Question

Hola Mates !! ❤️

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80° then ∠ POA is equal to.

(A) 50°

(B) 60°

(C) 70°

(D)80°.​

Answer 1

$\huge\mathfrak\red{Answer:}$

Given:

• We have been given that two tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°.

To Find:

• We need to find ∠POA.

Solution:

Since it is given that O is the centre of circle and two tangents from point P to the circle are PA and PB.

Therefore, OA ⊥ AP and OB ⊥ BP.

∠OAP = ∠OBP = 90°

In Quadrilateral PAQB, we have

∠APB + ∠PAQ + ∠AOB + ∠PBO = 360°

=> 80° + 90° + ∠AOB + 90° = 360°

=> 260° + ∠AOB = 360°

=> ∠AOB = 360° - 260°

=> ∠AOB = 100°

Now, in right angled triangles OAP and OBP, we have

OP = OP [common]

∠OAP = ∠OBP [Each 90°]

OA = OB [Radii of same circle]

So, we can say that ΔOAP ≅ ΔOBP by RHS congruency criteria.

=> ∠POA = ∠POB [CPCT]

Therefore,

∠POA = 1/2∠AOB

= 1/2 × 100

= 50°

Therefore, the value of ∠POA is 50°.

Hence, option A is correct.

Answer 2

Answer:

saif baiyya , ? ? are u there ..i wanna talk with u bro.. plz u just inbox to my insta account twinlight queenbee625676. really miss u ..plz message as soon as possible..i wanna talk with u ❤ ❤