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
- ∠e
- ∠f
(a)
∠NVR = 180° - ∠e (Interior angles, VN//UP)
∠SRV = 180° - ∠e (Angles on a straight line)
∠SVR = 180° - ∠e (Isosceles triangle)
57° + 180° - ∠e + 180° - ∠e + ∠f + 25° = 180° (Angles on a straight line, LU)
57° + 180° + 180° + 25° - ∠e - ∠e + ∠f = 180°
442° - 2∠e + ∠f = 180°
2∠e - ∠f = 442° - 180°
2∠e - ∠f = 262°
∠f = 2∠e - 262° --- (1)
∠VST
= ∠VTS
= 2 x (180° - ∠e)
= 360° - 2∠e (Exterior angle of a triangle)
∠f = 180° - (360° - 2∠e) - (360° - 2∠e)
∠f = 180° - 360° + 2∠e - 360° + 2∠e
∠f = 180° - 360° - 360° + 2∠e + 2∠e
∠f = 4∠e - 540° (Angles sum of triangle)
∠f = 4∠e - 540° --- (2)
(2) = (1)
4∠e - 540° = 2∠e - 262°
4∠e - 2∠e= 540° - 262°
2∠e = 278°
∠e
= 278° ÷ 2
= 139°
(b)
From (1)
∠f
= 2∠e - 262°
= 278° - 262°
= 16°
Answer(s): (a) 139°; (b) 16°