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
- ∠h
- ∠i
(a)
∠GRK = 180° - ∠h (Interior angles, RG//PH)
∠LKR = 180° - ∠h (Angles on a straight line)
∠LRK = 180° - ∠h (Isosceles triangle)
50° + 180° - ∠h + 180° - ∠h + ∠i + 17° = 180° (Angles on a straight line, FP)
50° + 180° + 180° + 17° - ∠h - ∠h + ∠i = 180°
427° - 2∠h + ∠i = 180°
2∠h - ∠i = 427° - 180°
2∠h - ∠i = 247°
∠i = 2∠h - 247° --- (1)
∠RLN
= ∠RNL
= 2 x (180° - ∠h)
= 360° - 2∠h (Exterior angle of a triangle)
∠i = 180° - (360° - 2∠h) - (360° - 2∠h)
∠i = 180° - 360° + 2∠h - 360° + 2∠h
∠i = 180° - 360° - 360° + 2∠h + 2∠h
∠i = 4∠h - 540° (Angles sum of triangle)
∠i = 4∠h - 540° --- (2)
(2) = (1)
4∠h - 540° = 2∠h - 247°
4∠h - 2∠h= 540° - 247°
2∠h = 293°
∠h
= 293° ÷ 2
= 146.5°
(b)
From (1)
∠i
= 2∠h - 247°
= 293° - 247°
= 46°
Answer(s): (a) 146.5°; (b) 46°