We are given a circle with center $O$. A line is tangent to the circle at point $K$. Point $T$ is on the circle such that $\angle TOK = 30^\circ$. Point $S$ is located outside the circle such that $SK$ is tangent to the circle at $K$ and $TS$ intersects the circle at $T$. We are asked to find the measure of $\angle S$.

1. Problem Description

We are given a circle with center $O$. A line is tangent to the circle at point $K$. Point $T$ is on the circle such that $\angle TOK = 30^\circ$. Point $S$ is located outside the circle such that $SK$ is tangent to the circle at $K$ and $TS$ intersects the circle at $T$. We are asked to find the measure of $\angle S$.

2. Solution Steps

Since $SK$ is tangent to the circle at $K$, $\angle OKS = 90^\circ$.

Since $OT$ and $OK$ are radii of the circle, $OT = OK$. Therefore, $\triangle OTK$ is an isosceles triangle.

Thus, $\angle OTK = \angle OKT$.

In $\triangle OTK$, the sum of the angles is $180^\circ$. So, $\angle OTK + \angle OKT + \angle TOK = 180^\circ$.

Since $\angle OTK = \angle OKT$ and $\angle TOK = 30^\circ$, we have $2 \angle OTK + 30^\circ = 180^\circ$.

Then, $2 \angle OTK = 150^\circ$, so $\angle OTK = 75^\circ$.

Now, consider $\triangle TKS$. We have $\angle TKS = \angle OKS - \angle OKT = 90^\circ - 75^\circ = 15^\circ$.

We also have $\angle KTS = \angle OTK = 75^\circ$.

The sum of angles in $\triangle TKS$ is $180^\circ$. So, $\angle KTS + \angle TKS + \angle S = 180^\circ$.

$75^\circ + 15^\circ + \angle S = 180^\circ$.

$90^\circ + \angle S = 180^\circ$.

Therefore, $\angle S = 180^\circ - 90^\circ = 90^\circ$.

3. Final Answer

The measure of $\angle S$ is $90^\circ$.