - 2∠r= 540° - 253°
2∠r = 287°

∠r
= 287° ÷ 2
= 143.5°

(b)
From (1)
∠s
=  2∠r - 253°
= 287° - 253° 
= 34°

Answer(s): (a) 143.5°; (b) 34°