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

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

Explanation

Solution

In the problem we have to show that ΔABC\Delta ABC is a right-angled triangle. In the problem we have given an equation in trigonometric ratios which is cosA=sinBcosC\cos A=\sin B-\cos C. We will rearrange the terms in the above equation so that all the cos\cos terms are at one side. Now we will use one of the trigonometric identities that is cosA+cosB=2cosA+B2cosAB2\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=πA+B+C=\pi and simplify the equation to get the required result.

Formula Use:
1. cosA+cosB=2cosA+B2cosAB2\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{A-B}{2}.
2. sinθ=2sin(θ2)cos(θ2)\sin \theta =2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right).
3. cos(π2θ)=sinθ\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta .
4. A+B+C=πA+B+C=\pi .

Complete Step by Step Solution:
Given that, cosA=sinBcosC\cos A=\sin B-\cos C.
Now we will be shifting the cosC\cos C to LHS side, then we will get
cosA+cosC=sinB\Rightarrow \cos A+\cos C=\sin B.
We know that trigonometric identity that is cosA+cosB=2cosA+B2cosAB2\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 cosA+cosC\cos A+\cos C, as
cosA+cosC=2cos(A+C2)cos(AC2)\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
2cos(A+B2)cos(AC2)=sinB2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-C}{2} \right)=\sin B.
Considering the RHS side which is sinB\sin B. We know that sinθ=2sin(θ2)cos(θ2)\sin \theta =2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right).
Now we are writing sinB\sin B as sinB=2sin(B2)cos(B2)\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
2cos(A+C2)cos(AC2)=2sin(B2)cos(B2)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=πA+B+C=\pi
From the above value we can write A+C=πBA+C=\pi -B.
Now substituting this value in the above equation, then we will have
2cos(π2B2)cos(AC2)=2sin(B2)cos(B2)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
2sin(B2)cos(AC2)=2sin(B2)cos(B2) cos(AC2)=cos(B2) \begin{aligned} & 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{aligned}
Equating on both sides, then we will get
(AC2)=(B2) AC=B A=B+C \begin{aligned} & \left( \dfrac{A-C}{2} \right)=\left( \dfrac{B}{2} \right) \\\ & A-C=B \\\ & \Rightarrow A=B+C \\\ \end{aligned}
We have A+B+C=πA+B+C=\pi , so
A+B+C=π A+A=π 2A=π A=π2 \begin{aligned} & A+B+C=\pi \\\ & \Rightarrow A+A=\pi \\\ & \Rightarrow 2A=\pi \\\ & \Rightarrow A=\dfrac{\pi }{2} \\\ \end{aligned}
Hence the ΔABC\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 π2\dfrac{\pi }{2} as the right angle. In this problem we can observe that the angle A=π2A=\dfrac{\pi }{2}, so the right angle is present at vertex AA.