Question : PT is a tangent at the point R on a circle with centre O. SQ is a diameter, which when produced meets the tangent PT at P. If $\angle$SPT = 32$^\circ$, then what will be the measure of $\angle$QRP?
Option 1: $58^\circ$
Option 2: $30^\circ$
Option 3: $29^\circ$
Option 4: $32^\circ$
Correct Answer: $29^\circ$
Solution : Given, $\angle SPT = 32^\circ$ $OR$ is perpendicular to $PT$, so $\angle ORP = 90^\circ$ Let $\angle QRP = \theta$ As we know, the sum of two interior opposite angles of a triangle is equal to its exterior angle. So, $\angle OQR = 32^\circ + \theta$ ⇒ $\angle OQR = \angle ORQ = 32^\circ + \theta$ [same radii] Now, $\angle ORP = \angle ORQ + \angle QRP$ ⇒ $90^\circ = 32^\circ + \theta + \theta$ ⇒ $2\theta = 90^\circ – 32^\circ = 58^\circ$ ⇒ $\theta = \frac{58^\circ}{2} = 29^\circ$ So, $\angle QRP = 29^\circ$ Hence, the correct answer is $29^\circ$.
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