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