What is the value of $x$ ?

Let $\angle BAC=x$ . From, $\angle BAD+\angle ACD=145$ $\color{#D61F06}\implies$ $\angle BAD = 145 - \angle ACD$

Consider $\triangle ACD:$

$\angle CAD+\angle ACD=180-50=130$ $\color{#D61F06}\implies$ $\angle CAD = 130-\angle ACD$

From the figure,

$x=\angle BAD-\angle ACD$

$x=145-\angle ACD- (130-\angle ACD)$

$x=145-\angle ACD - 130 + \angle ACD$

$x=145-130=$ $\boxed{15^\circ}$