1.

In the given figure, 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»

In the given figure,

TP is the tangent from an external point T and

∠PBT = 30°

Now,

∠APB = 90° (Angle in a semicircle)

∠PBT = 30° (given)

So, ∠PAB = 90° – 30° = 60°

But, ∠PAT + ∠PAB = 180° (Linear pair)

∠PAT + 60° = 180°

∠PAT = 180° – 60° = 120°

Also, ∠APT = ∠PBA = 30° (Angles in the alternate segment)

In ∆PAT,

∠PTA = 180° – (120° + 30°) = 180° – 150° = 30°

PA = AT

In right ∆APB,

sin 30° = AP/AB

1/2 = AP/AB

AB = 2 AP

Since AP = AT

AB = 2 AT

or AB/AT = 2/1

AB:AT = 2:1

or BA:AT = 2:1. Hence Proved.


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