Question

With usual notations in $\triangle \text{ABC}$, if $\frac{\sin \text{A}}{\sin \text{C}} = \frac{\sin(\text{A} - \text{B})}{\sin(\text{B} - \text{C})}$, then $a^2, b^2, c^2$ are in

• A. Not in AP
• B. HP
• C. AP
• D. GP

Hint:

Whenever an identity simplifies to a symmetric squared term relation like $\sin^2\text{B} = 2\sin\text{A}\sin\text{C}\cos\text{B}$, it almost always indicates an underlying Arithmetic Progression for the squares of the sides ($2b^2 = a^2 + c^2$). Memorizing this standard identity trap can save massive amounts of expansion algebra!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question gives a trigonometric ratio connecting the angles of a triangle $\triangle \text{ABC}$. We need to determine the structural progression relationship (AP, GP, or HP) followed by the squares of its side lengths, $a^2, b^2, c^2$.

Step 2: Key Formula or Approach:
We will cross-multiply the terms, expand using standard sine compound angle formulas, and apply the Sine Rule ($\frac{a}{\sin \text{A}} = \frac{b}{\sin \text{B}} = \frac{c}{\sin \text{C}} = 2\text{R}$) along with the Cosine Rule to convert the angles into side lengths.

Step 3: Detailed Explanation:
Given the equation:
$$\sin \text{A} \sin(\text{B} - \text{C}) = \sin \text{C} \sin(\text{A} - \text{B})$$ Expand both sides using $\sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi$:
$$\sin \text{A}(\sin \text{B} \cos \text{C} - \cos \text{B} \sin \text{C}) = \sin \text{C}(\sin \text{A} \cos \text{B} - \cos \text{A} \sin \text{B})$$ $$\sin \text{A} \sin \text{B} \cos \text{C} - \sin \text{A} \cos \text{B} \sin \text{C} = \sin \text{A} \cos \text{B} \sin \text{C} - \cos \text{A} \sin \text{B} \sin \text{C}$$ Group identical terms together by shifting them to opposite sides:
$$\sin \text{A} \sin \text{B} \cos \text{C} + \cos \text{A} \sin \text{B} \sin \text{C} = 2 \sin \text{A} \cos \text{B} \sin \text{C}$$ Factor out $\sin \text{B}$ from the left side expression:
$$\sin \text{B}(\sin \text{A} \cos \text{C} + \cos \text{A} \sin \text{C}) = 2 \sin \text{A} \cos \text{B} \sin \text{C}$$ Using the angle sum identity, the interior term simplifies to $\sin(\text{A} + \text{C})$:
$$\sin \text{B} \sin(\text{A} + \text{C}) = 2 \sin \text{A} \cos \text{B} \sin \text{C}$$ Since $\text{A} + \text{B} + \text{C} = \pi \implies \text{A} + \text{C} = \pi - \text{B}$, we know that $\sin(\text{A} + \text{C}) = \sin \text{B}$:
$$\sin \text{B} \cdot \sin \text{B} = 2 \sin \text{A} \cos \text{B} \sin \text{C} \implies \sin^2 \text{B} = 2 \sin \text{A} \cos \text{B} \sin \text{C}$$ Now apply the Sine Rule ($\sin \text{A} \propto a, \ \sin \text{B} \propto b, \ \sin \text{C} \propto c$) and the Cosine Rule ($\cos \text{B} = \frac{c^2 + a^2 - b^2}{2ac}$):
$$b^2 = 2 \cdot a \cdot \left(\frac{c^2 + a^2 - b^2}{2ac}\right) \cdot c$$ The product terms $2ac$ cancel out perfectly in the numerator and denominator:
$$b^2 = c^2 + a^2 - b^2$$ $$2b^2 = a^2 + c^2$$ The relation $2b^2 = a^2 + c^2$ is the exact definition of an Arithmetic Progression (AP). Thus, $a^2, b^2, c^2$ are in AP, matching option (C).

Step 4: Final Answer:
The values $a^2, b^2, c^2$ are in AP, which corresponds to option (C).