In the figure, NEFK is a trapezium and triangles NKH and HNG are isosceles triangles. LD, LF and DF are straight lines. NK = NH = HG. Find
- ∠q
- ∠r
(a)
∠ENG = 180° - ∠q (Interior angles, NE//LF)
∠HGN = 180° - ∠q (Angles on a straight line)
∠HNG = 180° - ∠q (Isosceles triangle)
52° + 180° - ∠q + 180° - ∠q + ∠r + 23° = 180° (Angles on a straight line, DL)
52° + 180° + 180° + 23° - ∠q - ∠q + ∠r = 180°
435° - 2∠q + ∠r = 180°
2∠q - ∠r = 435° - 180°
2∠q - ∠r = 255°
∠r = 2∠q - 255° --- (1)
∠NHK
= ∠NKH
= 2 x (180° - ∠q)
= 360° - 2∠q (Exterior angle of a triangle)
∠r = 180° - (360° - 2∠q) - (360° - 2∠q)
∠r = 180° - 360° + 2∠q - 360° + 2∠q
∠r = 180° - 360° - 360° + 2∠q + 2∠q
∠r = 4∠q - 540° (Angles sum of triangle)
∠r = 4∠q - 540° --- (2)
(2) = (1)
4∠q - 540° = 2∠q - 255°
4∠q - 2∠q= 540° - 255°
2∠q = 285°
∠q
= 285° ÷ 2
= 142.5°
(b)
From (1)
∠r
= 2∠q - 255°
= 285° - 255°
= 30°
Answer(s): (a) 142.5°; (b) 30°