Question

The lengths of the sides of the triangle are$\alpha -\beta ,\alpha +\beta $ and $\sqrt{3{{\alpha }^{2}}+{{\beta }^{2}}}$,$(\alpha >\beta >0)$. Its largest angle is
A. $\frac{3\pi }{4}$
B. $\frac{\pi }{2}$
C. $\frac{2\pi }{3}$
D. $\frac{5\pi }{6}$

Answer 1

Hint: To solve this question we will take each of the given lengths of the sides of the triangle as values of $a,b,c$. We will determine which side is the largest and then the angle opposite to it will be the largest angle. Then we will substitute the values of sides $a,b,c$in the cosine rule with the angle we have to find and then we will derive the value of the angle which will be the largest angle of the triangle.
 Formula used:
The expansion formulas are:
$\begin{align}
  & {{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & (a+b)(a-b)={{a}^{2}}-{{b}^{2}}
\end{align}$
Cosine rule:
\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\].
Complete step-by-step solution:
We are given a triangle having sides $\alpha -\beta ,\alpha +\beta $ and $\sqrt{3{{\alpha }^{2}}+{{\beta }^{2}}}$and we have to find the value of its largest angle when $(\alpha >\beta >0)$.
As three sides are given so,
$\begin{align}
  & a=\alpha -\beta \\
 & b=\alpha +\beta \\
 & c=\sqrt{3{{\alpha }^{2}}+{{\beta }^{2}}} \\
\end{align}$
Here the length of the side $c$ is the largest so the angle opposite to it will be the largest angle. Hence, angle $C$is the largest angle.
We will choose the cosine rule with angle $C$.
We will now substitute the values of $a,b$ and $c$in the cosine rule \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\].
\[{{\left( \sqrt{3{{\alpha }^{2}}+{{\beta }^{2}}} \right)}^{2}}={{(\alpha -\beta )}^{2}}+{{(\alpha +\beta )}^{2}}-2(\alpha -\beta )(\alpha +\beta )\cos C\]
We will now use the formulas of expansion to open the brackets,
\[\begin{align}
  & 3{{\alpha }^{2}}+{{\beta }^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta +{{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -2({{\alpha }^{2}}-{{\beta }^{2}})\cos C \\
 & 3{{\alpha }^{2}}+{{\beta }^{2}}=2{{\alpha }^{2}}+2{{\beta }^{2}}-2({{\alpha }^{2}}-{{\beta }^{2}})\cos C \\
 & {{\alpha }^{2}}-{{\beta }^{2}}=-2({{\alpha }^{2}}-{{\beta }^{2}})\cos C \\
 & \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{-2({{\alpha }^{2}}-{{\beta }^{2}})}=\cos C \\
 & -\frac{1}{2}=\cos C
\end{align}\]
The value of cosine $\frac{-1}{2}$ is at angle\[\frac{2\pi }{3}\] so,
 \[\begin{align}
  & \cos \frac{2\pi }{3}=\cos C \\
 & C=\frac{2\pi }{3}
\end{align}\]
The largest angle of the triangle is \[C=\frac{2\pi }{3}\] when the lengths of the sides of the triangle are$\alpha -\beta ,\alpha +\beta $ and $\sqrt{3{{\alpha }^{2}}+{{\beta }^{2}}}$and $(\alpha >\beta >0)$. Hence the correct option is (C).
Note:
We have used a theorem of triangle to select the largest angle of the triangle in the above solution according to which “ if two sides of a triangle are not equal then the angle opposite to the longer side is greater or larger”. Hence angle $C$ opposite to side $c$ which is the largest side of the triangle will be the largest angle.