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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.
उत्तर
AB is the chord passing through the center
So, AB is the diameter
Since, angle in a semicircle is a right angle
∴ ∠ APB = 90°
By using alternate segment theorem
We have ∠APB = ∠PAT =30°
Now, in ΔAPB
∠BAP + ∠APB + ∠BAP = 180° (Angle sum property of triangle)
⇒ ∠BAP =180° - 90° -30° = 60°
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60° = 30° +∠PTA
⇒ ∠PTA = 60°- 30°= 30°
We know that sides opposite to equal angles are equal
∴ AP = AT
In right triangle ABP
sin ∠ABP = `(AP)/(BA)`
⇒ sin 30° = `(AT)/(BA)`
⇒ `1/2 =(AT)/(BA)`
∴ BA : AT = 2 :1
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