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. ∠n
2. ∠p

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

57° + 180° - ∠n + 180° - ∠n + ∠p + 17° = 180° (Angles on a straight line, EN)
57° + 180° + 180° + 17° - ∠n - ∠n + ∠p  = 180°
434° - 2∠n + ∠p = 180°
2∠n - ∠p = 434° - 180° 
2∠n - ∠p = 254°

∠p = 2∠n - 254° --- (1)

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

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

∠p = 4∠n - 540° --- (2)

(2) = (1)
4∠n - 540° = 2∠n - 254°
4∠n - 2∠n= 540° - 254°
2∠n = 286°

∠n
= 286° ÷ 2
= 143°

(b)
From (1)
∠p
=  2∠n - 254°
= 286° - 254° 
= 32°

Answer(s): (a) 143°; (b) 32°