Question

In a $\Delta ABC$, if $\cos A=\dfrac{\sin B}{2\sin C}$, show that the triangle is isosceles.

Answer 1

Hint: We will be using a solution of triangles to solve the problem. We will be using sine rule to find the value of sinB and sinC also we will be using cosine rule to find the value of cosA in terms of the side of the triangles then we will use these value in the equation given to us to and simplify the equation to have the result that the two sides of the triangle are equal.

Complete Step-by-Step solution:
We have been given a $\Delta ABC$ and $\cos A=\dfrac{\sin B}{2\sin C}$, we have to show that the triangle is isosceles.
Now, we will draw a triangle ABC.

Now, we will apply sine rule in $\Delta ABC$, which states that the sides of any triangle is proportional to the sine of the angle opposite to them.
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k............\left( 1 \right)$
Where k is any constant. So, we get $\dfrac{\sin b}{\sin c}=\dfrac{bk}{ck}=\dfrac{b}{c}$ from (1).
Also, now applying cosine rule in $\Delta ABC$ for angle A we have,
$\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}................\left( 2 \right)$
Now, we have been given in question that $\cos A=\dfrac{\sin B}{2\sin C}...........\left( 3 \right)$
Now, substituting equation (1) and (2) in (3), we have,
$\begin{align}
  & \dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=\dfrac{b}{2c} \\
 & \dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2b}=\dfrac{b}{2} \\
\end{align}$
On cross multiplying we have,
$\begin{align}
  & {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=\dfrac{2{{b}^{2}}}{2} \\
 & {{b}^{2}}+{{c}^{2}}-{{a}^{2}}={{b}^{2}} \\
 & {{c}^{2}}={{a}^{2}} \\
 & \Rightarrow c=a \\
\end{align}$
Since, c = a in $\Delta ABC$ it is proved that $\Delta ABC$ is isosceles if $\cos A=\dfrac{\sin B}{2\sin C}$.

Note: To solve these type of questions it is important to notice that we have been using a fact that if two sides of a triangle are equal then the triangle is isosceles in nature also we have used cosine and sine rule to relate sine and cosine with the sides of the triangles.