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
- ∠k
- ∠m
(a)
∠HSL = 180° - ∠k (Interior angles, SH//RK)
∠NLS = 180° - ∠k (Angles on a straight line)
∠NSL = 180° - ∠k (Isosceles triangle)
53° + 180° - ∠k + 180° - ∠k + ∠m + 16° = 180° (Angles on a straight line, GR)
53° + 180° + 180° + 16° - ∠k - ∠k + ∠m = 180°
429° - 2∠k + ∠m = 180°
2∠k - ∠m = 429° - 180°
2∠k - ∠m = 249°
∠m = 2∠k - 249° --- (1)
∠SNP
= ∠SPN
= 2 x (180° - ∠k)
= 360° - 2∠k (Exterior angle of a triangle)
∠m = 180° - (360° - 2∠k) - (360° - 2∠k)
∠m = 180° - 360° + 2∠k - 360° + 2∠k
∠m = 180° - 360° - 360° + 2∠k + 2∠k
∠m = 4∠k - 540° (Angles sum of triangle)
∠m = 4∠k - 540° --- (2)
(2) = (1)
4∠k - 540° = 2∠k - 249°
4∠k - 2∠k= 540° - 249°
2∠k = 291°
∠k
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠m
= 2∠k - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°