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
- ∠v
- ∠w
(a)
∠KTN = 180° - ∠v (Interior angles, TK//SL)
∠PNT = 180° - ∠v (Angles on a straight line)
∠PTN = 180° - ∠v (Isosceles triangle)
54° + 180° - ∠v + 180° - ∠v + ∠w + 17° = 180° (Angles on a straight line, HS)
54° + 180° + 180° + 17° - ∠v - ∠v + ∠w = 180°
431° - 2∠v + ∠w = 180°
2∠v - ∠w = 431° - 180°
2∠v - ∠w = 251°
∠w = 2∠v - 251° --- (1)
∠TPR
= ∠TRP
= 2 x (180° - ∠v)
= 360° - 2∠v (Exterior angle of a triangle)
∠w = 180° - (360° - 2∠v) - (360° - 2∠v)
∠w = 180° - 360° + 2∠v - 360° + 2∠v
∠w = 180° - 360° - 360° + 2∠v + 2∠v
∠w = 4∠v - 540° (Angles sum of triangle)
∠w = 4∠v - 540° --- (2)
(2) = (1)
4∠v - 540° = 2∠v - 251°
4∠v - 2∠v= 540° - 251°
2∠v = 289°
∠v
= 289° ÷ 2
= 144.5°
(b)
From (1)
∠w
= 2∠v - 251°
= 289° - 251°
= 38°
Answer(s): (a) 144.5°; (b) 38°