Question

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

Answer 1

Hint: To solve this question, we will take the given equation and use the formula of \[\cos A+\cos B\]in it and simplify. We will then use the angle sum property of the triangle and derive an equation for $A+B$and substitute the equation in simplified equation and get a quadratic equation as a result. Using a discriminant formula, we will simplify the quadratic equation and get an equation for angles $A$ and $B$. Again we will solve the given equation using angles $B$ and $C$. With the resultant relation derived from $A$,$B$ and $C$, we will determine the type of the triangle.
Formula used:
\[\cos A+\cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}\]
$\cos A=1-2{{\sin }^{2}}\frac{A}{2}$
\[\frac{1+\cos A}{2}={{\cos }^{2}}A\]
Complete step-by-step solution:
We are given a triangle $ABC$ such that $\cos A+\cos B+\cos C=\frac{3}{2}$ and we have to determine the nature of the triangle.
We will take the given equation $\cos A+\cos B+\cos C=\frac{3}{2}$and use the formula of \[\cos A+\cos B\]in it.
$2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+\cos C=\frac{3}{2}....(ii)$
Now we will use the angle sum property of the triangle and derive the equation for $A+B$.
$\begin{align}
  & A+B+C=\pi \\
 & A+B=\pi -C....(i)
\end{align}$
We will now substitute the equation (i) in equation (ii).
\[\begin{align}
  & 2\cos \frac{\pi -C}{2}\cos \frac{A-B}{2}+\cos C=\frac{3}{2} \\
 & 2\cos \left( \frac{\pi }{2}-\frac{C}{2} \right)\cos \frac{A-B}{2}+\cos C=\frac{3}{2} \\
 & 2\sin \left( \frac{C}{2} \right)\cos \frac{A-B}{2}+\cos C=\frac{3}{2} \\
 & 4\sin \left( \frac{C}{2} \right)\cos \frac{A-B}{2}+2\cos C=3
\end{align}\]
We will now use the formula $\cos A=1-2{{\sin }^{2}}\frac{A}{2}$.
\[\begin{align}
  & 4\sin \left( \frac{C}{2} \right)\cos \frac{A-B}{2}+2\left( 1-2{{\sin }^{2}}\frac{C}{2} \right)=3 \\
 & 4\sin \left( \frac{C}{2} \right)\cos \frac{A-B}{2}+2-4{{\sin }^{2}}\frac{C}{2}=3 \\
 & 4{{\sin }^{2}}\frac{C}{2}-4\sin \left( \frac{C}{2} \right)\cos \frac{A-B}{2}+3-2=0 \\
 & 4{{\sin }^{2}}\frac{C}{2}-4\sin \left( \frac{C}{2} \right)\cos \frac{A-B}{2}+1=0
\end{align}\]
As the equation formed is a quadratic equation, then the angle $C$ will have the real values, that is the quadratic equation will have real roots and hence $\sin \frac{C}{2}$ will have real values. Now we know that the condition for the real roots is,
$D\ge 0$, where $D$ is discriminant and its value is $D={{b}^{2}}-4ac$.
$\begin{align}
  & {{b}^{2}}-4ac\ge 0 \\
 & {{\left( -4\cos \frac{A-B}{2} \right)}^{2}}-4\times 4\times 1\ge 0 \\
 & 16{{\cos }^{2}}\frac{A-B}{2}-16\ge 0 \\
 & 8\left( 2{{\cos }^{2}}\frac{A-B}{2}-2 \right)\ge 0 \\
 & 2{{\cos }^{2}}\frac{A-B}{2}-2\ge 0 \\
 & 2{{\cos }^{2}}\frac{A-B}{2}\ge 2
\end{align}$
We can write \[2{{\cos }^{2}}\frac{A-B}{2}\] as \[1+\cos (A-B)\] using formula \[\frac{1+\cos A}{2}={{\cos }^{2}}A\].
\[\begin{align}
  & 1+\cos (A-B)\ge 2 \\
 & \cos (A-B)\ge 1
\end{align}\]
As the interval of the values for cosine is $\left[ -1,1 \right]$, we will consider,
$\begin{align}
  & \cos (A-B)=1 \\
 & \cos (A-B)=\cos 0 \\
 & A-B=0 \\
 & A=B
\end{align}$
We will again solve by taking angles $B+C=\pi -A$ and $\cos B+\cos C$in the given equation $\cos A+\cos B+\cos C=\frac{3}{2}$.
Similarly we get an equation $B=C$.
Now with $A=B$and $B=C$, we can say that $A=B=C$. So the triangle will be an equilateral triangle.
The triangle$ABC$ such that$\cos A+\cos B+\cos C=\frac{3}{2}$is an equilateral triangle. Hence the correct option is (B).
Note:
Isosceles triangle is a type of triangle in which two sides and the angles opposite to both the sides are equal. Right angled triangle is a triangle in which all the three sides are unequal and one angle is ${{90}^{0}}$. And in the equilateral triangle all the three sides are equal and all the three angles are equal with magnitude of ${{60}^{0}}$.