In the figure, TKLR is a trapezium and triangles TRP and PTN are isosceles triangles. SH, SL and HL are straight lines. TR = TP = PN. Find
- ∠k
- ∠m
(a)
∠KTN = 180° - ∠k (Interior angles, TK//SL)
∠PNT = 180° - ∠k (Angles on a straight line)
∠PTN = 180° - ∠k (Isosceles triangle)
51° + 180° - ∠k + 180° - ∠k + ∠m + 24° = 180° (Angles on a straight line, HS)
51° + 180° + 180° + 24° - ∠k - ∠k + ∠m = 180°
435° - 2∠k + ∠m = 180°
2∠k - ∠m = 435° - 180°
2∠k - ∠m = 255°
∠m = 2∠k - 255° --- (1)
∠TPR
= ∠TRP
= 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 - 255°
4∠k - 2∠k= 540° - 255°
2∠k = 285°
∠k
= 285° ÷ 2
= 142.5°
(b)
From (1)
∠m
= 2∠k - 255°
= 285° - 255°
= 30°
Answer(s): (a) 142.5°; (b) 30°