Question

In the given fig. OP is equal to the diameter of the circle with centre O.Prove that ∆ ABP is an equilateral triangle.

Answer 1

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∆.

HOPE THIS WILL HELP YOU..

Answer 2

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