Question

In a triangle ABC, with usual notations, if $\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0,$ then $4si{n}^{2}A+24si{n}^{2}B+36si{n}^{2}C$ is equal to

• A. $48$
• B. $50$
• C. $44$
• D. $34$

Answer 1

The correct option is A $48$

$\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0$

using $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ property.

$\Rightarrow \left|\begin{array}{ccc}1& a& b\\ 0& c-a& a-b\\ 0& b-a& c-b\end{array}\right|=0$

Expanding along first column.

$\Rightarrow 1\left[\right(c-a\left)\right(c-b)-(b-a\left)\right(a-b\left)\right]+0=0$

$\Rightarrow c^2-ac-bc+ab-(ab-a^2-b^2+ab)=0$

$\Rightarrow c^2-ac-bc+ab-ab+a^2+b^2-ab=0$

$\Rightarrow a^2+b^2+c^2-ab-bc-ca=0$

using identify

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$

$\Rightarrow \frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$

It us only possible if $a+b=c$

$\Rightarrow$ Triangle is equilateral

$\Rightarrow \angle A=\angle B=\angle C={60}^{\circ }$

$\Rightarrow 4{\sin }^{2}A+24{\sin }^{2}B+36{\sin }^{2}c$

$=4{\sin }^2{60}^{\circ }+24{\sin }^2{60}^{\circ }+36{\sin }^2{60}^{\circ }$

$=64{\sin }^2{60}^{\circ }=64\times {\left(\frac{\sqrt{3}}{2}\right)}^2$

$=64\times \frac{3}{4}$

$=48$