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

Question

Anwesha Barman · Answer

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 POA =2 PBA=2 x 30 o = 60 o [ at centre is twice the subtended on the circle.] BPA = 90 o [ in semicircle.] PAB = 30 o [ by sum prop. in PAB ] In P OA,180 - ( O + A) = P => P =60 o Therefore, POA is euilateral. Hence, PA=PO=AP =r In OPT, O =60 o , P = 90 o [tangent perpendicular to radius] By sum prop., T = 30 o 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....... :)

Yash Chourasiya · Answer

Dear Student

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 -

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

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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

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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

2 Answers

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