Partially Inscribed Triangle

Construct a radius $BO$ then it says $\angle BOD = 2\angle BCD$ , here denotes as $\angle BCD = x, \angle BOD = 2x$ .

Then easily found that $\angle ADO=106^\circ$ , and $\angle ABO=90^\circ$ , then using the sum of internal angle of quadrilateral gives us $90^\circ+78^\circ+106^\circ+2x=360^\circ\longrightarrow x=\boxed{43^\circ}$ .

We will begin by drawing the chord $BC$ and another tangent from $D$ to meet the original tangent $AB$ at point $E$ .

According to the Alternate Segment Theorem , $\angle ABD = \angle BCA = x$ , as shown in blue.

Also, by Alternate Segment Theorem , $\angle EDB = \angle BCA = x$ , regarding $ED$ as the tangent.

Alternatively, since the lengths of tangents of external point are equal, $BE = ED$ , making $BED$ an isosceles triangle, and so $\angle EDB = \angle BCA = x$ .

Then because the radius $OD$ is perpendicular to $DE$ , $\angle BDO = 90^\circ - x$ .

We can then summarize the equation:

$\angle BAD + \angle ABD = \angle BDO + \angle ODC$

$78 + x = (90 - x) + 74$

$2x = 90 + 74 - 78 = 86$

Thus, $x = \boxed{43 ^\circ}$ .