If in a triangle ABC, \(\frac{{2\cos A}}{a} + \frac{{\cos B}}{b} + \frac{{2\cos C}}{c} = \frac{a}{{bc}} + \frac{b}{{ca}}\) then the value of the angle A is
Given:
\(\frac{{2\cos A}}{a} + \frac{{\cos B}}{b} + \frac{{2 \cos C}}{c} = \frac{a}{{bc}} + \frac{b}{{ca}}\)
Calculation:
\(\frac{{2\cos A}}{a} + \frac{{\cos B}}{b} + \frac{{2 \cos C}}{c} = \frac{a}{{bc}} + \frac{b}{{ca}}\)
After rearranging the above equation,
⇒ \(\frac{{2bc\cos A + ac\cos B + 2ab\cos C}}{{abc}} = \frac{{{a^2} + {b^2}}}{{abc}}\)
⇒ 2bc cos A + ac cos B + 2ab cos C = a2 + b2
⇒ bc cos A + bc cos A + ac cos B + ab cos C + ab cos C = a2 + b2
⇒ bc cos A + b (c cos A + a cos C) + a (c cos B + b cos C) = a2 + b2
⇒ bc cos A + b2 + a2 = a2 + b2
⇒ bc cos A = 0
⇒ cos A = 0 = \(\cos \frac{\pi }{2}\)
Then,
A = π/2
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