Question

From a point P,two tangents PA and PB are drawn to a circle with centre O.If OP=diameter of the circle,show that triangle APB is equilateral.

Answer 1

AP is the tangent to the circle.

∴ OA ⊥ AP  (Radius is perpendicular to the tangent at the point of contact)

⇒ ∠OAP = 90º 

In Δ OAP,

sin ∠OPA = OA/OP = r/2r [Diameter of the circle]

∴ sin ∠OPA = 1/2 = sin 30º

⇒ ∠OPA =  30º

Similarly, it can he prayed that ∠OPB = 30
How, LAPB = LOPP + LOPB = 30° + 30° = 60° 
In APPB, 
PA = PB [lengths &tangents drawn from an external point to circle are equal] 
⇒ ∠PAB = ∠PBA --- (1) [Equal sides have equal angles apposite to them] 
∠PAB + ∠PBA + ∠APB = 180° [Angle sum property] 

∠PAB + ∠PBA + ∠APB = 180° - 60° [Using (1)] 
⇒ 2∠PAB = 120°

⇒ ∠PAB = 60° 
From (1) and (2) 
∠PAB = ∠PBA = ∠APB = 60°
APPB is an equilateral triangle. 

There's an alternate method too:

∠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

 

A different approach:

P is any unknown points by which two tangents PA and PB drawn in circle at point A and B .

 

if we join the point A to O and B to O then we see PAOB is a quadrilateral .

where

angle OAP = angle OBP =90° ( due to AP and BP is tangent on circle )

 

know,

from property of quadrilateral ,

angle OAP + angleOBP + angle APB + angle AOB = 360°

angle AOB + angle APB =360°-(90°+90°)=180°.

 

according to question ,

PO = diameter =2r

 

know, join the point A and B

then two ∆ form ∆AOB and ∆APB

from ∆OAP , this is right angle ∆

let angle APO =∅

then, use sin∅

sin∅ = OA/OP =r/2r = 1/2

sin∅ =1/2 =sin30°

∅ = 30°

 

similarly from ∆OBP , this is also right angle traingle.

let angle BPO =ß

then

sinß =OB/OP =r/2r =1/2 =sin30°

ß =30°

 

hence, angle APB =30° + 30° =60°

we know,

tengents are equal length

e.g angle both angle ( angle ABP and angle BAP ) also equal

let @ each of that angles

now,

from ∆APB ,

angle ABP +angle BAP +angle APB =180°

@ + @ +60° = 180°

@ =60°

hence,

@ = @ =∅ = 60°

all angles of triangle are equal

so, ∆APB is an equilateral ∆

 

Hope This Helps :)