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
- ∠m
- ∠n
(a)
∠NVR = 180° - ∠m (Interior angles, VN//UP)
∠SRV = 180° - ∠m (Angles on a straight line)
∠SVR = 180° - ∠m (Isosceles triangle)
55° + 180° - ∠m + 180° - ∠m + ∠n + 23° = 180° (Angles on a straight line, LU)
55° + 180° + 180° + 23° - ∠m - ∠m + ∠n = 180°
438° - 2∠m + ∠n = 180°
2∠m - ∠n = 438° - 180°
2∠m - ∠n = 258°
∠n = 2∠m - 258° --- (1)
∠VST
= ∠VTS
= 2 x (180° - ∠m)
= 360° - 2∠m (Exterior angle of a triangle)
∠n = 180° - (360° - 2∠m) - (360° - 2∠m)
∠n = 180° - 360° + 2∠m - 360° + 2∠m
∠n = 180° - 360° - 360° + 2∠m + 2∠m
∠n = 4∠m - 540° (Angles sum of triangle)
∠n = 4∠m - 540° --- (2)
(2) = (1)
4∠m - 540° = 2∠m - 258°
4∠m - 2∠m= 540° - 258°
2∠m = 282°
∠m
= 282° ÷ 2
= 141°
(b)
From (1)
∠n
= 2∠m - 258°
= 282° - 258°
= 24°
Answer(s): (a) 141°; (b) 24°