please answer Q. 27
Answers
Answer:
Step-by-step explanation:
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
Link_--https://www.askiitians.com/forums/10-grade-maths/o-is-the-centre-of-the-circle-and-tp-is-the-tangen_142651.htm