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
- ∠p
- ∠q
(a)
∠HSL = 180° - ∠p (Interior angles, SH//RK)
∠NLS = 180° - ∠p (Angles on a straight line)
∠NSL = 180° - ∠p (Isosceles triangle)
51° + 180° - ∠p + 180° - ∠p + ∠q + 20° = 180° (Angles on a straight line, GR)
51° + 180° + 180° + 20° - ∠p - ∠p + ∠q = 180°
431° - 2∠p + ∠q = 180°
2∠p - ∠q = 431° - 180°
2∠p - ∠q = 251°
∠q = 2∠p - 251° --- (1)
∠SNP
= ∠SPN
= 2 x (180° - ∠p)
= 360° - 2∠p (Exterior angle of a triangle)
∠q = 180° - (360° - 2∠p) - (360° - 2∠p)
∠q = 180° - 360° + 2∠p - 360° + 2∠p
∠q = 180° - 360° - 360° + 2∠p + 2∠p
∠q = 4∠p - 540° (Angles sum of triangle)
∠q = 4∠p - 540° --- (2)
(2) = (1)
4∠p - 540° = 2∠p - 251°
4∠p - 2∠p= 540° - 251°
2∠p = 289°
∠p
= 289° ÷ 2
= 144.5°
(b)
From (1)
∠q
= 2∠p - 251°
= 289° - 251°
= 38°
Answer(s): (a) 144.5°; (b) 38°