Question

Given $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ for a $\Delta ABC$ with usual notation. If $\frac{\cos A}{\alpha }=\frac{\cos B}{\beta }=\frac{\cos C}{\gamma }$, then the ordered triad $(\alpha ,\beta ,\gamma )$ has a value :

• A. $(3,4,5)$
• B. $(19,7,25)$
• C. $(7,19,25)$
• D. $(5,12,13)$

Answer 1

The correct option is C $(7,19,25)$

$\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\lambda$

$\therefore$ $b+c=11\lambda \cdots \left(1\right)c+a=12\lambda \cdots \left(2\right)a+b=13\lambda \cdots \left(3\right)$

$\Rightarrow a+b+c=18\lambda \cdots \left(4\right)$
Now subtract $\left(1\right)$ from $\left(4\right)$, we get
$a=7\lambda \Rightarrow b=6\lambda ,c=5\lambda$

By cosine rule,
$\cos A=\frac{36{\lambda }^2+25{\lambda }^2-49{\lambda }^2}{2\times 6\lambda \times 5\lambda }=\frac{1}{5}$

$\cos B=\frac{25{\lambda }^2+49{\lambda }^2-36{\lambda }^2}{2\times 5\lambda \times 7\lambda }=\frac{19}{35}$

$\cos C=\frac{49{\lambda }^2+36{\lambda }^2-25{\lambda }^2}{2\times 7\lambda \times 6\lambda }=\frac{5}{7}$

$\because \frac{\cos A}{\alpha }=\frac{\cos B}{\beta }=\frac{\cos C}{\gamma }$
$\Rightarrow \frac{1}{5\alpha }=\frac{19}{35\beta }=\frac{5}{7\gamma }=k\therefore \alpha =\frac{1}{5k}=\frac{7}{35k},\beta =\frac{19}{35k},$
and $\gamma =\frac{5}{7k}=\frac{25}{35k}$

So, $(\alpha ,\beta ,\gamma ):(7,19,25)$