[Source text badly corrupted at this item — the printed condition involving $\sin(A-B)$, $\sin(B-?)$ and side $c$ could not be reliably reconstructed from the extracted characters.] ... then show that $a^2$, $b^2$, $c^2$ are in A.P.

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This Miscellaneous Exercise item's printed condition (something of the form $\sin(A-B)/\ldots=\sin(B-\,?)/c$ ) was extracted from a heavily scrambled region of the source PDF, and the exact relationship intended could not be confidently distinguished from several plausible-looking but mathematically inconsistent readings (several candidate reconstructions were tried and rejected because they turned out to be identities that hold for every triangle, not special conditions implying $a^2,b …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

Substitute $\cos C$ and $\cos B$ from the cosine rule and simplify.

By the cosine rule:

$\cos C=\frac{a^2+b^2-c^2}{2ab}, \qquad \cos B=\frac{a^2+c^2-b^2}{2ac}$

LHS $=a(b\cos C-c\cos B) = ab\cos C - ac\cos B$

$ab\cos C = ab\cdot\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2-c^2}{2}$

$ac\cos B = ac\cdot\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2-b^2}{2}$

…

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CBSE 2023Set ANNUAL2 marksMCQ
$\triangle ABC$
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Compare with the cosine rule $b^2=a^2+c^2-2ac\cos B$ .

By the cosine rule, $b^2=a^2+c^2-2ac\cos B \Rightarrow a^2+c^2-b^2=2ac\cos B$ .

…

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CBSE 2022Set ANNUAL2 marksMCQ
$\triangle ABC$
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Compare with the Cosine Rule formula for $\cos B$ .

By the Cosine Rule: $\cos B = \dfrac{a^2+c^2-b^2}{2ac}$

Given $c^2+a^2-b^2=ac$ : …

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CBSE 2017Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

Substitute $\cos B$ and $\cos C$ from the cosine rule and simplify.

By the cosine rule:

$\cos C = \dfrac{a^2+b^2-c^2}{2ab}, \qquad \cos B = \dfrac{a^2+c^2-b^2}{2ac}$

$a(b\cos C - c\cos B) = a\left[b\cdot\dfrac{a^2+b^2-c^2}{2ab} - c\cdot\dfrac{a^2+c^2-b^2}{2ac}\right]$

$= a\left[\dfrac{a^2+b^2-c^2}{2a} - \dfrac{a^2+c^2-b^2}{2a}\right] = \dfrac{1}{2}\Big[(a^2+b^2-c^2)-(a^2+c^2-b^2)\Big]$

…

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Q.[Source text badly corrupted at this item — the printed condition involving $\sin(A-B)$ , $\sin(B-?)$ and side $c$ could not be reliably reconstructed from the extracted characters.] ... then show that $a^2$ , $b^2$ , $c^2$ are in A.P.

Concept understanding — Cosine Rule

Cosine Rule

Where it comes from

How it is used

Q.[Source text badly corrupted at this item — the printed condition involving sin⁡(A−B)\sin(A-B)sin(A−B), sin⁡(B−?)\sin(B-?)sin(B−?) and side ccc could not be reliably reconstructed from the extracted characters.] ... then show that a2a^2a2, b2b^2b2, c2c^2c2 are in A.P.

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Q.[Source text badly corrupted at this item — the printed condition involving $\sin(A-B)$ , $\sin(B-?)$ and side $c$ could not be reliably reconstructed from the extracted characters.] ... then show that $a^2$ , $b^2$ , $c^2$ are in A.P.

Unlock everything free for 14 days