[tex]\text{AD is a bisector of angle BAC, so}\\\\\angle DAC = \angle BAD = 17^{\circ}~ \text{and}~ \angle BAC = 17^{\circ} + 17^{\circ}=34^{\circ}\\\\\text{In}~ \triangle ADB\\\\~~~~~~~\tan \angle BAD = \dfrac{DB}{AB}\\\\\implies DB = AB\tan \angle BAD \\\\~~~~~~~~~~~~=21 \tan 17^{\circ}\\\\[/tex]
[tex]\text{In}~ \triangle ABC\\\\~~~~~~~\tan \angle BAC = \dfrac{CB}{AB}\\\\\implies BC = AB \tan \angle BAC\\\\\implies DB+CD = AB\tan \angle BAC\\\\\implies CD = AB \tan \angle BAC - DB\\\\\implies CD=21\tan 34^{\circ}-21 \tan 17^{\circ}\\\\\implies x\approx8 ~ \text{cm}[/tex]
