Question

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

Answer 1

∠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

hope it helps!!!!

Answer 2

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 ∆