Question

In $\Delta ABC$, if $\cos A=\sin B-\cos C$, then show that it is a right-angled triangle.

Answer 1

Hint: In the problem we have to show that $\Delta ABC$ is a right-angled triangle. In the problem we have given an equation in trigonometric ratios which is $\cos A=\sin B-\cos C$. We will rearrange the terms in the above equation so that all the $\cos $ terms are at one side. Now we will use one of the trigonometric identities that is $\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$. In the above equation we will use the sum of interior angles of the trigonometry which is $A+B+C=\pi $ and simplify the equation to get the required result.

Formula Use:
1. $\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$.
2. $\sin \theta =2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)$.
3. $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $.
4. $A+B+C=\pi $.

Complete Step by Step Solution:
Given that, $\cos A=\sin B-\cos C$.
Now we will be shifting the $\cos C$ to LHS side, then we will get
$\Rightarrow \cos A+\cos C=\sin B$.
We know that trigonometric identity that is $\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$.
Using the above identity, we can write the value of $\cos A+\cos C$, as
$\cos A+\cos C=2\cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)$.
Now substituting this value in the given equation, then we will have
$2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=\sin B$.
Considering the RHS side which is $\sin B$. We know that $\sin \theta =2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)$.
Now we are writing $\sin B$ as $\sin B=2\sin \left( \dfrac{B}{2} \right)\cos \left( \dfrac{B}{2} \right)$.
Substituting the above expression in RHS, then we will have
$2\cos \left( \dfrac{A+C}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\sin \left( \dfrac{B}{2} \right)\cos \left( \dfrac{B}{2} \right)$.
We know that the sum of the interior angles in the triangle as $A+B+C=\pi $
From the above value we can write $A+C=\pi -B$.
Now substituting this value in the above equation, then we will have
$2\cos \left( \dfrac{\pi }{2}-\dfrac{B}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\sin \left( \dfrac{B}{2} \right)\cos \left( \dfrac{B}{2} \right)$.
 Simplifying the above expression, then we will get
$\begin{align}
  & 2\sin \left( \dfrac{B}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=2\sin \left( \dfrac{B}{2} \right)\cos \left( \dfrac{B}{2} \right) \\
 & \Rightarrow \cos \left( \dfrac{A-C}{2} \right)=\cos \left( \dfrac{B}{2} \right) \\
\end{align}$
Equating on both sides, then we will get
$\begin{align}
  & \left( \dfrac{A-C}{2} \right)=\left( \dfrac{B}{2} \right) \\
 & A-C=B \\
 & \Rightarrow A=B+C \\
\end{align}$
We have $A+B+C=\pi $, so
$\begin{align}
  & A+B+C=\pi \\
 & \Rightarrow A+A=\pi \\
 & \Rightarrow 2A=\pi \\
 & \Rightarrow A=\dfrac{\pi }{2} \\
\end{align}$
Hence the $\Delta ABC$ is a right-angled triangle.

Note:
In this problem they have only asked to prove the triangle as a right-angled triangle. So, we have followed the above procedure. If they have asked to find where the right angle is in the triangle. Then we can observe the angles of the triangle and write the angle which is equal to $\dfrac{\pi }{2}$ as the right angle. In this problem we can observe that the angle $A=\dfrac{\pi }{2}$, so the right angle is present at vertex $A$.