In the figure given, what is $\angle CBA$?

    (A)${30^ \circ }$
    (B)${45^ \circ }$
    (C)${50^ \circ }$
    (D)${60^ \circ }$

Hint: Sum of all the angles of a quadrilateral is ${360^ \circ }$
All the angles of a triangle is having sum equal to ${180^ \circ }$
If there are two angles at a point on a straight line then, total sum of both the angles is ${180^ \circ }$
The opposite angles of two intersecting lines are equal to each other.

Complete step-by-step answer: In this question we are given with:
$\begin{gathered}
  \angle ARQ = {30^ \circ } \\
\end{gathered} $
At point $C$, $AR$ is a straight line.
Therefore,
$\angle ACB + \angle BCR = {180^ \circ }$
$ \Rightarrow $${75^ \circ } + \angle BCR = {180^ \circ }$
     $\angle BCR = {180^ \circ } - {75^ \circ }$
    We get, $\angle BCR = {105^ \circ }$
Also
$\angle BCR = \angle QCR = {105^ \circ }$ ( Same angle)
Now in $\vartriangle CRQ$, $\angle CRQ = {30^ \circ }$
We know that sum of all the angles of a triangle is ${180^ \circ }$
Therefore,
$\angle CQR + \angle CRQ + \angle QCR = {180^ \circ }$
Now, $\angle CQR + {30^ \circ } + {105^ \circ } = {180^ \circ }$ ($\angle BCR = \angle QCR = {105^ \circ }$,Same angle)
          $\angle CQR = {180^ \circ } - {135^ \circ }$
So we get, $\angle CQR = {45^ \circ }$$....1$
At point $Q$, $PR$ is a straight line therefore,
$\angle CQR + \angle CQP = {180^ \circ }$
Now, from equation $1$
$\angle CQR = {45^ \circ }$
$\therefore $ ${45^ \circ } + \angle CQP = {180^ \circ }$
     $\angle CQP = {180^ \circ } - {45^ \circ }$
$ \Rightarrow $ $\angle CQP = {135^ \circ }$
Now, taking the $APQC$ is quadrilateral and its all the corners touch the circle.
Therefore sum of the opposite angles is always equal to ${180^ \circ }$
Hence, $\angle CQP + \angle PAC = {180^ \circ }$
             ${135^ \circ } + \angle PAC = {180^ \circ }$
$ \Rightarrow $ $\angle PAC = {180^ \circ } - {135^ \circ }$
We get, $\angle PAC = {45^ \circ }$
Now by taking triangle $\vartriangle ABC$
We know sum of all the angles of a triangle is ${180^ \circ }$
So, $\angle BAC + \angle ACB + \angle CBA = {180^ \circ }$
$ \Rightarrow $ ${45^ \circ } + {75^ \circ } + \angle CBA = {180^ \circ }$
         $\angle CBA = {180^ \circ } - {120^ \circ }$
We get, $\angle CBA = {60^ \circ }$
So, $\angle CBA = {60^ \circ }$
So option D is the correct option.

Note: In this question one should make mistakes while putting values of angles and assigning names to angles. Also one should have to take care while calculating values of unknown angles. One should have to remember all the properties of a triangle, quadrilaterals etc.