From a point p two tangents pa and pb

Asked by admin @ 17/08/2021 in Math viewed by 278 People

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.

Answered by admin @ 17/08/2021

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 ∆

From a point p two tangents pa and pb

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