Question

In triangle $ABC,$ if $si{n}^{2}A+si{n}^{2}B+si{n}^{2}C=2$, then the triangle is

• A. Right angle but need not be isosceles
• B. Right angled and isosceles
• C. Isosceles but need not be right angled
• D. Equilateral

Answer 1

The correct option is A

Right angle but need not be isosceles

Explanation for the correct options:

Determine the type of triangle:

$$\begin{gathered} {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2\mathbf{\left[}\mathbf{\text{Given}}\mathbf{\right]} \\ \Rightarrow si{n}^{2}A+si{n}^{2}B+si{n}^{2}C-2=0 \\ \Rightarrow 1-si{n}^{2}A+1-si{n}^{2}B+si{n}^{2}C=0 \\ \Rightarrow {\cos }^{2}A+{\cos }^{2}B–{\sin }^{2}C=0 \\ \Rightarrow 2{\cos }^{2}A+2{\cos }^{2}B–2{\sin }^{2}C=0\mathbf{[}\mathbf{\text{Multiplying by on}}\mathbf{}\mathbf{2}\mathbf{\text{both sides}}\mathbf{]} \\ \Rightarrow (1+\cos 2A)+(1+\cos 2B)–2(1–{\cos }^{2}C)=0 \\ \Rightarrow 2+\cos 2A+\cos 2B–2+2{\cos }^{2}C=0 \\ \Rightarrow \cos 2A+\cos 2B+2{\cos }^{2}C=0\mathbf{\left[}\mathbf{}\mathbf{cos}\mathbf{2}\mathbf{A}\mathbf{+}\mathbf{}\mathbf{cos}\mathbf{2}\mathbf{B}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{cos}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{}\mathbf{cos}\mathbf{(}\mathbf{A}\mathbf{–}\mathbf{B}\mathbf{)}\mathbf{}\mathbf{by}\mathbf{}\mathbf{trigonometric}\mathbf{}\mathbf{identity}\mathbf{\right]} \\ \Rightarrow 2\cos (A+B)\cos (A–B)+2{\cos }^{2}C=0 \\ \Rightarrow \cos (A+B)\cos (A–B)+{\cos }^{2}C=0 \\ \Rightarrow \cos (\pi –C)\cos (A–B)+{\cos }^{2}C=0\mathbf{[}\mathbf{\because }\mathbf{\text{In a triangle}}\mathbf{}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{\pi }\mathbf{]} \\ \Rightarrow -\cos C\cos (A–B)+{\cos }^{2}C=0 \\ \Rightarrow \cos C\cos (A–B)–\cos C\cos C=0 \\ \Rightarrow \cos C\cos (A–B)–\cos C\cos [\pi -(A+B\left)\right]=0\mathbf{[}\mathbf{\because }\mathbf{\text{In a triangle}}\mathbf{}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{\pi }\mathbf{]} \\ \Rightarrow \cos C\cos (A–B)+\cos C\cos (A+B)=0 \\ \Rightarrow \cos C\left[\cos \right(A–B)+\cos (A+B\left)\right]=0 \\ \Rightarrow \cos C\left(2\cos A\cos B\right)=0 \end{gathered}$$

This shows that, one of the angles $A,B\text{or}C$ must be$90$degrees.

Now, let us assume $A=\frac{\mathrm{\pi }}{2}=90°$ , then, $B+C=\frac{\mathrm{\pi }}{2}$

$$\begin{gathered} \Rightarrow {\sin }^{2}A={\sin }^2\frac{\mathrm{\pi }}{2} \\ =1 \\ \Rightarrow {\sin }^{2}B+{\sin }^{2}C={\sin }^2\left[\frac{\mathrm{\pi }}{2}–C\right]+{\sin }^{2}C \\ ={\cos }^{2}C+{\sin }^{2}C \\ =1 \\ \therefore {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2 \end{gathered}$$

So this is a right-angled triangle but need not necessarily be isosceles.

Hence, option A is the correct answer.