Question

In a $\Delta ABC$, if $a\cos A=b\cos B$, show that the triangle is either isosceles or right – angled.

Answer 1

Hint: We will be using the concept of solution of triangles to solve the problem. We will be using sine rule to relate the sides of the triangle a and b with the angle of the triangle ABC then we will use the trigonometric identity that $\sin 2A=2\sin A\cos A$ to convert the equation in terms of sinA and sinB then finally we will use the identity that $\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$ to deduce that the $\cos \left( A+B \right)=0$ to prove that the triangle is right angle.

Complete Step-by-Step solution:
We have been given that $a\cos A=b\cos B$, we have to prove that the triangle is either isosceles or right angled.
We will first draw a triangle ABC and name its sides a, b, c.

Now, we know that according to sine rule,
$\begin{align}
  & \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k \\
 & \Rightarrow \dfrac{\sin A}{k}=a.........\left( 1 \right) \\
 & \Rightarrow \dfrac{\sin B}{k}=b.........\left( 2 \right) \\
\end{align}$
Now, we have been given that,
$a\cos A=b\cos B..........\left( 3 \right)$
We will substitute (1) & (2) in (3)
$\begin{align}
  & \dfrac{\sin A}{k}\cos A=\dfrac{\sin B}{k}\cos B \\
 & \sin A\cos A=\sin B\cos B \\
\end{align}$
Now, we will multiply by 2 on both sides,
$2\sin A\cos A=2\sin B\cos B$
Now, we know that $\sin 2A=2\sin A\cos A$ using this identity we have,
$\begin{align}
  & \sin 2A=\sin 2B \\
 & \Rightarrow \sin 2A-\sin 2B=0 \\
\end{align}$
Now, we know that $\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$ using this, we have,
$\begin{align}
  & 2\sin \left( \dfrac{2A-2B}{2} \right)\cos \left( \dfrac{2A+2B}{2} \right)=0 \\
 & 2\sin \left( A-B \right)\cos \left( A+B \right)=0 \\
 & \sin \left( A-B \right)\cos \left( A+B \right)=0 \\
\end{align}$
Now, either,
$\sin \left( A-B \right)=0\ or\ \cos \left( A+B \right)=0$
So, we have,
$\sin \left( A-B \right)=0$
We know that $\sin \theta =0$, therefore,
A – B = 0
A = B
This shows that the triangle is isosceles. Also, we have,
$\cos \left( A+B \right)=0$
We know that $\cos \left( \dfrac{\pi }{2} \right)=0$ therefore, $A+B=\dfrac{\pi }{2}.......\left( 4 \right)$
Also, in $\Delta ABC$ by using angle sum property of triangle we have $A+B+C=\pi ..........\left( 5 \right)$
Now, we will use (4) in (5),
$\begin{align}
  & \dfrac{\pi }{2}+C=\pi \\
 & C=\pi -\dfrac{\pi }{2} \\
 & C=\dfrac{\pi }{2} \\
\end{align}$
This shows that the triangle is right angle but since either $\sin \left( A-B \right)=0\ or\ \cos \left( A+B \right)=0$, we have either $\Delta ABC$ is isosceles or right angled.
Hence Proved.

Note: In these types of questions it is important to note that we have used sine and cosine rule to relate the sides of the triangle with the cosines and sine of the angle also we have used the fact that a triangle is right angle if one side of the triangle is 90 degree. To further simplify the solution it is required that one remembers important trigonometric identities like, $\sin C-\sin D=2\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)$.