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
- ∠w
- ∠x
(a)
∠HSL = 180° - ∠w (Interior angles, SH//RK)
∠NLS = 180° - ∠w (Angles on a straight line)
∠NSL = 180° - ∠w (Isosceles triangle)
51° + 180° - ∠w + 180° - ∠w + ∠x + 18° = 180° (Angles on a straight line, GR)
51° + 180° + 180° + 18° - ∠w - ∠w + ∠x = 180°
429° - 2∠w + ∠x = 180°
2∠w - ∠x = 429° - 180°
2∠w - ∠x = 249°
∠x = 2∠w - 249° --- (1)
∠SNP
= ∠SPN
= 2 x (180° - ∠w)
= 360° - 2∠w (Exterior angle of a triangle)
∠x = 180° - (360° - 2∠w) - (360° - 2∠w)
∠x = 180° - 360° + 2∠w - 360° + 2∠w
∠x = 180° - 360° - 360° + 2∠w + 2∠w
∠x = 4∠w - 540° (Angles sum of triangle)
∠x = 4∠w - 540° --- (2)
(2) = (1)
4∠w - 540° = 2∠w - 249°
4∠w - 2∠w= 540° - 249°
2∠w = 291°
∠w
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠x
= 2∠w - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°