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
- ∠a
- ∠b
(a)
∠GRK = 180° - ∠a (Interior angles, RG//PH)
∠LKR = 180° - ∠a (Angles on a straight line)
∠LRK = 180° - ∠a (Isosceles triangle)
54° + 180° - ∠a + 180° - ∠a + ∠b + 25° = 180° (Angles on a straight line, FP)
54° + 180° + 180° + 25° - ∠a - ∠a + ∠b = 180°
439° - 2∠a + ∠b = 180°
2∠a - ∠b = 439° - 180°
2∠a - ∠b = 259°
∠b = 2∠a - 259° --- (1)
∠RLN
= ∠RNL
= 2 x (180° - ∠a)
= 360° - 2∠a (Exterior angle of a triangle)
∠b = 180° - (360° - 2∠a) - (360° - 2∠a)
∠b = 180° - 360° + 2∠a - 360° + 2∠a
∠b = 180° - 360° - 360° + 2∠a + 2∠a
∠b = 4∠a - 540° (Angles sum of triangle)
∠b = 4∠a - 540° --- (2)
(2) = (1)
4∠a - 540° = 2∠a - 259°
4∠a - 2∠a= 540° - 259°
2∠a = 281°
∠a
= 281° ÷ 2
= 140.5°
(b)
From (1)
∠b
= 2∠a - 259°
= 281° - 259°
= 22°
Answer(s): (a) 140.5°; (b) 22°