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 From a point P, 2 tangents PA & PB are drawn to a circle with centre O.If OP=diameter of the circle show that triangle APB is equilateral.

Asked by Annie (student) , on 23/2/11

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Hi Rida!
Here are the answers to your question.
The given information can be represented using a figure as
⇒ ∠OPA = 30°
Similarly, it can be proved that ∠OPB = 30°.
Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°
In ∆PAB,
PA = PB  [lengths of tangents drawn from an external point to a circle are equal]
⇒∠PAB = ∠PBA     ...(1)   [Equal sides have equal angles opposite to them]
∠PAB + ∠PBA + ∠APB = 180°    [Angle sum property]
⇒∠PAB + ∠PAB = 180° – 60° = 120°  [Using (1)]
⇒2∠PAB = 120°
⇒∠PAB = 60°      ...(2)
From (1) and (2)
∠PAB = ∠PBA = ∠APB = 60°
∴ ∆PAB is an equilateral triangle.
Hope! This will help you.

Posted by gopal.mohanty... (student) on 24/2/11

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More Answers


 sir is it necessary to use trigo in this??? i mean the tangents ARE ALWAYS inclined wid each othe at 60degreees rite? so can 't v write dis simply???????

Posted by Divya Kaur (student) on 10/3/11


no ...............m sorry ....................itz not always at 60 degrees ! it can vary  !!

Posted by Afzal Husain (student) on 10/3/11

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