011-40705070  or  
Call me
Download our Mobile App
Select Board & Class
  • Select Board
  • Select Class

 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

Become Expert
Score more in Math
Start Now with Video Lessons, Sample Papers, Revision Notes & more for Class-X - CBSE 



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.....

This conversation is already closed by Expert

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)

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

Posted by Afzal Husain(student)

Ask a QuestionHave a doubt? Ask our expert and get quick answers.
Show me more questions