Question

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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 segments joining the points of the contact to the centre.

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Answer 1

Step-by-step explanation:

Given:

• A circle withe centre O.
• Two tangents are drawn from an external point.

To Prove:

• ∠APB + ∠BOA = 180°

Proof: Let in the circle with centre O. We have

• AP = BP {tangents}
• OA = OB {radius}

As we know that the tangents to a circle is perpendicular to the radius through the point of contact.

Therefore,

PA ⟂ OA

∴ ∠OAP = 90°

PB ⟂ OB

∴ ∠OBP = 90°

$\implies{\rm }$ ∠OBP + ∠OAP

$\implies{\rm }$ 90° + 90°

$\implies{\rm }$ 180°

Now in quadrilateral AOBP.

• Sum of all angles of a quadrilateral is 360°

➯ ∠OAP + ∠OBP + ∠AOB + ∠BPA = 360°

➯ 180° + ∠AOB + ∠BPA = 360°

➯ ∠AOB + ∠BPA = 360° – 180°

➯ ∠AOB + ∠BPA = 180°

Supplementary angles measure 180°.

$\large\bold{\texttt {Proved }}$

Answer 2

Answer:

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⠀⠀⠀⠀$\rule{100}{1}$

Draw a circle with center O and take a external point P. PA and PB are the tangents.

As radius of the circle is perpendicular to the tangent.

OA ⊥ PA

Similarly OB ⊥ PB

• ∠OBP = 90°
• ∠OAP = 90°

$\underline{\bigstar\:\textsf{In Quadrilateral OAPB, sum of all interior angles :}}$

⇒ ∠OAP + ∠OBP + ∠BOA + ∠APB = 360°

⇒ 90° + 90° + ∠BOA + ∠APB = 360°

⇒ 180° + ∠BOA + ∠APB = 360°

⇒ ∠BOA + ∠APB = 360° - 180°

⇒ ∠BOA + ∠APB = 180°

It proves the angle between the two tangents drawn from an external point to a circle supplementary to the angle subtented by the line segment.