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
- ∠i
- ∠j
(a)
∠NVR = 180° - ∠i (Interior angles, VN//UP)
∠SRV = 180° - ∠i (Angles on a straight line)
∠SVR = 180° - ∠i (Isosceles triangle)
56° + 180° - ∠i + 180° - ∠i + ∠j + 16° = 180° (Angles on a straight line, LU)
56° + 180° + 180° + 16° - ∠i - ∠i + ∠j = 180°
432° - 2∠i + ∠j = 180°
2∠i - ∠j = 432° - 180°
2∠i - ∠j = 252°
∠j = 2∠i - 252° --- (1)
∠VST
= ∠VTS
= 2 x (180° - ∠i)
= 360° - 2∠i (Exterior angle of a triangle)
∠j = 180° - (360° - 2∠i) - (360° - 2∠i)
∠j = 180° - 360° + 2∠i - 360° + 2∠i
∠j = 180° - 360° - 360° + 2∠i + 2∠i
∠j = 4∠i - 540° (Angles sum of triangle)
∠j = 4∠i - 540° --- (2)
(2) = (1)
4∠i - 540° = 2∠i - 252°
4∠i - 2∠i= 540° - 252°
2∠i = 288°
∠i
= 288° ÷ 2
= 144°
(b)
From (1)
∠j
= 2∠i - 252°
= 288° - 252°
= 36°
Answer(s): (a) 144°; (b) 36°