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O is the centre of the circle and TP is the tangent to the circle from an external point T. If angle PBT = 30 degree , prove that BA : AT = 2 : 1

O is the centre of the circle and TP is the tangent to the circle from an external point T. If angle PBT = 30 degree , prove that BA : AT = 2 : 1
 

Grade:9

2 Answers

Anwesha Barman
38 Points
5 years ago
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 30= 60o  [\angle at centre is twice the \angle subtended on the circle.]
\angleBPA = 90o [ \angle in semicircle.]
\rightarrow\anglePAB = 30[ 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. …............proved.
Hope it helps....... :)
 
 
Yash Chourasiya
askIITians Faculty 256 Points
9 months ago
Dear Student

643-1942_Untitled.png
AB is the chord passing through the center
So, AB is the diameter
Since, angle in a semicircle is a right angleRead more on Sarthaks.com -643-343_1.PNG

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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