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AB is a diameter of a circle with centre o. There is an external point T and TP is a tangent to circle from T and if angle PBT =30degree. Prove that BA : AT = 2:1

AB is a diameter of a circle with centre o. There is an external point T and TP is a tangent to circle from T and if angle PBT =30degree. Prove that BA : AT = 2:1

Grade:10

1 Answers

Arun
25750 Points
6 years ago
Dear student
 
Hi, I’m not able to upload the diagram ….anyway, here’s the solution.
Construction : Join P and O to get PO
Proof :   Let Radius = r  => BO=PO=AO=r
 \anglePOA =2 \anglePBA=2 x 30o = 60o  [\angle at centre is twice the \angle subtended on the circle.]
\angleBPA = 90o [ \angle in semicircle.]
\rightarrow\anglePAB = 30o [ by \anglesum prop. in \Delta PAB ]
In\DeltaPOA,180 - (\angleO + \angleA) =\angleP  => \angle P =60o
 Therefore,   \DeltaPOA is euilateral.
Hence, PA=PO=AP =r
 In \DeltaOPT, \angleO =60o , \angleP = 90o [tangent perpendicular to radius]
By \angle sum prop., \angleT = 30o
On comparing, \DeltaBPA \cong\DeltaTPO { 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
 
Regards
Arun (askIITians forum expert)

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