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
- ∠g
- ∠h
(a)
∠KTN = 180° - ∠g (Interior angles, TK//SL)
∠PNT = 180° - ∠g (Angles on a straight line)
∠PTN = 180° - ∠g (Isosceles triangle)
48° + 180° - ∠g + 180° - ∠g + ∠h + 23° = 180° (Angles on a straight line, HS)
48° + 180° + 180° + 23° - ∠g - ∠g + ∠h = 180°
431° - 2∠g + ∠h = 180°
2∠g - ∠h = 431° - 180°
2∠g - ∠h = 251°
∠h = 2∠g - 251° --- (1)
∠TPR
= ∠TRP
= 2 x (180° - ∠g)
= 360° - 2∠g (Exterior angle of a triangle)
∠h = 180° - (360° - 2∠g) - (360° - 2∠g)
∠h = 180° - 360° + 2∠g - 360° + 2∠g
∠h = 180° - 360° - 360° + 2∠g + 2∠g
∠h = 4∠g - 540° (Angles sum of triangle)
∠h = 4∠g - 540° --- (2)
(2) = (1)
4∠g - 540° = 2∠g - 251°
4∠g - 2∠g= 540° - 251°
2∠g = 289°
∠g
= 289° ÷ 2
= 144.5°
(b)
From (1)
∠h
= 2∠g - 251°
= 289° - 251°
= 38°
Answer(s): (a) 144.5°; (b) 38°