Question

Given


\[\qquad \overline{PQ}\perp\overline{PS}\]
\[\qquad m \angle QPR = 7x - 9^\circ\]
\[\qquad m \angle RPS = 4x + 22^\circ\]
Find
\[m\angle QPR\]:
A right angle QPS. The sum of angle QPR and angle RPS equals angle QPS.
\[P\]
\[Q\]
\[S\]
\[R\]
\[{}^{\circ}\]

Answer 1

Given that line segments \(\overline{PQ}\) and \(\overline{PS}\) are perpendicular, we know that \(m \angle QPS = 90^\circ\).

According to the problem, we have:

\[
m \angle QPR = 7x - 9^\circ
\]
\[
m \angle RPS = 4x + 22^\circ
\]

We also know that the sum of angles \(m \angle QPR\) and \(m \angle RPS\) equals \(90^\circ\):

\[
m \angle QPR + m \angle RPS = m \angle QPS
\]

Substituting the expressions for the angles into the equation gives:

\[
(7x - 9) + (4x + 22) = 90
\]

Now, let's combine like terms:

\[
7x + 4x - 9 + 22 = 90
\]

This simplifies to:

\[
11x + 13 = 90
\]

Next, we will isolate \(11x\) by subtracting \(13\) from both sides:

\[
11x = 90 - 13
\]
\[
11x = 77
\]

Now, dividing both sides by \(11\) gives:

\[
x = \frac{77}{11} = 7
\]

With \(x\) found, we can now substitute it back into the expression for \(m \angle QPR\):

\[
m \angle QPR = 7x - 9
\]

Substituting \(x = 7\):

\[
m \angle QPR = 7(7) - 9
\]
\[
m \angle QPR = 49 - 9
\]
\[
m \angle QPR = 40^\circ
\]

Thus, the measure of angle \(QPR\) is

\[
\boxed{40^\circ}
\]