Find the value of x in the adjoining figure, if it is given that PR and QS are diameters of the circle.

38°

The correct answer is Option (3) → 38°

Step-by-step Solution

1. Analyze the Circle Properties:

Since PR and QS are both diameters of the circle, they intersect at the center of the circle, point O. This means that $OP$, $OQ$, $OR$, and $OS$ are all radii of the circle. Therefore:

$OP = OQ = OR = OS$

2. Identify the Triangle Type:

In triangle $\triangle SOR$, since $OS = OR$, it is an isosceles triangle. Similarly, in triangle $\triangle POQ$, since $OP = OQ$, it is also an isosceles triangle.

3. Use Angle Properties:

• Angles $\angle POS$ and $\angle QOR$ are vertically opposite angles, so they are equal.

• In $\triangle SOR$, the base angles opposite the equal sides are equal. However, we are given $\angle ORQ = 38^\circ$.

• Since $OQ = OR$, triangle $\triangle OQR$ is isosceles. This means the base angles are equal:

$\angle OQR = \angle ORQ = 38^\circ$

4. Find the Central Angle:

In $\triangle OQR$, the sum of angles is $180^\circ$:

$\angle QOR = 180^\circ - (38^\circ + 38^\circ)$

$\angle QOR = 180^\circ - 76^\circ = 104^\circ$

5. Relate to $x$:

• As mentioned, $\angle POS$ and $\angle QOR$ are vertically opposite:

$\angle POS = 104^\circ$

• In triangle $\triangle POS$, since $OP = OS$, it is an isosceles triangle. Therefore, the base angles are equal:

$\angle OPS = \angle OSP = x^\circ$

• Summing the angles in $\triangle POS$:

$x + x + 104^\circ = 180^\circ$

$2x = 180^\circ - 104^\circ$

$2x = 76^\circ$

$x = 38^\circ$