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
- ∠w
- ∠x
(a)
∠GRK = 180° - ∠w (Interior angles, RG//PH)
∠LKR = 180° - ∠w (Angles on a straight line)
∠LRK = 180° - ∠w (Isosceles triangle)
56° + 180° - ∠w + 180° - ∠w + ∠x + 23° = 180° (Angles on a straight line, FP)
56° + 180° + 180° + 23° - ∠w - ∠w + ∠x = 180°
439° - 2∠w + ∠x = 180°
2∠w - ∠x = 439° - 180°
2∠w - ∠x = 259°
∠x = 2∠w - 259° --- (1)
∠RLN
= ∠RNL
= 2 x (180° - ∠w)
= 360° - 2∠w (Exterior angle of a triangle)
∠x = 180° - (360° - 2∠w) - (360° - 2∠w)
∠x = 180° - 360° + 2∠w - 360° + 2∠w
∠x = 180° - 360° - 360° + 2∠w + 2∠w
∠x = 4∠w - 540° (Angles sum of triangle)
∠x = 4∠w - 540° --- (2)
(2) = (1)
4∠w - 540° = 2∠w - 259°
4∠w - 2∠w= 540° - 259°
2∠w = 281°
∠w
= 281° ÷ 2
= 140.5°
(b)
From (1)
∠x
= 2∠w - 259°
= 281° - 259°
= 22°
Answer(s): (a) 140.5°; (b) 22°