In the figure, LDEH is a trapezium and triangles LHG and GLF are isosceles triangles. KC, KE and CE are straight lines. LH = LG = GF. Find
- ∠n
- ∠p
(a)
∠DLF = 180° - ∠n (Interior angles, LD//KE)
∠GFL = 180° - ∠n (Angles on a straight line)
∠GLF = 180° - ∠n (Isosceles triangle)
52° + 180° - ∠n + 180° - ∠n + ∠p + 15° = 180° (Angles on a straight line, CK)
52° + 180° + 180° + 15° - ∠n - ∠n + ∠p = 180°
427° - 2∠n + ∠p = 180°
2∠n - ∠p = 427° - 180°
2∠n - ∠p = 247°
∠p = 2∠n - 247° --- (1)
∠LGH
= ∠LHG
= 2 x (180° - ∠n)
= 360° - 2∠n (Exterior angle of a triangle)
∠p = 180° - (360° - 2∠n) - (360° - 2∠n)
∠p = 180° - 360° + 2∠n - 360° + 2∠n
∠p = 180° - 360° - 360° + 2∠n + 2∠n
∠p = 4∠n - 540° (Angles sum of triangle)
∠p = 4∠n - 540° --- (2)
(2) = (1)
4∠n - 540° = 2∠n - 247°
4∠n - 2∠n= 540° - 247°
2∠n = 293°
∠n
= 293° ÷ 2
= 146.5°
(b)
From (1)
∠p
= 2∠n - 247°
= 293° - 247°
= 46°
Answer(s): (a) 146.5°; (b) 46°