In the figure, SHKP is a trapezium and triangles SPN and NSL are isosceles triangles. RG, RK and GK are straight lines. SP = SN = NL. Find
- ∠h
- ∠i
(a)
∠HSL = 180° - ∠h (Interior angles, SH//RK)
∠NLS = 180° - ∠h (Angles on a straight line)
∠NSL = 180° - ∠h (Isosceles triangle)
49° + 180° - ∠h + 180° - ∠h + ∠i + 20° = 180° (Angles on a straight line, GR)
49° + 180° + 180° + 20° - ∠h - ∠h + ∠i = 180°
429° - 2∠h + ∠i = 180°
2∠h - ∠i = 429° - 180°
2∠h - ∠i = 249°
∠i = 2∠h - 249° --- (1)
∠SNP
= ∠SPN
= 2 x (180° - ∠h)
= 360° - 2∠h (Exterior angle of a triangle)
∠i = 180° - (360° - 2∠h) - (360° - 2∠h)
∠i = 180° - 360° + 2∠h - 360° + 2∠h
∠i = 180° - 360° - 360° + 2∠h + 2∠h
∠i = 4∠h - 540° (Angles sum of triangle)
∠i = 4∠h - 540° --- (2)
(2) = (1)
4∠h - 540° = 2∠h - 249°
4∠h - 2∠h= 540° - 249°
2∠h = 291°
∠h
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠i
= 2∠h - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°