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Question

If in a â–³ABC,cosA+cosB+cosC=32, then the triangle is

A
equilateral
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B
isosceles
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C
right angle
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D
none of these
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Solution

The correct option is A equilateral
Putting cosA=b2+c2a22bc etc. in the given relation, we get
b2+c2a22bc+c2+a2b22bc+a2+b2c22ab32=0

a(b2+c2a2)+b(c2+a2b2)+c(a2+b2c2)=3abc
a(b2+c2)+b(c2+a2)+c(a2+b2)=a3+b3+c3+3abc
Now, subtract 6abc from each side,
a(bc)2+b(ca)2+c(ab)2=a3+b3+c33abc
=(a+b+c)(a2+b2+c2abbcca)
Multiply both sides by 2
2a(bc)2+2b(ca)2+2c(ab)2=(a+b+c)[(bc)2+(ca)2+(ab)2]

(bc)2(b+ca)+(ca)2(c+ab)+(ab)2(a+bc)=0 ...(1)

In a triangle, b+ca>0 etc. hence (1) will hold good if each factor is zero.
Now, (bc)2(b+ca)=0 gives bc=0 since b+ca0.

Similarly, ca=0 and ab=0.
a=b=c. Hence, the triangle is equilateral.

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