Question

If in a triangle $ABC$ , $\cos A + \cos B + \cos C = \dfrac{3}{2}$ , then the triangle is
A.Equilateral
B.Isosceles
C.Right angle triangle
D.None of these

Answer 1

In the given question , we consider a triangle with sides $a,b\& c$ , now in any triangle the cosine of any of the angle say $\left( A \right)$ is given by $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$ , where $a,b\& c$ are the sides of the triangle . So, using these formulas we will find the value of $\cos B$ and $\cos C$ , then solve it accordingly .

Complete step by step solution:
Given : $\cos A + \cos B + \cos C = \dfrac{3}{2}$
Now using the formula of $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$ we get ,
$\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + \cos B + \cos C = \dfrac{3}{2}$ ,
Now using the formula for $\cos B = \dfrac{{{c^2} + {a^2} - {b^2}}}{{2bc}}$ we get ,
$\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}} + \cos C = \dfrac{3}{2}$
Now using the formula for $\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$ , we get
$\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = \dfrac{3}{2}$
Now shifting $\dfrac{3}{2}$ on LHS , we get
$\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} - \dfrac{3}{2} = 0$
Taking $2abc$ as the LCM we get ,
$\dfrac{{a\left( {{b^2} + {c^2} - {a^2}} \right) + b\left( {{c^2} + {a^2} - {b^2}} \right) + c\left( {{a^2} + {b^2} - {c^2}} \right) - 3abc}}{{2abc}} = 0$
On simplifying we get ,
$a\left( {{b^2} + {c^2} - {a^2}} \right) + b\left( {{c^2} + {a^2} - {b^2}} \right) + c\left( {{a^2} + {b^2} - {c^2}} \right) = 3abc$
Now multiplying $a,b\& c$ with ${a^2},{b^2}\& {c^2}$ respectively inside the bracket , we get
$a\left( {{b^2} + {c^2}} \right) - {a^3} + b\left( {{c^2} + {a^2}} \right) - {b^3} + c\left( {{a^2} + {b^2}} \right) - {c^3} = 3abc$
Now taking the cube terms on RHS we get ,
$a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right) = {a^3} + {b^3} + {c^3} + 3abc$
Now adding $- 6abc$ on both sides we get ,
$a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right) - 6abc = {a^3} + {b^3} + {c^3} + 3abc - 6abc$
On simplifying by splitting the $- 6abc$ into $- 2abc$ three times we get ,
$a{b^2} + a{c^2} - 2abc + b{c^2} + b{a^2} - 2abc + c{a^2} + c{b^2} - 2abc = {a^3} + {b^3} + {c^3} - 3abc$
Now taking $a,b\& c$ common we get ,
$a\left( {{b^2} + {c^2} - 2bc} \right) + b\left( {{c^2} + {a^2} - 2ac} \right) + c\left( {{a^2} + {b^2} - 2ab} \right) = {a^3} + {b^3} + {c^3} - 3abc$
Now using the algebraic identities we get ,
$a{\left( {b - c} \right)^2} + b{\left( {c - a} \right)^2} + c{\left( {a - b} \right)^2} = {a^3} + {b^3} + {c^3} - 3abc$
Now using the identity of ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$ , we get
$a{\left( {b - c} \right)^2} + b{\left( {c - a} \right)^2} + c{\left( {a - b} \right)^2} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$
Now multiplying by $2$ on both sides we get ,
$2a{\left( {b - c} \right)^2} + 2b{\left( {c - a} \right)^2} + 2c{\left( {a - b} \right)^2} = \left( {a + b + c} \right)\left( {2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca} \right)$
On splitting the square terms oh RHS we get ,
$2a{\left( {b - c} \right)^2} + 2b{\left( {c - a} \right)^2} + 2c{\left( {a - b} \right)^2} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ca} \right)$
Now using the algebraic identities we get ,
$2a{\left( {b - c} \right)^2} + 2b{\left( {c - a} \right)^2} + 2c{\left( {a - b} \right)^2} = \left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$
On simplifying we get ,
$2a{\left( {b - c} \right)^2} + 2b{\left( {c - a} \right)^2} + 2c{\left( {a - b} \right)^2} = \left( {a + b + c} \right){\left( {a - b} \right)^2} + \left( {a + b + c} \right){\left( {b - c} \right)^2} + \left( {a + b + c} \right){\left( {c - a} \right)^2}$
Now subtracting the like terms form RHS with LHS , we get
$\left( {a + b + c - 2c} \right){\left( {a - b} \right)^2} + \left( {a + b + c - 2a} \right){\left( {b - c} \right)^2} + \left( {a + b + c - 2b} \right){\left( {c - a} \right)^2} = 0$
On solving we get ,
$\left( {a + b - c} \right){\left( {a - b} \right)^2} + \left( {b + c - a} \right){\left( {b - c} \right)^2} + \left( {a + c - b} \right){\left( {c - a} \right)^2} = 0$
Now in the above expression the terms $\left( {a + b - c} \right)$ , $\left( {b + c - a} \right)$ and $\left( {a + c - b} \right)$ represents the sum two sides minus third side , we know that in a triangle sum of two sides is always greater than third side, therefore these term will not be equal to zero , so the whole square terms will be zero ,
${\left( {a - b} \right)^2} = 0$ , ${\left( {b - c} \right)^2} = 0$ and ${\left( {c - a} \right)^2} = 0$
On solving we get ,
$a = b$ , $b = c$ and $c = b$
Therefore , we get
$a = b = c$
Therefore , the triangle is equilateral.
So, the correct answer is “Option A”.

Note: In question like these , never try for given option , like do not solve for equilateral or isosceles or any other given option , solve the related to trigonometry and triangle with the cosine formulas for different angles , then apply the condition for different triangle. All sides are equal and represent an equilateral triangle and also note that an equilateral triangle is also an equiangular triangle.