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
- ∠x
- ∠y
(a)
∠DLF = 180° - ∠x (Interior angles, LD//KE)
∠GFL = 180° - ∠x (Angles on a straight line)
∠GLF = 180° - ∠x (Isosceles triangle)
55° + 180° - ∠x + 180° - ∠x + ∠y + 23° = 180° (Angles on a straight line, CK)
55° + 180° + 180° + 23° - ∠x - ∠x + ∠y = 180°
438° - 2∠x + ∠y = 180°
2∠x - ∠y = 438° - 180°
2∠x - ∠y = 258°
∠y = 2∠x - 258° --- (1)
∠LGH
= ∠LHG
= 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 - 258°
4∠x - 2∠x= 540° - 258°
2∠x = 282°
∠x
= 282° ÷ 2
= 141°
(b)
From (1)
∠y
= 2∠x - 258°
= 282° - 258°
= 24°
Answer(s): (a) 141°; (b) 24°