Answer:
![64^\circ](https://img.qammunity.org/2021/formulas/mathematics/high-school/pigg5r9tso19ihf3c2vjyp0lzq134nuccl.png)
Explanation:
Please refer to the attached figure for the labeling and construction in the given figure:
Given that minor angle of arc AB is
.
Or in other words, we can say that angle subtended on the center O by the arc is
.
Now, PA and PB are the tangents so, if we join the center of circle O with A and B, the angles formed are right angles.
i.e.
![\angle PAO = 90^\circ\\\angle PBO = 90^\circ](https://img.qammunity.org/2021/formulas/mathematics/high-school/588pd5em4pq2w9akjss8z26v65zkr2z3n7.png)
Now, we know that sum of internal angles of a quadrilateral is equal to
.
Here, we have the quadrilateral AOPB.
![\therefore \angle PAO +\angle PBO+\angle AOB +\angle APB=360^\circ\\\Rightarrow 90+90+116+x=360^\circ\\\Rightarrow x = 360 - 180 - 116\\\Rightarrow x = 64^\circ](https://img.qammunity.org/2021/formulas/mathematics/high-school/pq892hrn490fvb2tmcr73aj3b63t7k273f.png)
Hence, the correct answer is:
![x = 64^\circ](https://img.qammunity.org/2021/formulas/mathematics/high-school/79oldep8vro6l3ktdmv2a5txmeiysh3i9b.png)