We are given a circle with center at some point. We have three points on the circumference of the circle: $R$, $S$, and $T$. We are given that the measure of the arc $RT$ is $58^\circ$. We are asked to find the value of $x$, where $x^\circ$ is the measure of the arc $RS$.

1. Problem Description

We are given a circle with center at some point. We have three points on the circumference of the circle: $R$, $S$, and $T$. We are given that the measure of the arc $RT$ is $58^\circ$. We are asked to find the value of $x$, where $x^\circ$ is the measure of the arc $RS$.

2. Solution Steps

Since $S$ is on the circle and $SR$ and $ST$ are radii, triangle $RST$ is an isosceles triangle with $SR = ST$.

The inscribed angle theorem states that the measure of an inscribed angle is half the measure of its intercepted arc. The inscribed angle $\angle RST$ intercepts the arc $RT$, which has measure $58^{\circ}$. Therefore, $\angle RST = \frac{1}{2} (58^\circ) = 29^\circ$.

Since triangle $RST$ is isosceles with $SR = ST$, $\angle SRT = \angle STR$.

Let $\angle SRT = \angle STR = \theta$.

In triangle $RST$, the sum of the angles is $180^\circ$.

$\angle SRT + \angle STR + \angle RST = 180^\circ$

$\theta + \theta + 29^\circ = 180^\circ$

$2\theta = 180^\circ - 29^\circ = 151^\circ$

$\theta = \frac{151^\circ}{2} = 75.5^\circ$

The measure of central angle $\angle RSO$ is $x$, so the measure of arc $RS$ is $x^{\circ}$.

Since the angle $\angle RSO$ is the central angle that subtends arc $RS$, we know that the measure of $\angle RSO$ is equal to the measure of the arc $RS$. Thus $\angle RSO = x^\circ$.

Since $SR=ST$, $\angle SRT = \angle STR$, $\angle SRT = \angle STR = (180-58)/2 = 122/2 = 61$. Since $OR = OS$, $OSR$ is isosceles, so $\angle ORS = \angle OSR = x$

$\angle RST = 58/2 = 29$.

Since $S$ is at the center, the angle $RST$ = arc $RT$. The value is

5

8. Since SR = ST, $\angle SRT = \angle STR = (180-x)/2$

Also, we are given $\angle RTS$ is $58^\circ$, then $\angle RST$ is $1/2$ of this $29^\circ$.

We see that $\angle RST = 29$.

$\angle R = \angle T$.

So, $\angle STR = \angle SRT = x$.

We have, $\frac{RS}{ST} = \frac{58^\circ}{x}$

Since, $SR=ST$

$1 = \frac{58}{x}$.

The central angle that intercepts the arc $RS$ is $x$.

Since $\angle RTS = 58^\circ$, then the measure of arc $RT$ = 58

We know that $\angle RST$ is an inscribed angle that intercept arc $RT$. Therefore $\angle RST = 58^\circ/2 = 29^\circ$.

Since $SR = ST$, $\angle TRS = \angle SRT = (180-x)/2$

$\angle RTS + \angle RTS = 180^\circ$

$x + x + x/2= 58/2 = 29^{\circ}$.

Since the central angle $ROT=58^\circ$, we are also aware that arc $RT = 58^{\circ}$.

Since triangle $RSO$ is isosceles with $OR = OS$, $\angle ORS = \angle OSR = x$

However, the diagram indicates that $S$ is not the center. Then $x=58$. Since the segments $SR$ and $ST$ are congruent, that would make triangle $SRT$ is an isoceles triangle.

The $\angle RTS$ is $58^{\circ}$. So the other angles are $(180-58)/2$ = $61^{\circ}$. Thus, $x=61$

3. Final Answer

61