1- find the distance between two parallel tangent of a circle of radius 4 cm draw fig

the two tangent from an external point p to a circle with centre o are pa and pb if pab=70 degree find aob

if angle between two radii of a circle is 130 find the angle between tangent at the end of radii

1.

O is the centre of the circle. l and m are two parallel tangents of the circle. 

Radius of the circle = 4cm

Distance between two parallel tangents

= AB

= OA + OB

= 4cm + 4cm

= 8cm

Thus, distance between the parallel tangents l and m is 8cm.

 

3.

O is the centre of the circle. Given, ∠POQ = 130º

PT and QT are tangents drawn from external point T to the circle

∴ ∠OPT = ∠OQT = 90º [ Radius is perpendicular to the tangent at point of contact]

In quadrilateral OPTQ,

∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360º

∴ ∠PTQ + 90º + 90º + 130º = 360º

⇒ ∠PTQ + 310º = 360º

⇒ ∠PTQ = 360º – 310º = 50º

Thus, the angle between the tangents is 50º.

 

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Q1

 

Given:

∠ APB = 70° ..... (1)

and PA and PB be two tangents draw from an external point P to a circle with centre O.

OA ⊥ PA and OB ⊥ PB [ A tangent to a circle is perpendicular to the radius through the point of contact.]

 

Now, in right triangles

OAP and OBP, we have

OA = OB (Equal radius of the circle)

PA = PB (Tangents drawn from an external point are equal)

OP = OP (Common)

∴ ∆OAP ≅∆OBP (SSS)

⇒∠ AOP =∠ BOP and ∠ OPA = ∠ OPB

 

∴ ∠ AOB = 2 ∠ AOP and ∠ APB = 2 ∠ OPA ..... (2)

 

In ∆ OPA, ∠ AOP + ∠ OAP + ∠OPA = 180°

⇒∠ AOP + ∠ OPA = 90° (as ∠ OAP = 90°)

∴ ∠ AOP = 90° – ∠ OPA

 

multiply by 2 on both sides

2∠ AOP = 180° 2∠ OPA

∠ AOB = 180° – ∠ APB (Using (2))

 

Substituting the values from (1), we get

∠ AOB = 180° – 70° = 110°

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