solve 26 iind part please
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Hii dear here is your answer.....
Construction : Join P and O to get PO
Proof : Let Radius = r => BO=PO=AO=r
POA =2 PBA=2 x 30o = 60o [ at centre is twice the  subtended on the circle.]
BPA = 90o [  in semicircle.]
PAB = 30o [ by sum prop. in  PAB ]
InPOA,180 - (O + A) =P =>  P =60o
Therefore, POA is equilateral.
Hence, PA=PO=AP =r
In OPT, O =60o , P = 90o [tangent perpendicular to radius]
By  sum prop., T = 30o
On comparing, BPA TPO { above angles and PO=PA=r} byASA test
By CPCT, BA=TO
BO+OA=OA+TA
2r=r + TA
TA=r
BA/AT = 2r/r =2/1 = 2:1. …............proved.
hope it helps you☺️☺️☺️
Construction : Join P and O to get PO
Proof : Let Radius = r => BO=PO=AO=r
POA =2 PBA=2 x 30o = 60o [ at centre is twice the  subtended on the circle.]
BPA = 90o [  in semicircle.]
PAB = 30o [ by sum prop. in  PAB ]
InPOA,180 - (O + A) =P =>  P =60o
Therefore, POA is equilateral.
Hence, PA=PO=AP =r
In OPT, O =60o , P = 90o [tangent perpendicular to radius]
By  sum prop., T = 30o
On comparing, BPA TPO { above angles and PO=PA=r} byASA test
By CPCT, BA=TO
BO+OA=OA+TA
2r=r + TA
TA=r
BA/AT = 2r/r =2/1 = 2:1. …............proved.
hope it helps you☺️☺️☺️
Answered by
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Answer:
Construction : Join P and O to get PO
Proof : Let Radius = r => BO=PO=AO=r
POA =2 PBA=2 x 30o = 60o [ at centre is twice the  subtended on the circle.]
BPA = 90o [  in semicircle.]
PAB = 30o [ by sum prop. in  PAB ]
InPOA,180 - (O + A) =P =>  P =60o
Therefore, POA is equilateral.
Hence, PA=PO=AP =r
In OPT, O =60o , P = 90o [tangent perpendicular to radius]
By  sum prop., T = 30o
On comparing, BPA TPO { above angles and PO=PA=r} byASA test
By CPCT, BA=TO
BO+OA=OA+TA
2r=r + TA
TA=r
BA/AT = 2r/r =2/1 = 2:1. …............proved.
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