Question

In \(\Delta ABC\), with usual notations, if \(\sin B-\cos C=\cos A\), then

• A. \(a^{2},b^{2},c^{2}\) form an Arithmetic Progression
• B. \(b^{2},c^{2},a^{2}\) form a Geometric Progression
• C. \(a^{2}=b^{2}+c^{2}\)
• D. \(b^{2}=a^{2}+c^{2}\)

Hint:

Rewrite as sin B = cos A + cos C and use sum-to-product to show that A equals 90 degrees.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Convert the condition to \(\sin B=\cos A+\cos C\) and use the sum-to-product formula.

Step 2: Simplify
\(\cos A+\cos C=2\cos\dfrac{A+C}{2}\cos\dfrac{A-C}{2}=2\sin\dfrac B2\cos\dfrac{A-C}{2}\)
Also \(\sin B=2\sin\dfrac B2\cos\dfrac B2\).

Step 3: Compare
Cancelling \(2\sin\dfrac B2\) (non-zero) gives \(\cos\dfrac B2=\cos\dfrac{A-C}{2}\), so \(B=A-C\), that is \(A=B+C\).

Step 4: Find the Angle
Since \(A+B+C=\pi\), \(2A=\pi\) and \(A=\dfrac\pi2\).
So \(a^{2}=b^{2}+c^{2}\), option (C).

Final Answer:
The triangle is right angled at A, so a squared equals b squared plus c squared. \[ \boxed{\text{(C) }a^{2}=b^{2}+c^{2}} \]