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
- ∠x
- ∠y
(a)
∠ENG = 180° - ∠x (Interior angles, NE//LF)
∠HGN = 180° - ∠x (Angles on a straight line)
∠HNG = 180° - ∠x (Isosceles triangle)
48° + 180° - ∠x + 180° - ∠x + ∠y + 16° = 180° (Angles on a straight line, DL)
48° + 180° + 180° + 16° - ∠x - ∠x + ∠y = 180°
424° - 2∠x + ∠y = 180°
2∠x - ∠y = 424° - 180°
2∠x - ∠y = 244°
∠y = 2∠x - 244° --- (1)
∠NHK
= ∠NKH
= 2 x (180° - ∠x)
= 360° - 2∠x (Exterior angle of a triangle)
∠y = 180° - (360° - 2∠x) - (360° - 2∠x)
∠y = 180° - 360° + 2∠x - 360° + 2∠x
∠y = 180° - 360° - 360° + 2∠x + 2∠x
∠y = 4∠x - 540° (Angles sum of triangle)
∠y = 4∠x - 540° --- (2)
(2) = (1)
4∠x - 540° = 2∠x - 244°
4∠x - 2∠x= 540° - 244°
2∠x = 296°
∠x
= 296° ÷ 2
= 148°
(b)
From (1)
∠y
= 2∠x - 244°
= 296° - 244°
= 52°
Answer(s): (a) 148°; (b) 52°