Question

in the given figure op is equal to the diameter of a circle with Centre O and P A and PB are the tangents prove that ABP is an equilateral triangle​

Answer 1

Answer:

Let radius of a Circle = r

GIVEN:

OA = OB = r, OP = diameter=  2r

In right angled ∆OAP,

Sin θ = AO/AP = r/2r = ½   [sin θ = P/H]

Sin θ = ½

Sin θ  = sin 30°

θ = 30°

∠APO = 30°

Similarly , ∠BPO = 30°

Hence, ∠APB = ∠APO + ∠BPO =30° + 30°= 60°.

∠OAP =∠OBP  90°

[We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.]

In quadrilateral OAPB,

∠OAP + ∠APB + ∠OBP + ∠AOB = 360°

[Sum of all interior angles of a quadrilateral is 360°]

90° + 60° + 90° + ∠AOB = 360°

240° + ∠AOB = 360°

∠AOB = 360° - 240° = 120°

In ∆AOB,

OA = OB     [ radius of the circle]

∠OAB = ∠OBA

[Angles opposite to equal sides of a ∆ are equal]

Let ∠OAB = ∠OBA  = x

∠OAB +∠AOB +  ∠OBA = 180°

[Sum of all angles in a triangle is 180°]

x + 120° + x = 180°

2x +120° = 180°

2x = 180° - 120°

2x = 60°

x = 60°/2= 30°

∠OAB = ∠OBA  = 30°

∠PAB = ∠OAP - ∠OAB  = 90° - 30°= 60°

Similarly, ∠PBA = 60°

In ∆ABP,

∠APB = ∠PAB = ∠PBA = 60°

In ∆APB all angles are of 60°, Hence, ∆APB is an equilateral∆.

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