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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