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
- ∠v
- ∠w
(a)
∠FPH = 180° - ∠v (Interior angles, PF//NG)
∠KHP = 180° - ∠v (Angles on a straight line)
∠KPH = 180° - ∠v (Isosceles triangle)
56° + 180° - ∠v + 180° - ∠v + ∠w + 18° = 180° (Angles on a straight line, EN)
56° + 180° + 180° + 18° - ∠v - ∠v + ∠w = 180°
434° - 2∠v + ∠w = 180°
2∠v - ∠w = 434° - 180°
2∠v - ∠w = 254°
∠w = 2∠v - 254° --- (1)
∠PKL
= ∠PLK
= 2 x (180° - ∠v)
= 360° - 2∠v (Exterior angle of a triangle)
∠w = 180° - (360° - 2∠v) - (360° - 2∠v)
∠w = 180° - 360° + 2∠v - 360° + 2∠v
∠w = 180° - 360° - 360° + 2∠v + 2∠v
∠w = 4∠v - 540° (Angles sum of triangle)
∠w = 4∠v - 540° --- (2)
(2) = (1)
4∠v - 540° = 2∠v - 254°
4∠v - 2∠v= 540° - 254°
2∠v = 286°
∠v
= 286° ÷ 2
= 143°
(b)
From (1)
∠w
= 2∠v - 254°
= 286° - 254°
= 32°
Answer(s): (a) 143°; (b) 32°