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
- ∠d
- ∠e
(a)
∠FPH = 180° - ∠d (Interior angles, PF//NG)
∠KHP = 180° - ∠d (Angles on a straight line)
∠KPH = 180° - ∠d (Isosceles triangle)
54° + 180° - ∠d + 180° - ∠d + ∠e + 21° = 180° (Angles on a straight line, EN)
54° + 180° + 180° + 21° - ∠d - ∠d + ∠e = 180°
435° - 2∠d + ∠e = 180°
2∠d - ∠e = 435° - 180°
2∠d - ∠e = 255°
∠e = 2∠d - 255° --- (1)
∠PKL
= ∠PLK
= 2 x (180° - ∠d)
= 360° - 2∠d (Exterior angle of a triangle)
∠e = 180° - (360° - 2∠d) - (360° - 2∠d)
∠e = 180° - 360° + 2∠d - 360° + 2∠d
∠e = 180° - 360° - 360° + 2∠d + 2∠d
∠e = 4∠d - 540° (Angles sum of triangle)
∠e = 4∠d - 540° --- (2)
(2) = (1)
4∠d - 540° = 2∠d - 255°
4∠d - 2∠d= 540° - 255°
2∠d = 285°
∠d
= 285° ÷ 2
= 142.5°
(b)
From (1)
∠e
= 2∠d - 255°
= 285° - 255°
= 30°
Answer(s): (a) 142.5°; (b) 30°