OA & OB a radii of this circle O & there are two tangents drawn from P namely PA & PB with OP equal to the diameter of circle. Prove ABP is an equilateral triangle
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Solution :- ∠OAP = 90°(PA and PB are the tangents to the circle.)
In ΔOPA,
sin ∠OPA = OA/OP = r/2r [OP is the diameter = 2*radius]
sin ∠OPA = 1 /2 = sin 30 ⁰
∠OPA = 30°
Similarly,
∠OPB = 30°.
∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°
In ΔPAB,
PA = PB (tangents from an external point to the circle)
⇒∠PAB = ∠PBA ............(1) (angles opp.to equal sides are equal)
∠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°
all angles are equal in an equilateral triangle.( 60 degrees)
ΔPAB is an equilateral triangle
In ΔOPA,
sin ∠OPA = OA/OP = r/2r [OP is the diameter = 2*radius]
sin ∠OPA = 1 /2 = sin 30 ⁰
∠OPA = 30°
Similarly,
∠OPB = 30°.
∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°
In ΔPAB,
PA = PB (tangents from an external point to the circle)
⇒∠PAB = ∠PBA ............(1) (angles opp.to equal sides are equal)
∠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°
all angles are equal in an equilateral triangle.( 60 degrees)
ΔPAB is an equilateral triangle
debadityadutta1:
thank you but can we do this question without applying the sine part
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