PT is a tangent at the point R on circle with centre O. SQ is a diameter, which when produced meets the tangent PT at P. If `angleSPT=32^@`, then what will be the measure of `angleQRP`?

Options

A. `32^(@)`
B. `58^(@)`
C. `30^(@)`
D. `29^(@)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Configuration
We have a circle with center O, a tangent PT at point R on the circle, and a diameter SQ. The tangent PT meets the extended diameter SQ at point P. We know that angle SPT = 32°.

### Step 2: Identify Key Angles
Since OR is a radius and PT is a tangent at point R, the angle between the radius (OR) and the tangent (PT) at the point of contact R is 90°. Therefore, we have:
\[ \angle ORP = 90° \]

### Step 3: Use Angle Sum Property in Triangle ORP
In triangle ORP, we have:
- \(\angle ORP = 90°\)
- \(\angle OPR = 32°\) (since angle SPT = angle OPR)

Using the angle sum property of triangles:
\[ \angle ROP + \angle OPR + \angle ORP = 180° \]
Substituting the known values:
\[ \angle ROP + 32° + 90° = 180° \]
\[ \angle ROP + 122° = 180° \]
\[ \angle ROP = 180° - 122° = 58° \]

### Step 4: Analyze Triangle OQR
In triangle OQR, since OQ and OR are both radii of the circle, they are equal:
\[ OR = OQ \]
This means triangle OQR is isosceles, and the angles opposite the equal sides (OR and OQ) are equal:
\[ \angle ORQ = \angle OQR = x \]

### Step 5: Use Angle Sum Property in Triangle OQR
Using the angle sum property in triangle OQR:
\[ \angle OQR + \angle ORQ + \angle ROP = 180° \]
Substituting the known values:
\[ x + x + 58° = 180° \]
\[ 2x + 58° = 180° \]
\[ 2x = 180° - 58° = 122° \]
\[ x = 61° \]

### Step 6: Find Angle QRP
Now, we can find angle QRP using the relationship:
\[ \angle QRP = \angle ORP - \angle ORQ \]
Substituting the known values:
\[ \angle QRP = 90° - 61° = 29° \]

### Final Answer
Thus, the measure of angle QRP is:
\[ \angle QRP = 29° \]