From an external point two tangents to a circle are drawn. The chord passing through the points of contact subtends an angle `72^@` at the centre. The angle between the tangents is

Options

A. `36^@`
B. `72^@`
C. `108^@`
D. `144^@`

Answer

To find the angle between the two tangents drawn from an external point to a circle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given information**:
- Let the circle have center O.
- Let the external point from which the tangents are drawn be P.
- The points of contact on the circle are A and B.
- The angle subtended by the chord AB at the center O is given as 72°.

2. **Understand the properties of tangents**:
- The radius at the point of contact is perpendicular to the tangent. Therefore, OA ⊥ PA and OB ⊥ PB.
- This means that angles OAP and OBP are both 90°.

3. **Create a quadrilateral**:
- Consider the quadrilateral OAPB formed by the center O, the external point P, and the points of contact A and B.

4. **Calculate the angles in the quadrilateral**:
- The sum of the angles in a quadrilateral is 360°.
- We have:
- Angle OAP = 90°
- Angle OBP = 90°
- Angle AOB = 72° (given)

5. **Set up the equation**:
- Let the angle between the tangents PA and PB be denoted as ∠APB.
- According to the property of quadrilaterals:
\[ \angle OAP + \angle OBP + \angle AOB + \angle APB = 360° \]
- Substituting the known values:
\[ 90° + 90° + 72° + \angle APB = 360° \]

6. **Simplify the equation**:
- Combine the known angles:
\[ 180° + 72° + \angle APB = 360° \]
- This simplifies to:
\[ 252° + \angle APB = 360° \]

7. **Solve for ∠APB**:
- Rearranging gives:
\[ \angle APB = 360° - 252° = 108° \]

8. **Conclusion**:
- The angle between the tangents PA and PB is 108°.

### Final Answer:
The angle between the tangents is **108°**.

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