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
- ∠p
- ∠q
(a)
∠ENG = 180° - ∠p (Interior angles, NE//LF)
∠HGN = 180° - ∠p (Angles on a straight line)
∠HNG = 180° - ∠p (Isosceles triangle)
55° + 180° - ∠p + 180° - ∠p + ∠q + 22° = 180° (Angles on a straight line, DL)
55° + 180° + 180° + 22° - ∠p - ∠p + ∠q = 180°
437° - 2∠p + ∠q = 180°
2∠p - ∠q = 437° - 180°
2∠p - ∠q = 257°
∠q = 2∠p - 257° --- (1)
∠NHK
= ∠NKH
= 2 x (180° - ∠p)
= 360° - 2∠p (Exterior angle of a triangle)
∠q = 180° - (360° - 2∠p) - (360° - 2∠p)
∠q = 180° - 360° + 2∠p - 360° + 2∠p
∠q = 180° - 360° - 360° + 2∠p + 2∠p
∠q = 4∠p - 540° (Angles sum of triangle)
∠q = 4∠p - 540° --- (2)
(2) = (1)
4∠p - 540° = 2∠p - 257°
4∠p - 2∠p= 540° - 257°
2∠p = 283°
∠p
= 283° ÷ 2
= 141.5°
(b)
From (1)
∠q
= 2∠p - 257°
= 283° - 257°
= 26°
Answer(s): (a) 141.5°; (b) 26°