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# If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of $80^{\circ}$, then $\angle&space;POA$ is equal to:A) $50^{\circ}$ B) $60^{\circ}$ C) $70^{\circ}$ D) $80^{\circ}$

CBSE/Class 10/Mathematics/UNIT IV: Geometry/Circles/NCERT Solutions/Exercise 10.2

It is given that PA and PB are tangents. Therefore, the radius drawn to these tangents will be perpendicular to the tangents. Thus, OA ⊥ PA and OB ⊥ PB ∠OBP = 90º ∠OAP = 90º In AOBP, Sum of all interior angles = 360 ∠OAP + ∠APB +∠PBO + ∠BOA = 360 90 + 80 +90º... read more

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90º

∠OAP = 90º

In AOBP,

Sum of all interior angles = 360

∠OAP + ∠APB +∠PBO + ∠BOA = 360

90 + 80 +90º +BOA = 360

∠BOA = 100

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)

A ↔ B, P ↔ P, O ↔ O

And thus, ∠POB = ∠POA

Hence,option (A) is correct.

Ex-army Public School Teacher 15 Years of Experience in Teaching Maths Under Cbse

A. 50

5 year teaching experience and IAS 2015 pre qualified

50

5 year teaching experience and IAS 2015 pre qualified

50

A)50 degree

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