Question

In any $\Delta ABC$, prove that $\dfrac{\cos A}{a}+\dfrac{\cos B}{b}+\dfrac{\cos C}{c}=\dfrac{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}{2abc}$.

Answer 1

Hint: We will be using the concept of trigonometry, solution of triangles to solve the problem. We will be using cosine rules to solve the problem. Use triangle diagrams to understand the concept easily.

Complete step-by-step answer:
We have been given a$\Delta ABC$ and we have to prove that $\dfrac{\cos A}{a}+\dfrac{\cos B}{b}+\dfrac{\cos C}{c}=\dfrac{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}{2abc}$.
Now, we will first draw a $\Delta ABC$ and label its sides a, b, c.

Now, we know that according to cosine rule,
$\begin{aligned} & \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}..........\left( 1 \right) \\\ & \cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}..........\left( 2 \right) \\\ & \cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}..........\left( 3 \right) \\\ \end{aligned}$
Now, we will take left – hand side and prove it to be equal to right hand side. Now, in L.H.S we have,
$\dfrac{\cos A}{a}+\dfrac{\cos B}{b}+\dfrac{\cos C}{c}$
Now, we will substitute the value of $\cos A,\cos B,\cos C$ from equation (1), (2), (3) in L.H.S. So, we get,

& \dfrac{\cos A}{a}+\dfrac{\cos B}{b}+\dfrac{\cos C}{c}=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2abc}+\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2abc}+\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2abc} \\\ & =\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}+{{a}^{2}}+{{c}^{2}}-{{b}^{2}}+{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2abc} \\\ \end{aligned}$$ Now, on further simplifying we get, $\dfrac{\cos A}{a}+\dfrac{\cos B}{b}+\dfrac{\cos C}{c}=\dfrac{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}{2abc}$ Since, L.H.S = R.H.S Hence, we have proved that $\dfrac{\cos A}{a}+\dfrac{\cos B}{b}+\dfrac{\cos C}{c}=\dfrac{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}{2abc}$. Note: To solve these types of questions one must remember cosine rule, sine rule to simplify the solution. Sine rule is: $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k$