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)
53° + 180° - ∠k + 180° - ∠k + ∠m + 20° = 180° (Angles on a straight line, HS)
53° + 180° + 180° + 20° - ∠k - ∠k + ∠m = 180°
433° - 2∠k + ∠m = 180°
2∠k - ∠m = 433° - 180°
2∠k - ∠m = 253°
∠m = 2∠k - 253° --- (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 - 253°
4∠k - 2∠k= 540° - 253°
2∠k = 287°
∠k
= 287° ÷ 2
= 143.5°
(b)
From (1)
∠m
= 2∠k - 253°
= 287° - 253°
= 34°
Answer(s): (a) 143.5°; (b) 34°