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
- ∠n
- ∠p
(a)
∠HSL = 180° - ∠n (Interior angles, SH//RK)
∠NLS = 180° - ∠n (Angles on a straight line)
∠NSL = 180° - ∠n (Isosceles triangle)
48° + 180° - ∠n + 180° - ∠n + ∠p + 15° = 180° (Angles on a straight line, GR)
48° + 180° + 180° + 15° - ∠n - ∠n + ∠p = 180°
423° - 2∠n + ∠p = 180°
2∠n - ∠p = 423° - 180°
2∠n - ∠p = 243°
∠p = 2∠n - 243° --- (1)
∠SNP
= ∠SPN
= 2 x (180° - ∠n)
= 360° - 2∠n (Exterior angle of a triangle)
∠p = 180° - (360° - 2∠n) - (360° - 2∠n)
∠p = 180° - 360° + 2∠n - 360° + 2∠n
∠p = 180° - 360° - 360° + 2∠n + 2∠n
∠p = 4∠n - 540° (Angles sum of triangle)
∠p = 4∠n - 540° --- (2)
(2) = (1)
4∠n - 540° = 2∠n - 243°
4∠n - 2∠n= 540° - 243°
2∠n = 297°
∠n
= 297° ÷ 2
= 148.5°
(b)
From (1)
∠p
= 2∠n - 243°
= 297° - 243°
= 54°
Answer(s): (a) 148.5°; (b) 54°