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# Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre

**Solution:**

- Let us consider O as the centre point of the circle.
- Let P be a point outside the circle from which two tangents PA and PB are drawn to the circle which touches the circle at point A and B respectively.
- Draw a line segment between points A and B such that it subtends ∠AOB at centre O of the circle.

According to Theorem 10.1: The tangent at any point of a circle is always perpendicular to the radius through the point of contact.

∴ ∠OAP = ∠OBP = 90° --- Equation (i)

In a quadrilateral, the sum of interior angles is 360°.

∴ In OAPB,

∠OAP + ∠APB + ∠PBO + ∠BOA = 360°

Using Equation (i), we can write the above equation as

90° + ∠APB + 90° + ∠BOA = 360°

∠APB + ∠BOA = 360° - 180°

∴ ∠APB + ∠BOA = 180°

Where,

∠APB = Angle between the two tangents PA and PB from external point P.

∠BOA = Angle subtended by the line segment AB joining the point of contacts at the centre.

Hence, proved the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 10

**Video Solution:**

## Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre

NCERT Solutions Class 10 Maths Chapter 10 Exercise 10.2 Question 10

**Summary:**

It has been proved that the angle between the two tangents drawn from an external point to a circle, that is, ∠APB is supplementary to the angle subtended by the line segment joining the point of contact at the centre, that is, ∠AOB. Thus, ∠APB + ∠BOA = 180°.

**☛ Related Questions:**

- Prove that the parallelogram circumscribing a circle is a rhombus.
- A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
- Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
- From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is(A) 7 cm(B) 12 cm(C) 15 cm(D) 24.5 cm

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