Question

In Figure 4, O is the centre of the circle and TP is the tangent to the
circle from an external point T. If  PBT = 30°, prove that
BA : AT = 2 : 1

Answer 1

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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 30o = 60o [\angle at centre is twice the \angle subtended on the circle.]

\BPA = 90o [ \angle in semicircle.]

\rightarrow\PAB = 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. …............proved.

Hope it helps....... :)

Answer 2

$\huge\boxed{\bf{\red{ANSWER}}}$

^_^^_^

Construction : Join P and O to get PO

Proof : Let Radius = r => BO=PO=AO=r

\POA =2 \PBA=2 x 30o = 60o [\angle at centre is twice the \angle subtended on the circle.]

\BPA = 90o [ \angle in semicircle.]

\rightarrow\PAB = 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. …............proved.

Hope it helps....... :)

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