Q189 / 270 · NDA 2023 (Phase 1) · Mathematics

Paragraph: In the triangle $A B C, a^2+b^2+c^2=a c+\sqrt{3} b c$
Question: What is the nature of the triangle?

  • A
    Equilateral
  • B
    Isosceles
  • Correct answer
    C
    Right angled triangle
  • D
    Scalene but not right angled

Solution

$a^2+b^2+c^2=a c+\sqrt{3} b c$ $\left(a^2+c^2-a c\right)+\left(b^2-\sqrt{3} b c\right)=0$ $\left[a^2+\left(\frac{c}{2}\right)^2-2 \cdot a\left(\frac{c}{2}\right)\right]+\left[b^2+\left(\frac{\sqrt{3} c}{2}\right)^2-2 \cdot b \cdot \frac{\sqrt{3} c}{2}\right]=0$ $\left(a-\frac{c}{2}\right)^2+\left(b-\frac{\sqrt{3} c}{2}\right)^2=0$ It is only possible if $a-\frac{c}{2}=0$ and $b-\frac{\sqrt{3} c}{2}=0$ $a=\frac{c}{2}$ and $b=\frac{\sqrt{3} c}{2}$ $a: b: c=\frac{c}{2}: \sqrt{3} \frac{c}{2}: c$ $=\frac{1}{2}: \frac{\sqrt{3}}{2}: 1$ since $\mathrm{a}=\mathrm{K} \sin \mathrm{A}$ $\mathrm{b}=\mathrm{k} \sin \mathrm{B}, \mathrm{C}=\mathrm{k} \sin \mathrm{c}$ $\sin A: \sin B: \sin C$ $=\sin 30^{\circ}: \sin 60^{\circ}: \sin 90^{\circ}$ $\therefore$ Angles of triangle are $30^{\circ}, 60^{\circ}$ and $90^{\circ}$ i.e. triangle is right angled.

Hence, the correct answer is option: 'C'.

Relevant Questions