PT is a tangent at the point R on circle with centre O. SQ is a diameter, which when produced meets the tangent PT at P. If ∠SPT = 32°, then what will be the measure of ∠QRP?

29°

Given, \(\angle\)SPT = \({32}^\circ\)

As we know,

OR is perpendicular to PT, so \(\angle\)ORP = \({90}^\circ\)

Suppose \(\angle\)QRP = \(\theta \)

As we know, sum of two interior opposite angles of a triangle is equal to its exterior angle.

\(\angle\)OQR = \({32}^\circ\) + \(\theta \)

\(\angle\)OQR = \(\angle\)ORQ = \({32}^\circ\) + \(\theta \) [same radii]

Now,

\(\angle\)ORP = \(\angle\)ORQ + \(\angle\)QRP

\({90}^\circ\) = \({32}^\circ\) + \(\theta \) + \(\theta \)

2\(\theta \) = \({90}^\circ\) - \({32}^\circ\) = \({58}^\circ\)

\(\theta \) = \(\frac{58}{2}\) = \({29}^\circ\)

Therefore, \(\angle\)QRP = \({29}^\circ\).