Question

In a triangle if $si{n}^{2}A+si{n}^{2}B+si{n}^{2}C=2$, then triangle is always -

• A. Equilateral
• B. Isosceles
• C. Right angled triangle
• D. Obtuse angled triangle

Answer 1

The correct option is C

Right angled triangle

The given equation is -

$\begin{array}{c}\Rightarrow si{n}^{2}A+si{n}^{2}B+si{n}^{2}C=2\\ \Rightarrow \left(1-co{s}^{2}A\right)+\left(1-co{s}^{2}B\right)+\left(1-co{s}^{2}C\right)=2\\ \Rightarrow co{s}^{2}A+co{s}^{2}B+co{s}^{2}C=1\\ \Rightarrow 2co{s}^{2}A+2co{s}^{2}B+2co{s}^{2}C=2\\ \Rightarrow -1=(2co{s}^{2}A-1)+(2co{s}^{2}B-1)+(2co{s}^{2}C-1)\\ \Rightarrow -1=\left\{cos2A+cos2B\right\}+cos2C\\ \Rightarrow -1=2cos\left(A+B\right)\ast cos\left(A-B\right)+cos2(A+B)\left[2C=360-2\left(A+B\right)\right]\\ \Rightarrow -1=2cos\left(A+B\right)\ast cos\left(A-B\right)+2co{s}^2(A+B)-1\\ \Rightarrow 0=2cos\left(A+B\right)\ast cos\left(A-B\right)+2co{s}^2(A+B)\end{array}$

$\begin{array}{c}\Rightarrow 0=2cos\left(A+B\right)\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\\ \Rightarrow 0=cos\left(A+B\right)\ast cosA\ast cosB\end{array}$

Means at least one of the three terms in the right hand side should be equal to 0, so either A + B = 90${}^{\circ }$ or A = 90${}^{\circ }$, or B = 90${}^{\circ }$.

Therefore, it's a right triangle.