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
- ∠y
- ∠z
(a)
∠NVR = 180° - ∠y (Interior angles, VN//UP)
∠SRV = 180° - ∠y (Angles on a straight line)
∠SVR = 180° - ∠y (Isosceles triangle)
56° + 180° - ∠y + 180° - ∠y + ∠z + 17° = 180° (Angles on a straight line, LU)
56° + 180° + 180° + 17° - ∠y - ∠y + ∠z = 180°
433° - 2∠y + ∠z = 180°
2∠y - ∠z = 433° - 180°
2∠y - ∠z = 253°
∠z = 2∠y - 253° --- (1)
∠VST
= ∠VTS
= 2 x (180° - ∠y)
= 360° - 2∠y (Exterior angle of a triangle)
∠z = 180° - (360° - 2∠y) - (360° - 2∠y)
∠z = 180° - 360° + 2∠y - 360° + 2∠y
∠z = 180° - 360° - 360° + 2∠y + 2∠y
∠z = 4∠y - 540° (Angles sum of triangle)
∠z = 4∠y - 540° --- (2)
(2) = (1)
4∠y - 540° = 2∠y - 253°
4∠y - 2∠y= 540° - 253°
2∠y = 287°
∠y
= 287° ÷ 2
= 143.5°
(b)
From (1)
∠z
= 2∠y - 253°
= 287° - 253°
= 34°
Answer(s): (a) 143.5°; (b) 34°