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
- ∠g
- ∠h
(a)
∠ENG = 180° - ∠g (Interior angles, NE//LF)
∠HGN = 180° - ∠g (Angles on a straight line)
∠HNG = 180° - ∠g (Isosceles triangle)
55° + 180° - ∠g + 180° - ∠g + ∠h + 24° = 180° (Angles on a straight line, DL)
55° + 180° + 180° + 24° - ∠g - ∠g + ∠h = 180°
439° - 2∠g + ∠h = 180°
2∠g - ∠h = 439° - 180°
2∠g - ∠h = 259°
∠h = 2∠g - 259° --- (1)
∠NHK
= ∠NKH
= 2 x (180° - ∠g)
= 360° - 2∠g (Exterior angle of a triangle)
∠h = 180° - (360° - 2∠g) - (360° - 2∠g)
∠h = 180° - 360° + 2∠g - 360° + 2∠g
∠h = 180° - 360° - 360° + 2∠g + 2∠g
∠h = 4∠g - 540° (Angles sum of triangle)
∠h = 4∠g - 540° --- (2)
(2) = (1)
4∠g - 540° = 2∠g - 259°
4∠g - 2∠g= 540° - 259°
2∠g = 281°
∠g
= 281° ÷ 2
= 140.5°
(b)
From (1)
∠h
= 2∠g - 259°
= 281° - 259°
= 22°
Answer(s): (a) 140.5°; (b) 22°