Find x in the first circle.

Let's analyze the first circle. We are given that the angle \(\angle MSN = 40^{\circ}\). Also, we know that \(OM\) and \(ON\) are radii of the circle, and \(MN\) is a chord. The angle \(\angle MSN\) is an inscribed angle subtended by the arc \(MN\). The central angle \(\angle MON\) subtended by the same arc \(MN\) is twice the inscribed angle \(\angle MSN\). Therefore: $$\angle MON = 2 \cdot \angle MSN = 2 \cdot 40^{\circ} = 80^{\circ}$$ Since \(OM = ON\) (both are radii), triangle \(\triangle MON\) is an isosceles triangle. Therefore, \(\angle OMN = \angle ONM\). The sum of angles in triangle \(\triangle MON\) is \(180^{\circ}\), so: $$\angle OMN + \angle ONM + \angle MON = 180^{\circ}$$ Since \(\angle OMN = \angle ONM\), we can write: $$2 \cdot \angle OMN + 80^{\circ} = 180^{\circ}$$ $$2 \cdot \angle OMN = 180^{\circ} - 80^{\circ} = 100^{\circ}$$ $$\angle OMN = \frac{100^{\circ}}{2} = 50^{\circ}$$ Now, consider the quadrilateral \(OMSN\). The sum of angles in a quadrilateral is \(360^{\circ}\). Therefore: $$\angle MSN + \angle MNS + \angle NOM + \angle SMO = 360^{\circ}$$ However, this approach doesn't seem to directly lead to finding \(x\). Let's look at the inscribed angle subtended by the same arc. The angle \(x\) is an inscribed angle that intercepts the same arc as the angle \(\angle MSN = 40^{\circ}\). Thus, these angles must be equal if they both intercept the same arc. However, x is the angle \(\angle MNS\). Since triangle \(\triangle MON\) is an isosceles triangle, we can calculate \(\angle OMN = \angle ONM = 50^{\circ}\). Looking at \(\angle MNS = x\), it is formed by the chord \(MN\) and the side \(SN\). Therefore \(x = \angle MNS\). Since angle \(MNS\) is inscribed and intercepts arc \(MS\), and angle \(MSN\) is inscribed and intercepts arc \(MN\). The angle \(\angle SMN\) intercepting the arc \(SN\) is equal to 90 degrees, then the opposite angle \(x = \angle MNS\) also should be equal to 90. Considering triangle \(\triangle SMN\), since \(SM\) passes through the center of the circle \(O\), \(SM\) is a diameter of the circle, therefore \(\angle SNM\) is a right angle, and \(\angle SNM = 90^{\circ}\). Thus \(x = 90^{\circ}\). Answer: 90