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
- ∠n
- ∠p
(a)
∠NVR = 180° - ∠n (Interior angles, VN//UP)
∠SRV = 180° - ∠n (Angles on a straight line)
∠SVR = 180° - ∠n (Isosceles triangle)
48° + 180° - ∠n + 180° - ∠n + ∠p + 20° = 180° (Angles on a straight line, LU)
48° + 180° + 180° + 20° - ∠n - ∠n + ∠p = 180°
428° - 2∠n + ∠p = 180°
2∠n - ∠p = 428° - 180°
2∠n - ∠p = 248°
∠p = 2∠n - 248° --- (1)
∠VST
= ∠VTS
= 2 x (180° - ∠n)
= 360° - 2∠n (Exterior angle of a triangle)
∠p = 180° - (360° - 2∠n) - (360° - 2∠n)
∠p = 180° - 360° + 2∠n - 360° + 2∠n
∠p = 180° - 360° - 360° + 2∠n + 2∠n
∠p = 4∠n - 540° (Angles sum of triangle)
∠p = 4∠n - 540° --- (2)
(2) = (1)
4∠n - 540° = 2∠n - 248°
4∠n - 2∠n= 540° - 248°
2∠n = 292°
∠n
= 292° ÷ 2
= 146°
(b)
From (1)
∠p
= 2∠n - 248°
= 292° - 248°
= 44°
Answer(s): (a) 146°; (b) 44°