WAEC Mathematics 2014 Theory — Question 2

Model answer

Draw a line through B parallel to EF and GH. Corresponding angles: the part of ∠ABC nearest EF equals ∠AEF = 3x°. So the remaining part, ∠NBH = 120°−3x° (alternate angle to ∠GHB). Also, by vertically opposite/co-interior relations at H, using ∠CHG=7x°: ∠GHB=180°−7x°. Equating: 180−7x = 120−3x ⟹ 60=4x ⟹ x=15. ∠GHB = 120−3(15) = 120−45 = 75°.