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
- ∠h
- ∠i
(a)
∠NVR = 180° - ∠h (Interior angles, VN//UP)
∠SRV = 180° - ∠h (Angles on a straight line)
∠SVR = 180° - ∠h (Isosceles triangle)
48° + 180° - ∠h + 180° - ∠h + ∠i + 15° = 180° (Angles on a straight line, LU)
48° + 180° + 180° + 15° - ∠h - ∠h + ∠i = 180°
423° - 2∠h + ∠i = 180°
2∠h - ∠i = 423° - 180°
2∠h - ∠i = 243°
∠i = 2∠h - 243° --- (1)
∠VST
= ∠VTS
= 2 x (180° - ∠h)
= 360° - 2∠h (Exterior angle of a triangle)
∠i = 180° - (360° - 2∠h) - (360° - 2∠h)
∠i = 180° - 360° + 2∠h - 360° + 2∠h
∠i = 180° - 360° - 360° + 2∠h + 2∠h
∠i = 4∠h - 540° (Angles sum of triangle)
∠i = 4∠h - 540° --- (2)
(2) = (1)
4∠h - 540° = 2∠h - 243°
4∠h - 2∠h= 540° - 243°
2∠h = 297°
∠h
= 297° ÷ 2
= 148.5°
(b)
From (1)
∠i
= 2∠h - 243°
= 297° - 243°
= 54°
Answer(s): (a) 148.5°; (b) 54°