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)
52° + 180° - ∠x + 180° - ∠x + ∠y + 17° = 180° (Angles on a straight line, DL)
52° + 180° + 180° + 17° - ∠x - ∠x + ∠y = 180°
429° - 2∠x + ∠y = 180°
2∠x - ∠y = 429° - 180°
2∠x - ∠y = 249°
∠y = 2∠x - 249° --- (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 - 249°
4∠x - 2∠x= 540° - 249°
2∠x = 291°
∠x
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠y
= 2∠x - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°