HKL is an equilateral triangle, BCDE is a rectangle and KLFG is a trapezium. HKG and HLF are straight lines. If ∠s = 77°, find

1. ∠KGF
2. Sum of ∠n, ∠p, ∠q and ∠r.

(a)
∠HKL = 60° (Equilateral triangle)

∠KGF = ∠HKL = 60° (Corresponding angles)

(b)
∠BCD = 90° (Right angle)

∠CBE = 90° (Right angle)

∠t + ∠u 
= 180° - 90°
= 90° (Angles sum of triangle)

∠v + ∠w
= 180° - 90°
= 90° (Angles sum of triangle)

∠n + ∠s + ∠t = 180° (Angles on a straight line)

∠p + ∠u = 180° (Angles on a straight line)

∠q + ∠v = 180° (Angles on a straight line)

∠r + ∠w = 180° (Angles on a straight line)

Sum of ∠n, ∠p, ∠q, ∠r and ∠s
= (∠n + ∠p + ∠q + ∠r + ∠s + ∠t + ∠u + ∠v + ∠w) - (∠t + ∠u  + ∠v + ∠w)
= (4 x 180°) - (2 x 90°)
= 720° - 180°
= 540°

Sum of ∠n, ∠p, ∠q and ∠r
= 540° - 77° 
= 463°

Answer(s): (a) 60°; (b) 463°