Question

If for a triangle ABC,
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$
then value of ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$ equals

• A. 1/4
• B. 1/2
• C. 3/4
• D. 1

Answer 1

The correct option is C 3/4

In a triangle $ABC$
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$
$\to (a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|=0$
$\Rightarrow (a+b+c)[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]=0$
in a triangle $a+b+c\ne 0$
$\therefore a=b=c=60$
Hence, ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=\frac{3}{4}$