The angle between two tangents drawn from an external point to a circle with centre O is 50 degree

If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60°, then find the length of OP

Let PA and PB be the two tangents drawn to the circle with centre O and radius a such that ∠APB=60°

In ∆OPB and ∆OPA

OB = OA = a (Radii of the circle)

∠OBP = ∠OAP=90° (Tangents are perpendicular to radius at the point of contact)

BP = PA (Lengths of tangents drawn from an external point to the circle are equal)

So, ∆OPB ≌ ∆OPA (SAS Congruence Axiom)

∴ ∠OPB = ∠OPA=30° (CPCT)

Now,

In ∆OPB

`sin 30^@ = "OB"/"OP"`

`=> 1/2 = a/(OP)`

`=> OP= 2a`

Thus the length of OP is 2a

Concept: Tangent to a Circle

  Is there an error in this question or solution?

Solution:

Let's draw a figure as per the question.

The lengths of tangents drawn from an external point to a circle are equal.

A tangent at any point of a circle is perpendicular to the radius at the point of contact.

In ΔOAP and in ΔOBP

OA = OB (radii of the circle are always equal)

AP = BP (length of the tangents)

OP = OP (common)

Therefore, by SSS congruency ΔOAP ≅ ΔOBP

SSS congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

If two triangles are congruent then their corresponding parts are equal.

Hence,

∠POA = ∠POB

∠OPA = ∠OPB

Therefore, OP is the angle bisector of ∠APB and ∠AOB

Hence, ∠OPA = ∠OPB = 1/2 (∠APB )

= 1/2 × 80°

= 40°

By angle sum property of a triangle,

In ΔOAP

∠A + ∠POA + ∠OPA = 180°

OA ⊥ AP (Theorem 10.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact.)

Therefore, ∠A = 90°

90° + ∠POA + 40° = 180°

130° + ∠POA = 180°

∠POA = 180° - 130°

∠POA = 50°

Thus, option (A) 50° is the correct answer.

☛ Check: NCERT Solutions Class 10 Maths Chapter 10

Video Solution:

Maths NCERT Solutions Class 10 Chapter 10 Exercise 10.2 Question 3

Summary:

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to 50°.

☛ Related Questions:

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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:

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:

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visual curriculum

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If angle between two tangents drawn from a point P to a circle of radius a and centre O is 60∘ , then OP = a √3.