In the figure, NEFK is a trapezium and triangles NKH and HNG are isosceles triangles. LD, LF and DF are straight lines. NK = NH = HG. Find
- ∠v
- ∠w
(a)
∠ENG = 180° - ∠v (Interior angles, NE//LF)
∠HGN = 180° - ∠v (Angles on a straight line)
∠HNG = 180° - ∠v (Isosceles triangle)
49° + 180° - ∠v + 180° - ∠v + ∠w + 20° = 180° (Angles on a straight line, DL)
49° + 180° + 180° + 20° - ∠v - ∠v + ∠w = 180°
429° - 2∠v + ∠w = 180°
2∠v - ∠w = 429° - 180°
2∠v - ∠w = 249°
∠w = 2∠v - 249° --- (1)
∠NHK
= ∠NKH
= 2 x (180° - ∠v)
= 360° - 2∠v (Exterior angle of a triangle)
∠w = 180° - (360° - 2∠v) - (360° - 2∠v)
∠w = 180° - 360° + 2∠v - 360° + 2∠v
∠w = 180° - 360° - 360° + 2∠v + 2∠v
∠w = 4∠v - 540° (Angles sum of triangle)
∠w = 4∠v - 540° --- (2)
(2) = (1)
4∠v - 540° = 2∠v - 249°
4∠v - 2∠v= 540° - 249°
2∠v = 291°
∠v
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠w
= 2∠v - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°