Question

In $\Delta ABC,\angle C=2\angle A,$ and $AC=2BC,$ then the value of $\frac{a^2+b^2+c^2}{R^2}$ is ?

(where R is circumradius of triangle)

• A. $8$
• B. $6$
• C. $5$
• D. None of these

Answer 1

Answer: (A)

$8$

Given $\angle C=2\angle A$ and $b=2a$
Now using sine rule
$\frac{a}{\sin A}=\frac{c}{\sin C}=2R$ (i)
From first two $\frac{a}{\sin A}=\frac{c}{\sin 2A}=\frac{c}{2\sin A\cos A}$
$\Rightarrow \cos A=\frac{c}{2a}=\frac{c}{b}=\frac{b^2+c^2-a^2}{2bc}$ using cosine rule.
$\Rightarrow c^2+a^2=b^2$. Now from (i) $R^2=\frac{a^2}{4{\sin }^{2}a}=\frac{a^2}{4(1-c^2/b^2)}$
$\Rightarrow R^2=\frac{a^2{b}^2}{4(b^2-c^2)}=\frac{b^2}{4}$
$\therefore \frac{a^2+b^2+c^2}{R^2}=\frac{2b^2}{b^2/4}=8$