The sides PQ and PR of $\triangle PQR $ are produced to points S and T, respectively. The bisectors of $\angle SQR$ and $\angle TRQ$ meet at U. If $\angle QUR = 59°$, then the measure of $\angle P$ is:

62°

As we know,

\(\angle\)PQR + \(\angle\)RQS = 180

\(\angle\)RQS = 180 - \(\angle\)PQR

\(\angle\)RQS/2 = 90 - \(\angle\)PQR/2

\(\angle\)RQU = 90 - \(\angle\)PQR/2

Similarly,

\(\angle\)QRU = 90 - \(\angle\)PRQ/2

In \(\Delta \)QUR

\(\angle\)RQU + \(\angle\)QRU + \(\angle\)QUR = 180

\(\angle\)QUR = 180 - (\(\angle\)RQU + \(\angle\)QRU)

\(\angle\)QUR = 180 - [180 - (\(\angle\)PQR + \(\angle\)PRQ)/2]

\(\angle\)QUR = 180 - [180 -(180 - \(\angle\)QPR)/2]

59 = 90 - \(\angle\)QPR/2

\(\angle\)QPR/2 = 90 - 59 = 11

\(\angle\)QPR = 31 x 2 = 62

Therefore, \(\angle\)QPR = \({62}^\circ\).