Q189 / 270 · NDA 2017 (Phase 2) · Mathematics

In triangle ABC , if $\frac{\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}+\sin ^2 \mathrm{C}}{\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}+\cos ^2 \mathrm{C}}=2$ then the triangle is

  • Correct answer
    A
    right-angled
  • B
    equilateral
  • C
    isosceles
  • D
    obtuse-angled

Solution

Given, $\frac{\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}+\sin ^2 \mathrm{C}}{\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}+\cos ^2 \mathrm{C}}=2$ Let us take $\mathrm{A}=30^{\circ}, \mathrm{B}=60^{\circ}, \mathrm{C}=90^{\circ}$ $\begin{aligned} & \frac{\sin ^2 30^{\circ}+\sin ^2 60^{\circ}+\sin ^2 90^{\circ}}{\cos ^2 30^{\circ}+\cos ^2 60^{\circ}+\cos ^2 90^{\circ}} \\ & =\frac{\frac{1}{4}+\frac{3}{4}+1}{\frac{3}{4}+\frac{1}{4}+0}=\frac{1+1}{1}=\frac{2}{1}=2\end{aligned}$ So, the given triangle is right-angled triangle.

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

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