Question

Tangents PA n PB of circle with centre O are inclined at 60°. Find ratio of OA: OP: AP

Answer 1

Answer:

/_A = /_B = 90° [Tangents are perpendicular to the circle]

=> /_AOB = 360 - 180 - 60 [Angles sum in a quadrilateral]

=> /_AOB = 120°

/_AOP = 1/2(/_AOB) [OP bisects /_AOB)

=> /_AOP = 60°

Let OA = x

In Triangle AOP:

tan 60 = $\tt{\sqrt{3}}$ = AP/OA

=> AP = $\tt{x \sqrt{3}}$

sin 30 = 1/2 = OA/OP

=> OP = 2x

OA : OP : AP = x : 2x : $\tt{x \sqrt{3}}$

=> 1 : 2 : $\tt{\sqrt{3}}$

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Answer 2

Answer: Ratio of OA : OP : AP is 1 : 2 : √3.

As Tangents are perpendicular to the circle,

Angle A = Angle B = 90

We know that, Sum of angles of triangle is 180°.

Angle AOB = 360 - 180 - 60

=> Angle AOB = 120°

Now,

OP bisects Angle AOB

.°. Angle AOP = 1/2( Angle AOB)

=> Angle AOP = 60°

Let OA be x.

In ∆ AOP,

tan 60 = √3 = AP/OA

.°. AP = x√3

sin 30 = 1/2 = OA/OP

.°. OP = 2x

OA : OP : AP = x : 2x : √3

=> 1 : 2 : √3