Question

Let $\frac{\sin A}{\sin B} = \frac{\sin(A - C)}{\sin(C - B)}$, where A, B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :

• A. $a^2, b^2, c^2$ are in A.P.
• B. $b^2, c^2, a^2$ are in A.P.
• C. $c^2, a^2, b^2$ are in A.P.
• D. $b^2 - a^2 = a^2 + c^2$

Hint:

Whenever you see a relation between sines of angles in a triangle, use the sine rule to convert them into a relation between side lengths. Use $\sin(A\pm B)$ expansions or product-to-sum formulas to simplify.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a triangle, $A + B + C = \pi$. We can use trigonometric identities and the sine rule ($a/\sin A = b/\sin B = c/\sin C = 2R$) to relate the angles to the side lengths.
Step 2: Detailed Explanation:
Given: $\frac{\sin A}{\sin B} = \frac{\sin(A - C)}{\sin(C - B)}$.
Since $A, B, C$ are angles of a triangle, $A = \pi - (B+C)$ and $B = \pi - (A+C)$.
So $\sin A = \sin(B+C)$ and $\sin B = \sin(A+C)$.
Substituting these:
\[ \frac{\sin(B+C)}{\sin(A+C)} = \frac{\sin(A-C)}{\sin(C-B)} \]
\[ \sin(B+C) \sin(C-B) = \sin(A+C) \sin(A-C) \]
Using the identity $\sin(x+y)\sin(x-y) = \sin^2 x - \sin^2 y$:
\[ \sin^2 C - \sin^2 B = \sin^2 A - \sin^2 C \]
\[ 2 \sin^2 C = \sin^2 A + \sin^2 B \]
By Sine Rule, $\sin A = a/2R$, etc.:
\[ 2 \left( \frac{c}{2R} \right)^2 = \left( \frac{a}{2R} \right)^2 + \left( \frac{b}{2R} \right)^2 \]
\[ 2c^2 = a^2 + b^2 \]
This means $a^2, c^2, b^2$ are in Arithmetic Progression (A.P.), or equivalently, $b^2, c^2, a^2$ are in A.P.
Step 3: Final Answer:
The squares of the sides $b^2, c^2, a^2$ are in A.P.