If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 130º, then ∠ AOB is
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Answer:
In a OPA and OPB
OP=OP (common)
Angle PAO =Angle PBO (90°each )
OA=OB (equal radii)
By RHS criteria
OPA≈ OPB
therefore Angle 1 = Angle 2 (By C. P. C. T)
Angle 1 = Angle 2 = 1/2 of 80
Angle 1 = Angle 2 = 40°
Now , In POA, By ASP
40°+ Angle POA + 90° = 180°
130°+ Angle POA +90°= 180°
Angle POA =180°-130°
Angle POA = 50°
OPTION (A) IS CORRECT :- 50°
hope it is helpful for u
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