Question

In any $\Delta ABC$, prove that $\dfrac{c-b\cos A}{b-c\cos A}=\dfrac{\cos B}{\cos C}$.

Answer 1

Hint: We will be using the concept of solution of triangles to solve the problem. We will be using the projection formula of the triangle to relate the side c and b with the other side and cosine of the other angle of the triangle then we will further simplify the left hand side of the equation by substituting the side c and b of the triangle.

Complete Step-by-Step solution:
We have been given a$\Delta ABC$ and we have to prove that $\dfrac{c-b\cos A}{b-c\cos A}=\dfrac{\cos B}{\cos C}$.
Now, we will first draw a $\Delta ABC$ and label its sides a, b, c.

Now, we know that according to projection formula we have, in $\Delta ABC$,
$\begin{align}
  & a=b\cos C+c\cos B..........\left( 1 \right) \\
 & b=c\cos A+a\cos C..........\left( 2 \right) \\
 & c=a\cos B+b\cos A..........\left( 3 \right) \\
\end{align}$
Now, we will take the left – hand side of the problem and prove it to be equal to the right hand side. In L.H.S we have,
$\dfrac{c-b\cos A}{b-c\cos A}$
Now, we will substitute the value of c and b in L.H.S in the terms only which are not multiplied cosine from (2) and (3). So, we have
$\begin{align}
  & \dfrac{c-b\cos A}{b-c\cos A}-\dfrac{a\cos B+b\cos A-b\cos A}{c\cos A+a\cos C-c\cos A} \\
 & =\dfrac{a\cos B}{a\cos C} \\
\end{align}$
Now, we will cancel a in both numerator and denominator,
$\dfrac{c-b\cos A}{b-c\cos A}=\dfrac{\cos B}{\cos C}$
Since, L.H.S = R.H.S
Hence, proved that $\dfrac{c-b\cos A}{b-c\cos A}=\dfrac{\cos B}{\cos C}$.

Note: In these types of questions it is important to remember projection formulae. According to projection formulae.
In any triangle ABC,
$\begin{align}
  & a=b\cos C+c\cos B \\
 & b=c\cos A+a\cos C \\
 & c=a\cos B+b\cos A \\
\end{align}$