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
- ∠s
- ∠t
(a)
∠DLF = 180° - ∠s (Interior angles, LD//KE)
∠GFL = 180° - ∠s (Angles on a straight line)
∠GLF = 180° - ∠s (Isosceles triangle)
48° + 180° - ∠s + 180° - ∠s + ∠t + 24° = 180° (Angles on a straight line, CK)
48° + 180° + 180° + 24° - ∠s - ∠s + ∠t = 180°
432° - 2∠s + ∠t = 180°
2∠s - ∠t = 432° - 180°
2∠s - ∠t = 252°
∠t = 2∠s - 252° --- (1)
∠LGH
= ∠LHG
= 2 x (180° - ∠s)
= 360° - 2∠s (Exterior angle of a triangle)
∠t = 180° - (360° - 2∠s) - (360° - 2∠s)
∠t = 180° - 360° + 2∠s - 360° + 2∠s
∠t = 180° - 360° - 360° + 2∠s + 2∠s
∠t = 4∠s - 540° (Angles sum of triangle)
∠t = 4∠s - 540° --- (2)
(2) = (1)
4∠s - 540° = 2∠s - 252°
4∠s - 2∠s= 540° - 252°
2∠s = 288°
∠s
= 288° ÷ 2
= 144°
(b)
From (1)
∠t
= 2∠s - 252°
= 288° - 252°
= 36°
Answer(s): (a) 144°; (b) 36°