मराठी

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.

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प्रश्न

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.

बेरीज
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उत्तर

AB is the chord passing through the center.

So, AB is the diameter.

Since an angle in a semicircle is a right angle.

∴ ∠ APB = 90°

By using the alternate segment theorem.

We have ∠ APB = ∠ PAT = 30°

Now, in Δ APB

∠ BAP + ∠ APB + ∠ BAP = 180°   ...(Angle sum property of triangle)

∠ BAP = 180° − 90° − 30°

∠ BAP = 60°

Now, ∠ BAP + ∠ APT + ∠ PTA    ...(Exterior angle property)

60° = 30° + ∠ PTA

∠ PTA = 60° − 30°

∠ PTA = 30°

We know that sides opposite to equal angles are equal,

∴ AP = AT

In the right triangle ABP

sin ∠ ABP = `(AP)/(BA)`

sin 30° = `(AT)/(BA)`

`1/2 = (AT)/(BA)`

∴ BA : AT = 2 : 1

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पाठ 8: Circles - EXERCISE 8A [पृष्ठ ४९२]

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आर. एस. अग्रवाल Mathematics [English] Class 10
पाठ 8 Circles
EXERCISE 8A | Q 16. | पृष्ठ ४९२
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