If PQ is a tangent of a circle (touches the circle at Q) with centre…

The tangent at any point makes an angle of 90 degrees with the radius.

Therefore, \(\angle\)OQP = 90   (Since OQ is the radius and PQ is the tangent at Q.)

We need to find the value of \(\angle\)OPQ.

However, we know that the sum of the angles in a triangle equals 180 degrees.

So, we have \(\angle\)OPQ + \(\angle\)OQP + \(\angle\)POQ = 180

Substituting the given values into the equation, we get

\(\angle\)OPQ + 90 + 75 = 180

\(\angle\)OPQ = 180 - 90 - 75

= \(\angle\)OPQ = \({15}^\circ\)

Therefore, \(\angle\)OPQ is \({15}^\circ\).