In the diagram, $O$ is the center of the circle, and $\overline{PQ}$ and $\overline{RS}$ are tangents to the circle. Find the value of $(m+n)$.

1. Problem Description

In the diagram, $O$ is the center of the circle, and $\overline{PQ}$ and $\overline{RS}$ are tangents to the circle. Find the value of $(m+n)$.

2. Solution Steps

Let the point where the tangent $\overline{PQ}$ touches the circle be $A$, and the point where the tangent $\overline{RS}$ touches the circle be $B$. Since $\overline{PQ}$ and $\overline{RS}$ are tangents to the circle at $A$ and $B$ respectively, the radii $\overline{OA}$ and $\overline{OB}$ are perpendicular to the tangents at those points. Thus, $\angle OAP = 90^\circ$ and $\angle OBS = 90^\circ$.

Consider the quadrilateral formed by the points $A, O, B$, and the intersection of the tangents. Let the intersection be denoted by $C$. The angles in the quadrilateral $AOBC$ sum to $360^\circ$. We have $\angle OAP = 90^\circ$ and $\angle OBS = 90^\circ$, so $\angle AOB + \angle ACB = 360^\circ - 90^\circ - 90^\circ = 180^\circ$.

Since the measure of the central angle is twice the measure of the inscribed angle subtended by the same arc, we have $\angle AOB = 2n$.

Also, the sum of angles in the triangle with angle $m$ is 180 degrees. Since $\overline{OA}$ and $\overline{OB}$ are radii, the triangle with angle $m$ is an isosceles triangle, with two equal angles of measure $m$. Thus, $\angle AOB + m + m = 180^\circ$, or $\angle AOB + 2m = 180^\circ$.

Therefore, $2n + 2m = 180^\circ$, which simplifies to $n + m = 90^\circ$.

3. Final Answer

B. $90^\circ$