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
- ∠m
- ∠n
(a)
∠KTN = 180° - ∠m (Interior angles, TK//SL)
∠PNT = 180° - ∠m (Angles on a straight line)
∠PTN = 180° - ∠m (Isosceles triangle)
56° + 180° - ∠m + 180° - ∠m + ∠n + 15° = 180° (Angles on a straight line, HS)
56° + 180° + 180° + 15° - ∠m - ∠m + ∠n = 180°
431° - 2∠m + ∠n = 180°
2∠m - ∠n = 431° - 180°
2∠m - ∠n = 251°
∠n = 2∠m - 251° --- (1)
∠TPR
= ∠TRP
= 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 - 251°
4∠m - 2∠m= 540° - 251°
2∠m = 289°
∠m
= 289° ÷ 2
= 144.5°
(b)
From (1)
∠n
= 2∠m - 251°
= 289° - 251°
= 38°
Answer(s): (a) 144.5°; (b) 38°