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.

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.
Cheers!
 

  • 91

Sir, you are wonderful.

  • 7

 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???????

  • -1

u too r ryt !!

  • 0

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

  • 1
What are you looking for?