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
- ∠s
- ∠t
(a)
∠HSL = 180° - ∠s (Interior angles, SH//RK)
∠NLS = 180° - ∠s (Angles on a straight line)
∠NSL = 180° - ∠s (Isosceles triangle)
51° + 180° - ∠s + 180° - ∠s + ∠t + 16° = 180° (Angles on a straight line, GR)
51° + 180° + 180° + 16° - ∠s - ∠s + ∠t = 180°
427° - 2∠s + ∠t = 180°
2∠s - ∠t = 427° - 180°
2∠s - ∠t = 247°
∠t = 2∠s - 247° --- (1)
∠SNP
= ∠SPN
= 2 x (180° - ∠s)
= 360° - 2∠s (Exterior angle of a triangle)
∠t = 180° - (360° - 2∠s) - (360° - 2∠s)
∠t = 180° - 360° + 2∠s - 360° + 2∠s
∠t = 180° - 360° - 360° + 2∠s + 2∠s
∠t = 4∠s - 540° (Angles sum of triangle)
∠t = 4∠s - 540° --- (2)
(2) = (1)
4∠s - 540° = 2∠s - 247°
4∠s - 2∠s= 540° - 247°
2∠s = 293°
∠s
= 293° ÷ 2
= 146.5°
(b)
From (1)
∠t
= 2∠s - 247°
= 293° - 247°
= 46°
Answer(s): (a) 146.5°; (b) 46°