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
- ∠r
- ∠s
(a)
∠GRK = 180° - ∠r (Interior angles, RG//PH)
∠LKR = 180° - ∠r (Angles on a straight line)
∠LRK = 180° - ∠r (Isosceles triangle)
49° + 180° - ∠r + 180° - ∠r + ∠s + 16° = 180° (Angles on a straight line, FP)
49° + 180° + 180° + 16° - ∠r - ∠r + ∠s = 180°
425° - 2∠r + ∠s = 180°
2∠r - ∠s = 425° - 180°
2∠r - ∠s = 245°
∠s = 2∠r - 245° --- (1)
∠RLN
= ∠RNL
= 2 x (180° - ∠r)
= 360° - 2∠r (Exterior angle of a triangle)
∠s = 180° - (360° - 2∠r) - (360° - 2∠r)
∠s = 180° - 360° + 2∠r - 360° + 2∠r
∠s = 180° - 360° - 360° + 2∠r + 2∠r
∠s = 4∠r - 540° (Angles sum of triangle)
∠s = 4∠r - 540° --- (2)
(2) = (1)
4∠r - 540° = 2∠r - 245°
4∠r - 2∠r= 540° - 245°
2∠r = 295°
∠r
= 295° ÷ 2
= 147.5°
(b)
From (1)
∠s
= 2∠r - 245°
= 295° - 245°
= 50°
Answer(s): (a) 147.5°; (b) 50°