Answer :
Both tangents [tex]$\overline{QP}$[/tex] and [tex]$\overline{QR}$[/tex] create right angles with the radius drawn from the point of tangency. These two right angles combined form a straight angle, meaning [tex]$\angle PCQ = 180^\circ$[/tex]. Subtracting the known [tex]$\angle PCR$[/tex], we find [tex]\angle PQR[/tex] = 60°.
Since [tex]\overline{QP}[/tex] and [tex]\overline{QR}[/tex] are tangents to circle C, we know that [tex]\angle PQC[/tex] and [tex]\angle RQC[/tex] are right angles. Therefore, together they form [tex]\angle PCQ[/tex], which is a straight angle and so measures 180°. Since [tex]\angle PCR = 120^\circ[/tex], we have [tex]m\angle PCQ - m\angle PCR = m\angle PQR[/tex], so [tex]m\angle PQR = 180^\circ - 120^\circ = 60°[/tex].
You can also see this solution geometrically by constructing the circle's diameter through P and Q. Since a triangle's angle measures add up to 180°, we have m\angle QPR = 180° - (90° + 90°) = 0°. Therefore, [tex]\angle PQR[/tex] and [tex]\angle QPR[/tex] are supplementary angles, so [tex]m\angle PQR + m\angle QPR = 180^\circ[/tex], which means [tex]m\angle PQR = 180^\circ - m\angle QPR = 180^\circ - 0^\circ = 60°[/tex].
