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
- ∠x
- ∠y
(a)
∠FPH = 180° - ∠x (Interior angles, PF//NG)
∠KHP = 180° - ∠x (Angles on a straight line)
∠KPH = 180° - ∠x (Isosceles triangle)
57° + 180° - ∠x + 180° - ∠x + ∠y + 18° = 180° (Angles on a straight line, EN)
57° + 180° + 180° + 18° - ∠x - ∠x + ∠y = 180°
435° - 2∠x + ∠y = 180°
2∠x - ∠y = 435° - 180°
2∠x - ∠y = 255°
∠y = 2∠x - 255° --- (1)
∠PKL
= ∠PLK
= 2 x (180° - ∠x)
= 360° - 2∠x (Exterior angle of a triangle)
∠y = 180° - (360° - 2∠x) - (360° - 2∠x)
∠y = 180° - 360° + 2∠x - 360° + 2∠x
∠y = 180° - 360° - 360° + 2∠x + 2∠x
∠y = 4∠x - 540° (Angles sum of triangle)
∠y = 4∠x - 540° --- (2)
(2) = (1)
4∠x - 540° = 2∠x - 255°
4∠x - 2∠x= 540° - 255°
2∠x = 285°
∠x
= 285° ÷ 2
= 142.5°
(b)
From (1)
∠y
= 2∠x - 255°
= 285° - 255°
= 30°
Answer(s): (a) 142.5°; (b) 30°