In the figure, VNPT is a trapezium and triangles VTS and SVR are isosceles triangles. UL, UP and LP are straight lines. VT = VS = SR. Find

1. ∠x
2. ∠y

(a)
∠NVR = 180° - ∠x (Interior angles, VN//UP)
∠SRV = 180° - ∠x (Angles on a straight line)
∠SVR = 180° - ∠x (Isosceles triangle)

50° + 180° - ∠x + 180° - ∠x + ∠y + 19° = 180° (Angles on a straight line, LU)
50° + 180° + 180° + 19° - ∠x - ∠x + ∠y  = 180°
429° - 2∠x + ∠y = 180°
2∠x - ∠y = 429° - 180° 
2∠x - ∠y = 249°

∠y = 2∠x - 249° --- (1)

∠VST
= ∠VTS
= 2 x (180° - ∠x)
= 360° - 2∠x (Exterior angle of a triangle)

∠y = 180° - (360° - 2∠x) - (360° - 2∠x)
∠y = 180° - 360° + 2∠x - 360° + 2∠x 
∠y = 180° - 360° - 360° + 2∠x + 2∠x
∠y = 4∠x - 540° (Angles sum of triangle) 

∠y = 4∠x - 540° --- (2)

(2) = (1)
4∠x - 540° = 2∠x - 249°
4∠x - 2∠x= 540° - 249°
2∠x = 291°

∠x
= 291° ÷ 2
= 145.5°

(b)
From (1)
∠y
=  2∠x - 249°
= 291° - 249° 
= 42°

Answer(s): (a) 145.5°; (b) 42°