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
- ∠i
- ∠j
(a)
∠ENG = 180° - ∠i (Interior angles, NE//LF)
∠HGN = 180° - ∠i (Angles on a straight line)
∠HNG = 180° - ∠i (Isosceles triangle)
48° + 180° - ∠i + 180° - ∠i + ∠j + 19° = 180° (Angles on a straight line, DL)
48° + 180° + 180° + 19° - ∠i - ∠i + ∠j = 180°
427° - 2∠i + ∠j = 180°
2∠i - ∠j = 427° - 180°
2∠i - ∠j = 247°
∠j = 2∠i - 247° --- (1)
∠NHK
= ∠NKH
= 2 x (180° - ∠i)
= 360° - 2∠i (Exterior angle of a triangle)
∠j = 180° - (360° - 2∠i) - (360° - 2∠i)
∠j = 180° - 360° + 2∠i - 360° + 2∠i
∠j = 180° - 360° - 360° + 2∠i + 2∠i
∠j = 4∠i - 540° (Angles sum of triangle)
∠j = 4∠i - 540° --- (2)
(2) = (1)
4∠i - 540° = 2∠i - 247°
4∠i - 2∠i= 540° - 247°
2∠i = 293°
∠i
= 293° ÷ 2
= 146.5°
(b)
From (1)
∠j
= 2∠i - 247°
= 293° - 247°
= 46°
Answer(s): (a) 146.5°; (b) 46°