Question

two tangents PA and PB are drawn to a circle with centre O form an external point P. prove that angle APB =2

Answer 1

Step-by-step explanation:

FIGURE IS IN THE ATTACHMENT.

Given:

PA & PB are the tangents to a circle, with Centre O from a point P outside it.

To Prove:

∠APB = 2∠AOB

Proof:

Let ∠APB= x°

We know that the tangents to a circle from an external point are equal in length so PA= PB.

PA =PB

∠PBA = ∠PAB

[Angles opposite to the equal sides of a triangle are equal.]

∠APB+ ∠PBA +∠PAB= 180°

[Sum of the angles of a triangle is 180°]

x° + ∠PAB +∠PAB = 180°

[∠PBA = ∠PAB]

x° + 2∠PAB = 180°

∠PAB =½(180° - x°)

∠PAB =90° - x°/2

∠OAB +∠PAB=90°

∠OAB =90° - ∠PAB

∠OAB =90° - (90° - x°/2)

∠OAB =90° - 90° + x°/2

∠OAB = x°/2

∠OAB = ∠APB /2

∠OAB = 1/2∠APB

∠APB = 2∠OAB

HOPE THIS WILL HELP YOU...

Answer 2

Step-by-step explanation:

ZAP8 + 22PAB = 180 ° 1 PAB = ( 181® – ZAPB ) AND EN PAB = 90 ° 2 ZOAB 1 PAB = 91 % ZOAB = 91º – PAB ZOAB = 90 – 19.0 ( sp • _ 4423 ZOAB = 90 ° — 90 ° + 2 ZOAB = 2 APB = 2 / O4B