In the diagram above EFGH is a cyclic quadrilat... - JAMB Mathematics 1999 Question

∠EFH = ∠EGH(∠s in same segment)

= 34^o

∠HEG = ∠HFG(∠s in same segment)

= X

also ∠HFG = ∠EHF (alternate ∠s)

But ∠EHG + ∠EFG =180(opp sof a cyclic quad)

42 + x + 34 + x = 180

2x + 76 = 180

2x = 180 - 76

2x = 104

x = 52^o