A geometry problem by A Former Brilliant Member

the touching PQ , QR makes 90 degrees with the radius PO,RO then 360 -(QPO + QRO+POR) = PQR THEN 360-(120+90+90) = 60

It should be mentioned that O is the center of circle or OP is the radius because the figure itself is vague.

Consider the diagram. A tangent line to a circle makes a $90^\circ$ angle with the radius at the point of contact. The sum of the interior angles of a quadrilateral is $360^\circ$ . Thus,

$\angle PQR=360-90-90-120=$ $\boxed{60^\circ}$

Tangents make 90 degrees with circle and hence left out angle is 60 degrees

Angle POR = 120° ......(given) the tangent is always perpendicular to the radius, hence, Angle OPQ = 90°, Angle ORQ = 90°, so measure Angle PQR + OPQ + ORQ + PQR = 360...............(sum of measure of angles of a quadrilateral), hence, 120° + 90° + 90° + PQR = 360°, On solving this we get, measure angle PQR = 60°

PQ & RQ are tangents therefore OPQ= 90 & ORQ=90
there fore OPQR is a quadrilateral = 360
PQR= 360 - 120 - 90 - 90
PQR= 60