Properties of Triangles 1 Question 18
18. 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)
Analytical & Descriptive Questions
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Answer:
Correct Answer: 18. $c=\sqrt{6}, \angle B=45^{\circ}$ and $\angle A=75^{\circ}$
Solution:
- Given, $\frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{b c}+\frac{b}{c a}$
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
$$ \begin{gathered} \frac{2\left(b^{2}+c^{2}-a^{2}\right)}{2 a b c}+\frac{c^{2}+a^{2}-b^{2}}{2 a b c} \\ +\frac{2\left(a^{2}+b^{2}-c^{2}\right)}{2 a b c}=\frac{a}{b c}+\frac{b}{c a} \\ \Rightarrow \quad \frac{2\left(b^{2}+c^{2}-a^{2}\right)+c^{2}+a^{2}-b^{2}+2\left(a^{2}+b^{2}-c^{2}\right)}{2 a b c} \\ =\frac{a^{2}+b^{2}}{a b c} \\ \Rightarrow \quad 3 b^{2}+c^{2}+a^{2}=2 a^{2}+2 b^{2} \\ b^{2}+c^{2}=a^{2} \end{gathered} $$
Hence, the angle $A$ is $90^{\circ}$.
