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)
50° + 180° - ∠p + 180° - ∠p + ∠q + 20° = 180° (Angles on a straight line, DL)
50° + 180° + 180° + 20° - ∠p - ∠p + ∠q = 180°
430° - 2∠p + ∠q = 180°
2∠p - ∠q = 430° - 180°
2∠p - ∠q = 250°
∠q = 2∠p - 250° --- (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 - 250°
4∠p - 2∠p= 540° - 250°
2∠p = 290°
∠p
= 290° ÷ 2
= 145°
(b)
From (1)
∠q
= 2∠p - 250°
= 290° - 250°
= 40°
Answer(s): (a) 145°; (b) 40°