Prove that in a triangle other than an equilateral triangle, angle opposite the longest side is greater than 23 of a right angle.
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Given -- ΔABC other than equilateral triangle.
The construction is already given in the picture.So let me just establish the relation between the sides.
Let AB = AD = DB
ABD is an equilateral triangle since AB = AD = DB
∠ABD = ∠ADB = ∠BAD = 60°
Longest side= BC
Angle opposite to longest side = ∠BAC
Here it is clear that ∠BAC = ∠BAD + ∠CAD
So ∠BAC > ∠BAD
∠BAC > 60°
Hence proved!
Hope This Helps You augusta
-navya
The construction is already given in the picture.So let me just establish the relation between the sides.
Let AB = AD = DB
ABD is an equilateral triangle since AB = AD = DB
∠ABD = ∠ADB = ∠BAD = 60°
Longest side= BC
Angle opposite to longest side = ∠BAC
Here it is clear that ∠BAC = ∠BAD + ∠CAD
So ∠BAC > ∠BAD
∠BAC > 60°
Hence proved!
Hope This Helps You augusta
-navya
augustachristy:
thnx alot gurl
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