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