We are given a circle with two secants intersecting at point $P$ outside the circle. We are given $m \angle P = (3x)^\circ$, $m \stackrel{\frown}{QT} = 39^\circ$, and $m \stackrel{\frown}{RS} = (7x+33)^\circ$. We need to find the measure of $\angle P$ and the measure of arc $RS$ by substituting $x=6$.

1. Problem Description

We are given a circle with two secants intersecting at point $P$ outside the circle. We are given $m \angle P = (3x)^\circ$, $m \stackrel{\frown}{QT} = 39^\circ$, and $m \stackrel{\frown}{RS} = (7x+33)^\circ$. We need to find the measure of $\angle P$ and the measure of arc $RS$ by substituting $x=6$.

2. Solution Steps

The measure of an angle formed by two secants intersecting outside a circle is one-half the difference of the intercepted arcs. Therefore,

$m \angle P = \frac{1}{2} (m \stackrel{\frown}{RS} - m \stackrel{\frown}{QT})$

Substitute the given expressions:

$(3x)^\circ = \frac{1}{2} ((7x+33)^\circ - 39^\circ)$

Multiply both sides by 2:

$6x = 7x + 33 - 39$

$6x = 7x - 6$

Subtract $7x$ from both sides:

$-x = -6$

$x = 6$

We are given $x=6$ and need to find $m \angle P$ and $m \stackrel{\frown}{RS}$.

$m \angle P = (3x)^\circ = (3 \cdot 6)^\circ = 18^\circ$

$m \stackrel{\frown}{RS} = (7x+33)^\circ = (7 \cdot 6 + 33)^\circ = (42 + 33)^\circ = 75^\circ$

3. Final Answer

$m \angle P = 18^\circ$

$m \stackrel{\frown}{RS} = 75^\circ$