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
- ∠g
- ∠h
(a)
∠NVR = 180° - ∠g (Interior angles, VN//UP)
∠SRV = 180° - ∠g (Angles on a straight line)
∠SVR = 180° - ∠g (Isosceles triangle)
48° + 180° - ∠g + 180° - ∠g + ∠h + 16° = 180° (Angles on a straight line, LU)
48° + 180° + 180° + 16° - ∠g - ∠g + ∠h = 180°
424° - 2∠g + ∠h = 180°
2∠g - ∠h = 424° - 180°
2∠g - ∠h = 244°
∠h = 2∠g - 244° --- (1)
∠VST
= ∠VTS
= 2 x (180° - ∠g)
= 360° - 2∠g (Exterior angle of a triangle)
∠h = 180° - (360° - 2∠g) - (360° - 2∠g)
∠h = 180° - 360° + 2∠g - 360° + 2∠g
∠h = 180° - 360° - 360° + 2∠g + 2∠g
∠h = 4∠g - 540° (Angles sum of triangle)
∠h = 4∠g - 540° --- (2)
(2) = (1)
4∠g - 540° = 2∠g - 244°
4∠g - 2∠g= 540° - 244°
2∠g = 296°
∠g
= 296° ÷ 2
= 148°
(b)
From (1)
∠h
= 2∠g - 244°
= 296° - 244°
= 52°
Answer(s): (a) 148°; (b) 52°