We are given a circle with a tangent line $TP$ and a secant line $RP$. We are given the measure of angle $P$ as $(3x)^{\circ}$, the measure of arc $QT$ as $39^{\circ}$, and the measure of arc $RS$ as $(7x + 33)^{\circ}$. We are asked to find the value of $x$.

1. Problem Description

We are given a circle with a tangent line $TP$ and a secant line $RP$. We are given the measure of angle $P$ as $(3x)^{\circ}$, the measure of arc $QT$ as $39^{\circ}$, and the measure of arc $RS$ as $(7x + 33)^{\circ}$. We are asked to find the value of $x$.

2. Solution Steps

The measure of an angle formed by a tangent and a secant drawn from a point outside the circle is equal to one-half the difference of the intercepted arcs. In this case, angle $P$ is formed by tangent $TP$ and secant $RP$. The intercepted arcs are $RS$ and $QT$. Therefore, we have

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

Substituting the given values, we have

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

$3x = \frac{1}{2}(7x - 6)$

Multiplying both sides by 2, we get:

$6x = 7x - 6$

Subtracting $7x$ from both sides gives:

$-x = -6$

Multiplying both sides by $-1$, we get:

$x = 6$

3. Final Answer

$x = 6$