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?
Answer & explanation
Correct answer: option 4
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 = \({29}^\circ\).