In the figure, PFGL is a trapezium and triangles PLK and KPH are isosceles triangles. NE, NG and EG are straight lines. PL = PK = KH. Find
- ∠j
- ∠k
(a)
∠FPH = 180° - ∠j (Interior angles, PF//NG)
∠KHP = 180° - ∠j (Angles on a straight line)
∠KPH = 180° - ∠j (Isosceles triangle)
58° + 180° - ∠j + 180° - ∠j + ∠k + 24° = 180° (Angles on a straight line, EN)
58° + 180° + 180° + 24° - ∠j - ∠j + ∠k = 180°
442° - 2∠j + ∠k = 180°
2∠j - ∠k = 442° - 180°
2∠j - ∠k = 262°
∠k = 2∠j - 262° --- (1)
∠PKL
= ∠PLK
= 2 x (180° - ∠j)
= 360° - 2∠j (Exterior angle of a triangle)
∠k = 180° - (360° - 2∠j) - (360° - 2∠j)
∠k = 180° - 360° + 2∠j - 360° + 2∠j
∠k = 180° - 360° - 360° + 2∠j + 2∠j
∠k = 4∠j - 540° (Angles sum of triangle)
∠k = 4∠j - 540° --- (2)
(2) = (1)
4∠j - 540° = 2∠j - 262°
4∠j - 2∠j= 540° - 262°
2∠j = 278°
∠j
= 278° ÷ 2
= 139°
(b)
From (1)
∠k
= 2∠j - 262°
= 278° - 262°
= 16°
Answer(s): (a) 139°; (b) 16°