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
- ∠f
- ∠g
(a)
∠KTN = 180° - ∠f (Interior angles, TK//SL)
∠PNT = 180° - ∠f (Angles on a straight line)
∠PTN = 180° - ∠f (Isosceles triangle)
58° + 180° - ∠f + 180° - ∠f + ∠g + 24° = 180° (Angles on a straight line, HS)
58° + 180° + 180° + 24° - ∠f - ∠f + ∠g = 180°
442° - 2∠f + ∠g = 180°
2∠f - ∠g = 442° - 180°
2∠f - ∠g = 262°
∠g = 2∠f - 262° --- (1)
∠TPR
= ∠TRP
= 2 x (180° - ∠f)
= 360° - 2∠f (Exterior angle of a triangle)
∠g = 180° - (360° - 2∠f) - (360° - 2∠f)
∠g = 180° - 360° + 2∠f - 360° + 2∠f
∠g = 180° - 360° - 360° + 2∠f + 2∠f
∠g = 4∠f - 540° (Angles sum of triangle)
∠g = 4∠f - 540° --- (2)
(2) = (1)
4∠f - 540° = 2∠f - 262°
4∠f - 2∠f= 540° - 262°
2∠f = 278°
∠f
= 278° ÷ 2
= 139°
(b)
From (1)
∠g
= 2∠f - 262°
= 278° - 262°
= 16°
Answer(s): (a) 139°; (b) 16°