# 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

Arun
25757 Points
4 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
$\angle$POA =2 $\angle$PBA=2 x 30= 60o  [$\angle$ at centre is twice the $\angle$ subtended on the circle.]
$\angle$BPA = 90o [ $\angle$ in semicircle.]
$\rightarrow$$\angle$PAB = 30[ by $\angle$sum prop. in $\Delta$ PAB ]
In$\Delta$POA,180 - ($\angle$O + $\angle$A) =$\angle$P  => $\angle$ P =60o
Therefore,   $\Delta$POA is euilateral.
Hence, PA=PO=AP =r
In $\Delta$OPT, $\angle$O =60o , $\angle$P = 90o [tangent perpendicular to radius]
By $\angle$ sum prop., $\angle$T = 30o
On comparing, $\Delta$BPA $\cong$$\Delta$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

Regards