TP and TQ are the tangents from the external point T of a circle with centre O. If `angle OPQ = 30^(@)` then find the measure of `angle TQP`.

Answer

To solve the problem, we need to find the measure of angle TQP given that angle OPQ is 30 degrees. Here’s how we can approach the solution step by step:

### Step 1: Understand the Geometry
We have a circle with center O, and two tangents TP and TQ drawn from an external point T to points P and Q on the circle. By the properties of tangents, we know that OP and OQ are radii of the circle.

### Step 2: Identify Given Angles
We are given that angle OPQ = 30 degrees.

### Step 3: Identify Right Angles
Since OP and OQ are radii and TP and TQ are tangents, we know that:
- Angle OTP = 90 degrees (tangent to radius)
- Angle OTQ = 90 degrees (tangent to radius)

### Step 4: Use the Triangle Properties
In triangle OPQ, we can find angle OQT:
- Angle OPT + angle OPQ + angle OQP = 180 degrees (sum of angles in a triangle)
- We know angle OPT = 90 degrees and angle OPQ = 30 degrees.

So, we can write:
\[ 90 + 30 + \text{angle OQP} = 180 \]
This simplifies to:
\[ \text{angle OQP} = 180 - 120 = 60 \text{ degrees} \]

### Step 5: Relate Angles in Triangle TQP
Now, we need to find angle TQP. Since TP = TQ (tangents from the same external point), triangle TQP is isosceles. Therefore:
- Angle TQP = angle QTP.

### Step 6: Use the Angle Sum Property
In triangle TQP:
- Angle TQP + angle QTP + angle QPT = 180 degrees.

Since angle QPT = angle OQP = 60 degrees, we have:
\[ \text{angle TQP} + \text{angle TQP} + 60 = 180 \]
Let angle TQP = x:
\[ x + x + 60 = 180 \]
This simplifies to:
\[ 2x + 60 = 180 \]
Subtracting 60 from both sides gives:
\[ 2x = 120 \]
Dividing by 2 gives:
\[ x = 60 \]

### Conclusion
Thus, the measure of angle TQP is 60 degrees.