Given:
It is given that,
PQ ⊥ PS and
∠QPR = 7x-9
∠RPS = 4x+22
To find the value of ∠QPR.
Formula
As per the given problem PR lies between PQ and PS,
So,
∠QPR+∠RPS = 90°
Now,
Putting the values of ∠RPS and ∠QPR we get,
[tex]7x-9+4x+22 = 90[/tex]
or, [tex]11x = 90-22+9[/tex]
or, [tex]11x = 77[/tex]
or, [tex]x = \frac{77}{11}[/tex]
or, [tex]x = 7[/tex]
Substituting the value of [tex]x = 7[/tex] in ∠QPR we get,
∠QPR = [tex]7(7)-9[/tex]
or, ∠QPR = [tex]40^\circ[/tex]
Hence,
The value of ∠QPR is 40°.