In triangle ABC, if a:b:c = 7:8:9, then find cos A : cos B : cos C

Question 34 of 40

\triangle ABC

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2024Subjective· 4mImportance★★★★★

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Plug $a=7k,b=8k,c=9k$ into the cosine rule for each angle; the $k^2$ factors cancel, leaving a clean numeric ratio.

Given: In $\triangle ABC$ , $a:b:c = 7:8:9$ . Let $a=7k, b=8k, c=9k$ .

Step 1. Cosine rule: $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ .

$\cos A = \dfrac{64k^2+81k^2-49k^2}{2(8k)(9k)} = \dfrac{96k^2}{144k^2} = \dfrac{2}{3}$

Step 2. $\cos B = \dfrac{a^2+c^2-b^2}{2ac}$ :

$\cos B = \dfrac{49k^2+81k^2-64k^2}{2(7k)(9k)} = \dfrac{66k^2}{126k^2} = \dfrac{11}{21}$

Step 3. $\cos C = \dfrac{a^2+b^2-c^2}{2ab}$ :

$\cos C = \dfrac{49k^2+64k^2-81k^2}{2(7k)(8k)} = \dfrac{32k^2}{112k^2} = \dfrac{2}{7}$

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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 $\triangle ABC$ , if $a:b:c = 7:8:9$ , then find $\cos A : \cos B : \cos C$

Concept understanding — Cosine Rule

Cosine Rule

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Q.In △ABC\triangle ABC△ABC, if a:b:c=7:8:9a:b:c = 7:8:9a:b:c=7:8:9, then find cos⁡A:cos⁡B:cos⁡C\cos A : \cos B : \cos CcosA:cosB:cosC

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Q.In $\triangle ABC$ , if $a:b:c = 7:8:9$ , then find $\cos A : \cos B : \cos C$

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