Answer:
![64^\circ](https://tex.z-dn.net/?f=64%5E%5Ccirc)
Step-by-step 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://tex.z-dn.net/?f=%5Cangle%20PAO%20%3D%2090%5E%5Ccirc%5C%5C%5Cangle%20PBO%20%3D%2090%5E%5Ccirc)
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://tex.z-dn.net/?f=%5Ctherefore%20%5Cangle%20PAO%20%2B%5Cangle%20PBO%2B%5Cangle%20AOB%20%2B%5Cangle%20APB%3D360%5E%5Ccirc%5C%5C%5CRightarrow%2090%2B90%2B116%2Bx%3D360%5E%5Ccirc%5C%5C%5CRightarrow%20x%20%3D%20360%20-%20180%20-%20116%5C%5C%5CRightarrow%20x%20%3D%2064%5E%5Ccirc)
Hence, the correct answer is:
![x = 64^\circ](https://tex.z-dn.net/?f=x%20%3D%2064%5E%5Ccirc)