Find: c/(a + b) + b/(a + c) = ?

[Albert1]

$\triangle ABC$ has side lengths $a,b,c$

$\text{if}\,\, \angle A=60^o$

find :

$\dfrac {c}{a+b}+\dfrac {b}{a+c}=?$

 

[juantheron]

Re: find :c/(a+b) + b/(a+c) =?

Using Cosine formula $\displaystyle \cos (A) = \frac{b^2+c^2-a^2}{2bc}$

Now Given $A = 60^0$ . So $\displaystyle \frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$

So $b^2+c^2-a^2 = bc \Rightarrow b^2+c^2 = a^2+bc$

Now Given $\displaystyle \frac{c}{a+b}+\frac{b}{a+c} = \frac{c\cdot(a+c)+b\cdot (a+b)}{(a+b)\cdot (a+c)}$

So $\displaystyle = \frac{c^2+ac+b^2+ab}{a^2+ab+bc+ca} = \frac{a^2+ab+bc+ca}{a^2+ab+bc+ca} = 1$

Using $b^2+c^2 = a^2+bc$

 

[Albert1]

Re: find :c/(a+b) + b/(a+c) =?

Can someone find its value using geometry only ?

 

[Albert1]

Re: find :c/(a+b) + b/(a+c) =?

View attachment 1380

 

[CellCoree]

I would approach this problem by first recognizing that the given information involves a triangle with side lengths a, b, and c, and an angle of 60 degrees. I would then use my knowledge of geometry and trigonometry to solve for the unknown value of the expression $\dfrac {c}{a+b}+\dfrac {b}{a+c}$.

Using the Law of Cosines, I can determine that $c^2 = a^2 + b^2 - 2ab\cos 60^o$. Simplifying this equation, I get $c^2 = a^2 + b^2 - ab$. Similarly, I can use the Law of Sines to determine that $\dfrac{c}{\sin 60^o} = \dfrac{a}{\sin B}$. Since $\sin 60^o = \dfrac{\sqrt{3}}{2}$ and $\sin B = \dfrac{b}{c}$, I can substitute these values into the equation to get $\dfrac{2c}{\sqrt{3}} = \dfrac{a}{b}$. Rearranging this equation, I get $c = \dfrac{a\sqrt{3}}{2b}$.

Now, I can substitute this value of c into the original expression to get $\dfrac {\dfrac{a\sqrt{3}}{2b}}{a+b}+\dfrac {b}{a+\dfrac{a\sqrt{3}}{2b}}$. Simplifying this further, I get $\dfrac{\sqrt{3}}{2(a+b)}+\dfrac{2b}{2ab+a\sqrt{3}}$. Finally, I can combine these fractions to get $\dfrac{\sqrt{3}(2ab+a\sqrt{3}+2b^2)}{2(a+b)(2ab+a\sqrt{3})}$. This is the final answer for the expression $\dfrac {c}{a+b}+\dfrac {b}{a+c}$ when $\angle A = 60^o$.