In the figure, RGHN is a trapezium and triangles RNL and LRK are isosceles triangles. PF, PH and FH are straight lines. RN = RL = LK. Find
- ∠i
- ∠j
(a)
∠GRK = 180° - ∠i (Interior angles, RG//PH)
∠LKR = 180° - ∠i (Angles on a straight line)
∠LRK = 180° - ∠i (Isosceles triangle)
49° + 180° - ∠i + 180° - ∠i + ∠j + 22° = 180° (Angles on a straight line, FP)
49° + 180° + 180° + 22° - ∠i - ∠i + ∠j = 180°
431° - 2∠i + ∠j = 180°
2∠i - ∠j = 431° - 180°
2∠i - ∠j = 251°
∠j = 2∠i - 251° --- (1)
∠RLN
= ∠RNL
= 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 - 251°
4∠i - 2∠i= 540° - 251°
2∠i = 289°
∠i
= 289° ÷ 2
= 144.5°
(b)
From (1)
∠j
= 2∠i - 251°
= 289° - 251°
= 38°
Answer(s): (a) 144.5°; (b) 38°