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 + 15° = 180° (Angles on a straight line, HS)
54° + 180° + 180° + 15° - ∠v - ∠v + ∠w = 180°
429° - 2∠v + ∠w = 180°
2∠v - ∠w = 429° - 180°
2∠v - ∠w = 249°
∠w = 2∠v - 249° --- (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 - 249°
4∠v - 2∠v= 540° - 249°
2∠v = 291°
∠v
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠w
= 2∠v - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°