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If in a Delta ABC, cosA+ cosB + cosc =3/...

If in a `Delta ABC, cosA+ cosB + cosc =3/2.` Prove that `DeltaABC` is an equilateral triangle.

Text Solution

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Let a b, c are the sides of a `DeltaABC`
Given, `cos A+cosB+cos C=(3)/(2)`
`rArr(b^(2)+c^(2)-a^(2))/(2bc)+(a^(2)+c^(2)-b^(2))/(2ac) +(a^(2)+b^(2)-c^(2))/(2ab)=(3)/(2)`
`rArr ab^(2)+ac^(2)-a^(3)+ba^(2)+bc^(2)-b^(3)+bca^(2)+cb^(2)-c^(3)=3abc`
`rarr a(b-c)^(2)+b(c-a)^(2)+c(a-b)^(2)=((a+b+c))/(2)[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)]`
`rArr (a+b+c)(a-b)^(2)+(b+c-a)(b-c)^(2)+(c_a-b)(c-a)^(2)=0`
as we know, `+b-cgt0,b+c-age0, c+a-bgt0`]
`:.` Each term on the left of equation has positive coefficient multiplied by perfect square, each term must be separately zero.
`rArr a=b=c`
`:.` Triangle is an equilateral
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