Answer:
59°
Step-by-step explanation:
According to the given description, [tex] \angle QRS[/tex] is the exterior angle of [tex] \triangle PQR[/tex]
So, by exterior angle property of a triangle, we have:
[tex] m\angle QRS=m\angle RPQ+m\angle PQR\\\\
(10x - 1)\degree = (3x +17)\degree+(2x +12)\degree\\\\
(10x - 1)\degree = (3x +17+2x +12)\degree\\\\
(10x - 1)\degree = (5x +29)\degree\\\\
(10x - 1)= (5x +29)\\\\
10x - 5x = 29 + 1\\\\
5x = 30\\\\
x = \frac{30}{5}\\\\
x = 6\\\\
\because m\angle QRS = (10x - 1)\degree \\\\
\therefore m\angle QRS = (10\times 6- 1)\degree \\\\
\therefore m\angle QRS = (60- 1)\degree \\\\
\huge\orange{\boxed{\therefore m\angle QRS = 59\degree }}\\
[/tex]
