Question

If in a $△ABC,\cos A+\cos B+\cos C=\frac{3}{2}$, then the triangle is

• A. equilateral
• B. isosceles
• C. right angle
• D. none of these

Answer 1

The correct option is A equilateral

Putting $\cos A=\frac{b^2+c^2-a^2}{2bc}$ etc. in the given relation, we get

$\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2bc}+\frac{a^2+b^2-c^2}{2ab}-\frac{3}{2}=0$

$\Rightarrow a(b^2+c^2-a^2)+b(c^2+a^2-b^2)+c(a^2+b^2-c^2)=3abc$

$\Rightarrow a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)=a^3+b^3+c^3+3abc$

Now, subtract $6abc$ from each side,

$\therefore a{(b-c)}^2+b{(c-a)}^2+c{(a-b)}^2=a^3+b^3+c^3-3abc$

$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Multiply both sides by $2$

$\therefore 2a{(b-c)}^2+2b{(c-a)}^2+2c{(a-b)}^2=(a+b+c)[{(b-c)}^2+{(c-a)}^2+{(a-b)}^2]$

$\Rightarrow {(b-c)}^2(b+c-a)+{(c-a)}^2(c+a-b)+{(a-b)}^2(a+b-c)=0$ ...(1)

In a triangle, $b+c-a>0$ etc. hence (1) will hold good if each factor is zero.

Now, ${(b-c)}^2(b+c-a)=0$ gives $b-c=0$ since $b+c-a\ne 0.$

Similarly, $c-a=0$ and $a-b=0.$

$\therefore a=b=c.$ Hence, the triangle is equilateral.