Properties Of Triangles Ques 10
If in a $\triangle A B C$,
$ \frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{b c}+\frac{b}{c a} $
Then, the value of the $\angle A$ is degree.
(1993, 2M)
Show Answer
Answer:
Correct Answer: 10.$(90^{\circ})$
Solution:
- Given, $\frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{b c}+\frac{b}{c a}$ $\quad$ …….(i)
We know that, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
$ \begin{aligned} \cos B & =\frac{c^{2}+a^{2}-b^{2}}{2 a c} \\ \text { and } \quad \cos C & =\frac{a^{2}+b^{2}-c^{2}}{2 a b} \end{aligned} $
On putting these values in Eq. (i), we get
$\frac{2\left(b^{2}+c^{2}-a^{2}\right)}{2 a b c}+\frac{c^{2}+a^{2}-b^{2}}{2 a b c} $ $+\frac{2\left(a^{2}+b^{2}-c^{2}\right)}{2 a b c}=\frac{a}{b c}+\frac{b}{c a} $
$\Rightarrow \quad \frac{2\left(b^{2}+c^{2}-a^{2}\right)+c^{2}+a^{2}-b^{2}+2\left(a^{2}+b^{2}-c^{2}\right)}{2 a b c} $
$= \quad \frac{a^{2}+b^{2}}{a b c} $
$\Rightarrow \quad 3 b^{2}+c^{2}+a^{2}=2 a^{2}+2 b^{2} $
$b^{2}+c^{2}=a^{2}$
Hence, the angle $A$ is $90^{\circ}$.
BETA
