Solution:

According to the given question,

\[\angle CAB\text{ }=\text{ }{{75}^{o}}~and\angle CBA\text{ }=\text{ }{{50}^{o}}\]

In \[\vartriangle ABC\], by angle sum property we have

\[\angle ACB\text{ }=\text{ }{{180}^{o}}~\text{ }(\angle CBA\text{ }+\angle CAB)\]

\[=\text{ }{{180}^{o}}~\text{ }({{50}^{o}}~+\text{ }{{75}^{o}})\text{ }=\text{ }{{180}^{o}}-\text{ }{{125}^{o}}\]

\[=\text{ }{{55}^{o}}\]

And,

\[\angle ADB\text{ }=\angle ACB\text{ }=\text{ }{{55}^{o}}\]

[Angles subtended by the same chord on the circle are equal]

Now, taking ∆ABD

\[\angle DAB\text{ }+\angle ABD\text{ }+\angle ADB\text{ }=\text{ }{{180}^{o}}\]

\[\angle DAB\text{ }+\angle ABD\text{ }+\text{ }{{55}^{o}}~=\text{ }{{180}^{o}}\]

\[\angle DAB\text{ }+\angle ABD\text{ }=\text{ }{{180}^{o}}-\text{ }{{55}^{o}}\]

\[\angle DAB\text{ }+\angle ABD\text{ }=\text{ }{{125}^{o}}\]