Given,

AB divides ∠DAC in the ratio 1: 3

∠DAB: ∠BAC = 1: 3

∠DAC + ∠EAC = 180^o

∠DAC + 108^o = 180^o

∠DAC = 180^o – 108^o

= 72^o

∠DAB = * 72^o = 18^o

∠BAC = * 72^o = 54^o

In

∠DAB + ∠ADB + ∠ABD = 180^o

18^o + 18^o + ∠ABD = 180^o

36^o + ∠ABD = 180^o

∠ABD = 180^o – 36^o

= 144^o

∠ABD + ∠ABC = 180^o (Linear pair)

144^o + ∠ABC = 180^o

∠ABC = 180^o – 144^o

= 36^o

In

∠BAC + ∠ABC + ∠ACB = 180^o

54^o + 36^o + x = 180^o

90^o + x = 180^o

x = 180^o – 90^o

= 90^o

Thus, x = 90^o