Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In triangle $\triangle \; ABC$ , if $$2\frac{\cos A}{a} + \frac{\cos B}{b} + 2\frac{\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$$ find angle $A$.

This is a quiz bee problem sent to me by my friend in FB. He asked me if I can do a solution for it. Well I tried several ways but I am out of idea now. The answer is 90 degree but what he asked, and I am also asking it now, is the solution for it.

Thank you.

share|cite|improve this question
up vote 5 down vote accepted

Using $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ etc.,

we get, $$\frac2a\frac{b^2+c^2-a^2}{2bc}+\frac1b\frac{a^2+c^2-b^2}{2ac}+\frac2c\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2}{abc}$$

or, $$ 2(b^2+c^2-a^2)+(a^2+c^2-b^2)+2(a^2+b^2-c^2)=2(a^2+b^2)$$

or $b^2+c^2=a^2$ as $abc\ne0$ $a,b,c$ being the sides of triangle.

So, $\cos A=0\implies A=(2n+1)\frac\pi2 $ where $n$ is any integer.

As $0<A<\pi,A=\frac\pi2$

share|cite|improve this answer
I don't understand how the cosine law arrive to the quantity $b^2 + c^2 = a^2$. – Romel Verterra Nov 21 '12 at 12:36
@RomelVerterra, please find the edited answer – lab bhattacharjee Nov 21 '12 at 12:44
Would you please verify the original question.I think the coefficient of $\frac{\cos B}b$ was 1, but I could not find that version . – lab bhattacharjee Nov 21 '12 at 12:46
Thank you. And yes the coefficient of $\frac{cos B}{b}$ is $1$, not $2$. I corrected the problem. – Romel Verterra Nov 21 '12 at 12:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.