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
- ∠d
- ∠e
(a)
∠NVR = 180° - ∠d (Interior angles, VN//UP)
∠SRV = 180° - ∠d (Angles on a straight line)
∠SVR = 180° - ∠d (Isosceles triangle)
51° + 180° - ∠d + 180° - ∠d + ∠e + 18° = 180° (Angles on a straight line, LU)
51° + 180° + 180° + 18° - ∠d - ∠d + ∠e = 180°
429° - 2∠d + ∠e = 180°
2∠d - ∠e = 429° - 180°
2∠d - ∠e = 249°
∠e = 2∠d - 249° --- (1)
∠VST
= ∠VTS
= 2 x (180° - ∠d)
= 360° - 2∠d (Exterior angle of a triangle)
∠e = 180° - (360° - 2∠d) - (360° - 2∠d)
∠e = 180° - 360° + 2∠d - 360° + 2∠d
∠e = 180° - 360° - 360° + 2∠d + 2∠d
∠e = 4∠d - 540° (Angles sum of triangle)
∠e = 4∠d - 540° --- (2)
(2) = (1)
4∠d - 540° = 2∠d - 249°
4∠d - 2∠d= 540° - 249°
2∠d = 291°
∠d
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠e
= 2∠d - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°