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Since ∠CDA = 70 (from 180 - 110) and ∠CAD = 70, triangle ACD is an isosceles triangle and we have AC = CD.
Since ∠CAB = 110 (from 180 - 70) and ∠ABC = 35, we have ∠ACB = 35, triangle ABC is an isosceles triangle and we have AB = AC.
Thus we have AB = AC = CD.
Then we have x = 3 since AB = 3.

Therefore, A is the answer.
Answer: A

$$From\ \triangle ABC,\ length\ AB=3,\ \angle B=35^0$$
$$\triangle A=180^0-70^0\ =110^0\left(180^0=angle\ on\ a\ straight\ line\right)$$
$$\triangle C=180^0-\left(A^0+B^0\right)$$
$$\triangle C=180^0-\left(110^0+35^0\right)=35^0$$
Triangle ABC is an isosceles triangle with 2 equal angles (definitely 2 of its sides must also be equal)
Let AB=c, AC=b and BC=a
Using the sine rule
$$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$$
$$U\sin g;\ \frac{b}{SinB}=\frac{c}{SinC}=>\frac{b}{Sin35}=\frac{3}{Sin35}$$
$$b=3$$
From triangle ADC, length AC=3, Angle A=70 degrees
Angle D = 180 - 110 = 70
Angle C = 180 - (70+70) = 40
Triangle ADC is an isosceles triangle with 2 equal angles
Let AC=d, CD=x and AD=c
Using the sine rule;
$$\frac{d}{SinD}=\frac{x}{SinA}=\frac{c}{SinC}$$
$$U\sin g;\ \frac{d}{SinD}=\frac{x}{SinA}=>\frac{3}{Sin70}=\frac{x}{Sin70}$$
$$x=3$$

Answer = A