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
- ∠i
- ∠j
(a)
∠HSL = 180° - ∠i (Interior angles, SH//RK)
∠NLS = 180° - ∠i (Angles on a straight line)
∠NSL = 180° - ∠i (Isosceles triangle)
57° + 180° - ∠i + 180° - ∠i + ∠j + 22° = 180° (Angles on a straight line, GR)
57° + 180° + 180° + 22° - ∠i - ∠i + ∠j = 180°
439° - 2∠i + ∠j = 180°
2∠i - ∠j = 439° - 180°
2∠i - ∠j = 259°
∠j = 2∠i - 259° --- (1)
∠SNP
= ∠SPN
= 2 x (180° - ∠i)
= 360° - 2∠i (Exterior angle of a triangle)
∠j = 180° - (360° - 2∠i) - (360° - 2∠i)
∠j = 180° - 360° + 2∠i - 360° + 2∠i
∠j = 180° - 360° - 360° + 2∠i + 2∠i
∠j = 4∠i - 540° (Angles sum of triangle)
∠j = 4∠i - 540° --- (2)
(2) = (1)
4∠i - 540° = 2∠i - 259°
4∠i - 2∠i= 540° - 259°
2∠i = 281°
∠i
= 281° ÷ 2
= 140.5°
(b)
From (1)
∠j
= 2∠i - 259°
= 281° - 259°
= 22°
Answer(s): (a) 140.5°; (b) 22°