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
- ∠m
- ∠n
(a)
∠DLF = 180° - ∠m (Interior angles, LD//KE)
∠GFL = 180° - ∠m (Angles on a straight line)
∠GLF = 180° - ∠m (Isosceles triangle)
56° + 180° - ∠m + 180° - ∠m + ∠n + 23° = 180° (Angles on a straight line, CK)
56° + 180° + 180° + 23° - ∠m - ∠m + ∠n = 180°
439° - 2∠m + ∠n = 180°
2∠m - ∠n = 439° - 180°
2∠m - ∠n = 259°
∠n = 2∠m - 259° --- (1)
∠LGH
= ∠LHG
= 2 x (180° - ∠m)
= 360° - 2∠m (Exterior angle of a triangle)
∠n = 180° - (360° - 2∠m) - (360° - 2∠m)
∠n = 180° - 360° + 2∠m - 360° + 2∠m
∠n = 180° - 360° - 360° + 2∠m + 2∠m
∠n = 4∠m - 540° (Angles sum of triangle)
∠n = 4∠m - 540° --- (2)
(2) = (1)
4∠m - 540° = 2∠m - 259°
4∠m - 2∠m= 540° - 259°
2∠m = 281°
∠m
= 281° ÷ 2
= 140.5°
(b)
From (1)
∠n
= 2∠m - 259°
= 281° - 259°
= 22°
Answer(s): (a) 140.5°; (b) 22°