Question

If in a triangle, $a^2+b^2+c^2=ca+ab\sqrt{3}$ then the triangle is

• A. equilateral
• B. isosceles
• C. right angled with $A=90°,B=60°andC=30°$
• D. None of these

Answer 1

The correct option is C

right angled with $A=90°,B=60°andC=30°$

Explanation for correct option:

Step 1: Find the value of $a,b$ and $c$:

Given,

$a^2+b^2+c^2=ca+ab\sqrt{3}$

$$\begin{gathered} \Rightarrow a^2+b^2+c^2–ca–ab\sqrt{3}=0 \\ \Rightarrow {(a\frac{\sqrt{3}}{2}–b)}^2+{(\frac{a}{2}–c)}^2=0 \\ \Rightarrow (a\frac{\sqrt{3}}{2}–b)=0...\left(1\right) \end{gathered}$$

And

$\Rightarrow \begin{array}{rcl}& & (\frac{\mathrm{a}}{2}–\mathrm{c})\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 0\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}...\left(2\right)$

since

$\sqrt{3}a=2b=2\sqrt{3}c\left[\mathrm{from}1\&2\right]$

Let, $\sqrt{3}a=2b=2\sqrt{3}c=k$

Then$a=\frac{k}{\sqrt{3}},b=\frac{k}{2},c=\frac{k}{2\sqrt{3}}$

Step 2: Find the nature of triangle:

$\begin{array}{rcl}{b}^2+c^2& =& \frac{k^2}{4}+\frac{k^2}{12}\\ & =& \frac{4k^2}{12}\\ & =& \frac{k^2}{3}\\ & =& a^2\end{array}$

So $A=90°$

$\begin{array}{rcl}SinB& =& \frac{b}{a}\\ & =& \frac{\sqrt{3}}{2}\end{array}$

So,

$$\begin{gathered} B=60° \\ C=30° \end{gathered}$$

So, the triangle is right angled with$A=90°,B=60°andC=30°$

Hence, the correct option is C.