Question

Moar questions!

Answer 1

If in a triangle ABC,\[\dfrac{2\cos A}{a} + \frac{\cos B}{b} + \dfrac{2\cos C}{c} = \dfrac{a}{bc} + \dfrac{b}{ca}\]then find \(\angle A\).

Answer 2

@dan815

Answer 3

Use cosine law, and you're done!

I can write it as :

\[2bc \cos A + ac \cos B + 2ab \cos C = a^2 + b^2 \\
(b^2 + \cancel{c^2} - \cancel{a^2} + \cancel{a^2} + b^2 - \cancel{c^2}) + \cfrac{a^2 + c^2 - b^2}{2} = a^2 + b^2 \\
c^2 + a^2 - b^2 = 2(a^2 - b^2) \\
c^2 = a^2 - b^2 \\
a^2 = c^2 + b^2 \]

So, a is the length of hypotenuse. That is, \(\angle A = 90^{\circ}\)

Answer 4

Good, mathslover got it. :D