In the figure, PFGL is a trapezium and triangles PLK and KPH are isosceles triangles. NE, NG and EG are straight lines. PL = PK = KH. Find
- ∠k
- ∠m
(a)
∠FPH = 180° - ∠k (Interior angles, PF//NG)
∠KHP = 180° - ∠k (Angles on a straight line)
∠KPH = 180° - ∠k (Isosceles triangle)
49° + 180° - ∠k + 180° - ∠k + ∠m + 21° = 180° (Angles on a straight line, EN)
49° + 180° + 180° + 21° - ∠k - ∠k + ∠m = 180°
430° - 2∠k + ∠m = 180°
2∠k - ∠m = 430° - 180°
2∠k - ∠m = 250°
∠m = 2∠k - 250° --- (1)
∠PKL
= ∠PLK
= 2 x (180° - ∠k)
= 360° - 2∠k (Exterior angle of a triangle)
∠m = 180° - (360° - 2∠k) - (360° - 2∠k)
∠m = 180° - 360° + 2∠k - 360° + 2∠k
∠m = 180° - 360° - 360° + 2∠k + 2∠k
∠m = 4∠k - 540° (Angles sum of triangle)
∠m = 4∠k - 540° --- (2)
(2) = (1)
4∠k - 540° = 2∠k - 250°
4∠k - 2∠k= 540° - 250°
2∠k = 290°
∠k
= 290° ÷ 2
= 145°
(b)
From (1)
∠m
= 2∠k - 250°
= 290° - 250°
= 40°
Answer(s): (a) 145°; (b) 40°