Question

In fig Op is equal to diameter of the circle. Prove that abp is a equilateral triangle..

sry abt the figure but i dnt know how to upload the figure

But its a circle with two tangents AP and BP... and joining OP, OA and OB ( o is the centre of the circle... )

Answer 1

Hey....

Wr iz the picture...??

Answer 2

Step-by-step explanation:

PA and PB are the tangents to the circle.

∴ OA ⊥ PA

⇒ ∠OAP = 90°

In ΔOPA,

 sin ∠OPA = OA OP  =  r 2r   [Given OP is the diameter of the circle]

⇒ sin ∠OPA = 1 2 =  sin  30 ⁰

⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,  

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°

∴ ΔPAB is an equilateral triangle.