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
- ∠p
- ∠q
(a)
∠GRK = 180° - ∠p (Interior angles, RG//PH)
∠LKR = 180° - ∠p (Angles on a straight line)
∠LRK = 180° - ∠p (Isosceles triangle)
50° + 180° - ∠p + 180° - ∠p + ∠q + 24° = 180° (Angles on a straight line, FP)
50° + 180° + 180° + 24° - ∠p - ∠p + ∠q = 180°
434° - 2∠p + ∠q = 180°
2∠p - ∠q = 434° - 180°
2∠p - ∠q = 254°
∠q = 2∠p - 254° --- (1)
∠RLN
= ∠RNL
= 2 x (180° - ∠p)
= 360° - 2∠p (Exterior angle of a triangle)
∠q = 180° - (360° - 2∠p) - (360° - 2∠p)
∠q = 180° - 360° + 2∠p - 360° + 2∠p
∠q = 180° - 360° - 360° + 2∠p + 2∠p
∠q = 4∠p - 540° (Angles sum of triangle)
∠q = 4∠p - 540° --- (2)
(2) = (1)
4∠p - 540° = 2∠p - 254°
4∠p - 2∠p= 540° - 254°
2∠p = 286°
∠p
= 286° ÷ 2
= 143°
(b)
From (1)
∠q
= 2∠p - 254°
= 286° - 254°
= 32°
Answer(s): (a) 143°; (b) 32°