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
- ∠b
- ∠c
(a)
∠HSL = 180° - ∠b (Interior angles, SH//RK)
∠NLS = 180° - ∠b (Angles on a straight line)
∠NSL = 180° - ∠b (Isosceles triangle)
57° + 180° - ∠b + 180° - ∠b + ∠c + 15° = 180° (Angles on a straight line, GR)
57° + 180° + 180° + 15° - ∠b - ∠b + ∠c = 180°
432° - 2∠b + ∠c = 180°
2∠b - ∠c = 432° - 180°
2∠b - ∠c = 252°
∠c = 2∠b - 252° --- (1)
∠SNP
= ∠SPN
= 2 x (180° - ∠b)
= 360° - 2∠b (Exterior angle of a triangle)
∠c = 180° - (360° - 2∠b) - (360° - 2∠b)
∠c = 180° - 360° + 2∠b - 360° + 2∠b
∠c = 180° - 360° - 360° + 2∠b + 2∠b
∠c = 4∠b - 540° (Angles sum of triangle)
∠c = 4∠b - 540° --- (2)
(2) = (1)
4∠b - 540° = 2∠b - 252°
4∠b - 2∠b= 540° - 252°
2∠b = 288°
∠b
= 288° ÷ 2
= 144°
(b)
From (1)
∠c
= 2∠b - 252°
= 288° - 252°
= 36°
Answer(s): (a) 144°; (b) 36°