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Two radii OA And OB of a circle are inclined at 120. Tangent are drawn at A and B to the circle to intersect at C. Show that ABC is equilateral triangle



Angle A and B are 90°
angle 0 is 120°
so angle C = 60°[sum of interior angles of a quadrilateral is 360°]
now in triangle OAB
OA=OB (radii)
it means opp angles of these sides will be equal
angle OAB=angle OBA = x
x+x+angle O= 180°(ASP)
2x= 180-120°
now we know that AC=BC ( tangents)
opp angles will be equal
angle CAB=angleCBA
angleOBC-angle OBA=angle CBA
angleCBA = angle CAB= angle ACB=60°
in equilateral triangle all angles are of 60° as triangle ABC has so it proves it to be an equilateral triangle
0 0 0
Angle ACB = 360-(90+90+120)
OA =OB so angle OAB =OBA=30 .
so angle BAC=ABC=60
As all angles ae 60 in tr.ABC thus ABC is equilateral
0 0 0
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