Question

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

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

Answer 1

⭕️r=radius of the circle

  AO= r

  OP=2r

 angle OAP=90'

⭕️ in right triangle OAP

  sin Φ=AO/OP =r/2r =1/2

  Φ=30°

therefore angle APO=30°

angle APO =angle BPO =30°

angle APB =30°+30°=60°

PA =PB (TANGENTS THEOREM)

ANGLE PAB= ANGLE PBA

⭕️angle PAB+ angle PBA + angle APB =180'  (ANGLE SUM PROPERTY)

angle PAB+ angle PBA +60' =180°

angle PAB+ angle PBA =120°

angle PAB= angle PBA =60°

✔️✔️so we can say triangle APB  equilateral triangle as all anglas are 60°

$\huge\bf{\red{\mathfrak{thank \: you :)}}}$

Answer 2

Heyy mate ❤✌✌❤

Here's your Answer....

⤵️⤵️⤵️⤵️⤵️⤵️

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.
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