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
- ∠s
- ∠t
(a)
∠ENG = 180° - ∠s (Interior angles, NE//LF)
∠HGN = 180° - ∠s (Angles on a straight line)
∠HNG = 180° - ∠s (Isosceles triangle)
54° + 180° - ∠s + 180° - ∠s + ∠t + 23° = 180° (Angles on a straight line, DL)
54° + 180° + 180° + 23° - ∠s - ∠s + ∠t = 180°
437° - 2∠s + ∠t = 180°
2∠s - ∠t = 437° - 180°
2∠s - ∠t = 257°
∠t = 2∠s - 257° --- (1)
∠NHK
= ∠NKH
= 2 x (180° - ∠s)
= 360° - 2∠s (Exterior angle of a triangle)
∠t = 180° - (360° - 2∠s) - (360° - 2∠s)
∠t = 180° - 360° + 2∠s - 360° + 2∠s
∠t = 180° - 360° - 360° + 2∠s + 2∠s
∠t = 4∠s - 540° (Angles sum of triangle)
∠t = 4∠s - 540° --- (2)
(2) = (1)
4∠s - 540° = 2∠s - 257°
4∠s - 2∠s= 540° - 257°
2∠s = 283°
∠s
= 283° ÷ 2
= 141.5°
(b)
From (1)
∠t
= 2∠s - 257°
= 283° - 257°
= 26°
Answer(s): (a) 141.5°; (b) 26°