Answer
$AD \approx 5.25$ or, $AD=5.3$
Work Step by Step
Need to find the length of AD.
It can be seen that the triangle BDC is an equilateral triangle, so $BD=DC=12$. This means that $\angle {CBD}=\angle {BCD}=25^{\circ}$
$\angle {ACB}=\angle {BCD}+\angle {ACD}$
$\implies \angle {ACB}= 25^{\circ}+25^{\circ}=50^{\circ}$
Now, $\angle {A}=180^{\circ}-(\angle {B}+\angle {C})$
$\implies \angle {A}=180^{\circ}-75^{\circ}=105^{\circ}$
Next, it can be seen that the triangle ACD we can determine the length of the side AD by applying law of sines formula.
$\dfrac{\sin \angle{CAD}}{CD}=\dfrac{\sin \angle{ACD}}{AD}$
or, $\dfrac{\sin 105^{\circ}}{12}=\dfrac{\sin 25^{\circ}}{AD}$
Then, we have $AD=\dfrac{12 \sin 25^{\circ} }{\sin 105^{\circ}}$
This implies that $AD \approx 5.25$ or, $AD=5.3$
