The given figures shows a circle with centre O. Also , PQ = QR = RS and ` angle PTS = 75^(@)` Calculate:
(i) ` agnle POS `
(ii) ` angle QOR `
(iii) ` angle PQR `
` (##SEL_RKB_ICSE_MAT_X_C17_E03_004_Q01.png" width="80%">

Answer

To solve the problem step by step, we will calculate the angles as requested.

### Step 1: Understand the Given Information
We have a circle with center O. The chords PQ, QR, and RS are equal, and we are given that the angle PTS = 75 degrees.

### Step 2: Calculate Angle POS
Using the property of circles, we know that the angle at the center (angle POS) is twice the angle at the circumference (angle PTS).

\[ \text{Angle POS} = 2 \times \text{Angle PTS} = 2 \times 75^\circ = 150^\circ \]

### Step 3: Calculate Angle QOR
Since PQ = QR = RS, the angles subtended by these chords at the center (angle POQ, angle QOR, and angle ROS) are equal. Let each of these angles be denoted as X.

From the previous step, we know that:
\[ \text{Angle POS} = \text{Angle POQ} + \text{Angle QOR} + \text{Angle ROS} = 150^\circ \]
Since all three angles are equal:
\[ X + X + X = 150^\circ \implies 3X = 150^\circ \implies X = 50^\circ \]
Thus,
\[ \text{Angle QOR} = 50^\circ \]

### Step 4: Calculate Angle PQR
In triangle OPQ, since OP = OQ (both are radii of the circle), we have an isosceles triangle. Therefore, the angles at P and Q are equal. Let these angles be denoted as Y.

Using the triangle sum property:
\[ \text{Angle OPQ} + \text{Angle OQP} + \text{Angle POQ} = 180^\circ \]
Substituting the known values:
\[ Y + Y + 50^\circ = 180^\circ \implies 2Y = 130^\circ \implies Y = 65^\circ \]
Thus,
\[ \text{Angle PQR} = 65^\circ \]

### Summary of Results
(i) Angle POS = 150 degrees
(ii) Angle QOR = 50 degrees
(iii) Angle PQR = 65 degrees