In a triangle with sides a, b and c, \(\frac3{a+b+c}=\frac1{a+b} + \frac1{a+c}\).

Find the angle between the sides b and c, in degrees.

MathCuber Jan 1, 2020

#1**+2 **

**Hi MathCuber,**

I did this without any trig until the very end.

Just expand and simplify for starters.

Then compare what you have to the cosine rule and you should be able to work it out.

If you have questions then ask. Show or tell me what you have done though.

**This is a teaching answer. Please noone interfer with a more full answer.**

Melody Jan 1, 2020

#2**0 **

I expanded it to \(a^2-2ab-2ac-b^2-c^2-bc=2a^2\).

I don't know how to compare that to the law of cosines: can I simplify further?

MathCuber
Jan 2, 2020

#3**+1 **

That is not finished and it is not what I got.

Can you show your working so I can see where you went wrong?

(mine could be wrong too but it worked out nicely so it probbly isn't)

Did you start it one side at a time or what?

Melody
Jan 2, 2020

#6**0 **

I multiplied everything by (a+b+c)(a+b)(a+c) and simplified it to where there is only one term on one side that is squared because that is like the cosine rule.

MathCuber
Jan 2, 2020

#8**+1 **

ok that should have worked.

\(\frac3{a+b+c}=\frac1{a+b} + \frac1{a+c}\\ \frac3{a+b+c}(a+b+c)(a+b)(a+c)=\frac1{a+b}(a+b+c)(a+b)(a+c) + \frac1{a+c}(a+b+c)(a+b)(a+c)\\ 3(a+b)(a+c)=(a+b+c)(a+c) + (a+b+c)(a+b)\\ 3(a^2+ac+ab+bc)=a^2+ac+ab+bc+ca+c^2\quad +\quad a^2+ab+ab+b^2+ca+cb\\ 3a^2+3ac+3ab+3bc=2a^2+3ac+3ab+2bc+c^2+b^2\\ \)

You can take it from there. I doubt that you will get the same answer as you got before.

When you get the algebra right you can continue to the cosine rule relevance.

And yes, you do want to make it look as much like the cosine rule as possible.

Melody
Jan 2, 2020