The angle between two tangents drawn from an external point to a circle with centre O is 60 degree

Question 5 Circles - Exercise 9.2

Answer:

Tangent is always perpendicular to the radius at the point of contact.

Hence, ∠RPT = 90

If 2 tangents are drawn from an external point, then they are equally inclined to the line segment joining the centre to that point.

Consider the following figure,

From point P, two tangents are drawn.

Therefore, ∠PTO = ∠PRO = 90°

Since three angles of quadrilateral PROT are 90°, the fourth angle is also of 90°.

Therefore, PROT is a rectangle, but adjacent sides of reactangle are also equal. (OT = OR = a, and PT = PR)

Hence, PROT is a square of side a.

OP is a diagonal of square PROT, so, OP a√2.

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Text Solution

Solution : PA and PB are two tangents from an external point P such that ` " "angleAPB=60^(@)` `:." "angleOPA=angleOPB=30^(@)` (`because` tangents are equally inclined at the centre) Also,`" "angleOAP=90^(@)" "`(radius through point of contact is perpendicular to the tangent) <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NTN_MATH_X_C10_S01_003_S01.png" width="80%"> Now in right `DeltaOAP,` `sin30^(@)=(OA)/(OP)` `implies" "(1)/(2)=(a)/(OP)" "implies" "OP=2a` units

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)`