Triangle PQR is inscribed in a circle, center at O such that `angleQ = 70^@ and angleR = 55^@`. PT is perpendicular to QR and the line OP touches the side QR at the point S. Find `angleTPS`

Options

A. `15//2^@`
B. `15^@`
C. `20^@`
D. `30^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have triangle PQR inscribed in a circle with center O. We know:
- Angle Q = 70°
- Angle R = 55°
- PT is perpendicular to QR
- OP touches QR at point S

### Step 2: Find angle P
To find angle P, we use the fact that the sum of angles in a triangle is 180°.
\[ \text{Angle P} = 180° - \text{Angle Q} - \text{Angle R} \]
Substituting the values:
\[ \text{Angle P} = 180° - 70° - 55° = 55° \]

### Step 3: Analyze triangle PQT
In triangle PQT, we know:
- Angle PTQ = 90° (since PT is perpendicular to QR)
- Angle PQT = Angle Q = 70°

Now we can find angle QPT:
\[ \text{Angle QPT} = 180° - \text{Angle PTQ} - \text{Angle PQT} \]
Substituting the values:
\[ \text{Angle QPT} = 180° - 90° - 70° = 20° \]

### Step 4: Find angle POR
Angle POR is the central angle corresponding to angle PQR. By the property of circles, the central angle is twice the inscribed angle:
\[ \text{Angle POR} = 2 \times \text{Angle PQR} = 2 \times 70° = 140° \]

### Step 5: Use the isosceles triangle property
In triangle POR, OP = OR (both are radii of the circle). Therefore, angles OPR and ORP are equal. Let’s denote them as x:
\[ \text{Angle OPR} = \text{Angle ORP} = x \]
Using the triangle sum property:
\[ \text{Angle POR} + \text{Angle OPR} + \text{Angle ORP} = 180° \]
Substituting the known values:
\[ 140° + x + x = 180° \]
\[ 2x = 40° \implies x = 20° \]
Thus, Angle OPR = Angle ORP = 20°.

### Step 6: Find angle TPS
Now we will find angle TPS using the angles in triangle PTS:
\[ \text{Angle P} = \text{Angle QPT} + \text{Angle TPS} + \text{Angle SPR} \]
We know:
- Angle P = 55°
- Angle QPT = 20°
- Angle SPR = Angle OPR = 20°

Substituting the values:
\[ 55° = 20° + \text{Angle TPS} + 20° \]
\[ 55° = 40° + \text{Angle TPS} \]
\[ \text{Angle TPS} = 55° - 40° = 15° \]

### Final Answer
Thus, the angle TPS is:
\[ \text{Angle TPS} = 15° \]