Question

Two tangents PA and PB are drawn to a circle with the centre O from an external point P
Prove that < APB = 2 < OAB .


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Answer 1

●Question

■Two tangents PA and PB are drawn to a circle with the centre O from an external point P

■Prove that < APB = 2 < OAB .

Prove that < APB = 2 < OAB .

$\huge \bold{ \fbox \pink{required \: answer}}$

$\bold{ \underline{ \underline \red{answer}}} →$

●Given:

Let ∠APB= x°

●To find

Prove that < APB = 2 < OAB .

$\bold{ \underline{ \underline \red{solution}}} →$

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

●PA =PB

12. Two tangents PA and PB are drawn to a cirele with centre O from an external point P. Prove that [2016 LAPB2ZOAB

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

■Therefore, ∠APB = 2∠OAB