Find the angle B measure

Let $\angle B=\theta$

$\triangle ِADC$ is Isosceles triangle //since AD=AC

$\Rightarrow \angle ADC=\angle ACD={ 40 }^{ \circ }$

$\Rightarrow \frac { \sin { \left( \angle ADC \right) } }{ \overline { AC } } =\frac { \sin { \left( \angle DAC \right) } }{ \overline { DC } }$ // Sine Rule

$\Rightarrow \frac { \sin { \left( { 40 }^{ \circ } \right) } }{ \overline { AC } } =\frac { \sin { \left( { 100 }^{ \circ } \right) } }{ \overline { DC } } \Rightarrow \frac { \overline { DC } }{ \overline { AC } } =\frac { \sin { \left( { 100 }^{ \circ } \right) } }{ \sin { \left( { 40 }^{ \circ } \right) } }$

$\Rightarrow \frac { \overline { DC } }{ \overline { AC } } =\frac { \sin { \left( { 80 }^{ \circ } \right) } }{ \sin { \left( { 40 }^{ \circ } \right) } } =\frac { 2\sin { \left( { 40 }^{ \circ } \right) } \cos { \left( { 40 }^{ \circ } \right) } }{ \sin { \left( { 40 }^{ \circ } \right) } } =2\cos { \left( { 40 }^{ \circ } \right) } \rightarrow \left( 1 \right)$

In $\triangle ِBDC\quad \angle BDC={ 140 }^{ \circ }$ //supplementary angles

$\Rightarrow \angle BCD={ 40 }^{ \circ }-\theta$

$\angle ACB=\angle BCD+\angle ACD={ 40 }^{ \circ }-\theta +{ 40 }^{ \circ }={ 80 }^{ \circ }-\theta$

Apply Sine Rule for $\triangle ِABC$

$\frac { \sin { \left( \angle ACB \right) } }{ \overline { AB } } =\frac { \sin { \left( \angle B \right) } }{ \overline { AC } } \Rightarrow \frac { \sin { \left( { 80 }^{ \circ }-\theta \right) } }{ \overline { DC } } =\frac { \sin { \left( \theta \right) } }{ \overline { AC } }$

$\Rightarrow \frac { \overline { DC } }{ \overline { AC } } =\frac { \sin { \left( { 80 }^{ \circ }-\theta \right) } }{ \sin { \left( \theta \right) } } \Rightarrow \frac { \overline { DC } }{ \overline { AC } } =\frac { \cos { \left( { 10 }^{ \circ }+\theta \right) } }{ \sin { \left( \theta \right) } }$

$\Rightarrow \frac { \overline { DC } }{ \overline { AC } } =\frac { \cos { \left( \theta \right) } \cos { \left( { 10 }^{ \circ } \right) } -\sin { \left( \theta \right) } \sin { \left( { 10 }^{ \circ } \right) } }{ \sin { \left( \theta \right) } } =\cot { \left( \theta \right) } \cos { \left( { 10 }^{ \circ } \right) } -\sin { \left( { 10 }^{ \circ } \right) } \rightarrow \left( 2 \right)$

from (1) and (2)

$\cot { \left( \theta \right) } \cos { \left( { 10 }^{ \circ } \right) } -\sin { \left( { 10 }^{ \circ } \right) } =2\cos { \left( { 40 }^{ \circ } \right) }$

$\cot { \left( \theta \right) } =\frac { 2\cos { \left( { 40 }^{ \circ } \right) } +\sin { \left( { 10 }^{ \circ } \right) } }{ \cos { \left( { 10 }^{ \circ } \right) } }$

$\theta =\cot ^{ -1 }{ \left( \frac { 2\cos { \left( { 40 }^{ \circ } \right) } +\sin { \left( { 10 }^{ \circ } \right) } }{ \cos { \left( { 10 }^{ \circ } \right) } } \right) }$

$\theta ={ 90 }^{ \circ }-\tan ^{ -1 }{ \left( \frac { 2\cos { \left( { 40 }^{ \circ } \right) } +\sin { \left( { 10 }^{ \circ } \right) } }{ \cos { \left( { 10 }^{ \circ } \right) } } \right) } ={ 30 }^{ \circ }$ //since $\cot ^{ -1 }{ \left( \theta \right) } =\frac { \pi }{ 2 } -\tan ^{ -1 }{ \left( \theta \right) }$

Let AD=1, and BD=x. Use cos rule twice, then DC = 1.532. x=0.532. CB=1.9696. Use sin rule to find <B. sin B= sin 100/1.9696 = 0.5. B=30.