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Q.

$PA$ and $PB$ from a point $P$ to a circle with centre O are drawn so that $∠ APB = 80 °$ . The value of $∠ POA=$ [[1]].

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Detailed Solution

PA

and

PB

from a point

P

to a circle with centre O are drawn so that

∠ APB = 80 °

.The value of

∠ POA=

50 °

.
We know that the tangents drawn from an external point to a circle are equal and are also perpendicular to the radius at the point of contact.
We also know that the angle sum property of triangle states that the sum of all interior angle of a triangle

180 °

.
SSS congruent condition is true when three sides of both the triangles are equal.
Given is the following figure,
Question Image

Question Image

From

Δ O P A

and

Δ O P B

,

⇒ O A = O B

[Radii of the same side]

[Common side]

⇒ P A = P B

Using the SSS rule,

Δ O P A ≅ Δ O P B

⇒ ∠ A P O = ∠ B P O ⇒ ∠ A P O = 12 ∠ A P B ⇒ ∠ A P O = 40 °

The tangents drawn from an external point are perpendicular to the radius at the point of contact.

∴ ∠ O A P = 90 °

By the angle sum property in

Δ O A P

,

∴ ∠ A O P + ∠ O A P + ∠ A P O = 180 ° ⇒ ∠ A O P + 90 ° + 40 ° = 180 ° ⇒ ∠ A O P = 180 ° − 130 ° ⇒ ∠ A O P = 50 °

Therefore we get the value of

∠ P O A = 50 °

.

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