## Book: RS Aggarwal - Mathematics

### Chapter: 12. Circles

#### Subject: Maths - Class 10th

##### Q. No. 16 of Exercise 12A

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16
##### In the given figure, O is the center of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA : AT = 2 : 1.

In Given Figure, we have a circle with center O let the radius of circle be r.

Construction : Join OP

Now, In APB

ABP = 30°

APB = 90°

[Angle in a semicircle is a right angle]

By angle sum Property of triangle,

ABP + APB + PAB = 180

30° + 90° + PAB = 180

PAB = 60°

OP = OA = r [radii]

PAB = OPA = 60°

[Angles opposite to equal sides are equal]

By angle sum Property of triangle

OPA + OAP + AOP = 180°

60° + PAB + AOP = 180

60 + 60 + AOP = 180

AOP = 60°

As all angles of OPA equals to 60°, OPA is an equilateral triangle

So, we have, OP = OA = PA = r [1]

OPT = 90°

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

OPA + APT = 90

60 + APT = 90

APT = 30°

Also,

PAB + PAT = 180° [Linear pair]

60° + PAT = 180°

PAT = 120°

In APT

APT + PAT + PTA = 180°

30° + 120° + PTA = 180°

PTA = 30°

So,

We have

APT = PTA = 30°

AT = PA

[Sides opposite to equal angles are equal]

AT = r [From 1] [2]

Now,

AB = OA + OB = r + r = 2r [3]

From [2] and [3]

AB : AT = 2r : r = 2 : 1

Hence Proved !

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