Question

For a triangle ABC it is given that $cosA+cosB+cosC=\frac{3}{2}$ Prove that the triangle is equilateral

• A. True
• B. False

Answer 1

The correct option is A True

Let 'a', ‘b’ and ‘c’ be the sides of the triangle ABC.
It is given that $cosA+cosB+cosC=\frac{3}{2}$
Hence, $\frac{(b^2+c^2-a^2)}{2bc}+\frac{(a^2+c^2-b^2)}{2ac}+\frac{(b^2+b^2-c^2)}{2ab}=\frac{3}{2}$
Simplifying this we get.
$a{b}^2+a{c}^2-a^3+b{a}^2+b{c}^2-b^3+c{a}^2+c{b}^2-c^3=3abc$
Hence, $a(b-c{)}^2+b(c-a{)}^2+c(a-b{)}^2=\frac{(a+b+c)}{2}[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$
This gives
$(a+b-c)(a-b{)}^2+(b+c-a\left)\right(b-c{)}^2+(c+a-b)(c-a{)}^2=0$
(since we know that $(a+b-c)>0,(b+c-a)>0,(c+a-b)>0$)
On the left hand side, we have coefficient multiplied by the square of a number. Moreover, each coefficient is positive and hence for the sum to be zero, each term separately must be equal to zero. This means we must have
$(a+b-c)(a-b{)}^2=0=(b+c-a\left)\right(b-c{)}^2=(c+a-b)(c-a{)}^2$
This implies a = b = c
This implies that the triangle is equilateral.