In $\triangle ABC$, $ac\cos B - bc\cos A =$ ____________. (a) $a^2-b^2$ (b) $b^2-c^2$ (c) $c^2-a^2$ (d) $a^2-b^2-c^2$

In triangle ABC, accos B - bccos A = ______. (a) a – Class 12 Mathematics

Miscellaneous Exercise 3 · Q74

\triangle ABC

(a)

a^2-b^2

(b)

b^2-c^2

(c)

c^2-a^2

(d)

a^2-b^2-c^2

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$ac\cos B-bc\cos A=c(a\cos B-b\cos A)$ . From the Cosine Rule, $a\cos B=\dfrac{a^2+c^2-b^2}{2c}$ and $b\cos A=\dfrac{b^2+c^2-a^2}{2c}$ , so $a\cos B-b\cos A=\dfrac{(a^2+c^2-b^2)-(b^2+c^2-a^2)}{2c}=\dfrac{2a^2-2b^2}{2c}=\dfrac{ …

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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.In $\triangle ABC$ , $ac\cos B - bc\cos A =$ ____________.

Concept understanding — Cosine Rule

Cosine Rule

Where it comes from

How it is used

Q.In △ABC\triangle ABC△ABC, accos⁡B−bccos⁡A=ac\cos B - bc\cos A =accosB−bccosA= ____________.

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Q.In $\triangle ABC$ , $ac\cos B - bc\cos A =$ ____________.

Unlock everything free for 14 days