Solve the triangle in which $a = \sqrt{3}+1$, $b = \sqrt{3}-1$ and $C = 60^\circ$.

Solve the triangle in which a = √(3)+1, b = √(3)-1 and C = 60^°.

Miscellaneous Exercise 3 · Q102

a = \sqrt{3}+1

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$a-b=2$ , $a+b=2\sqrt3$ . By Napier's Analogy, $\tan\dfrac{A-B}{2}=\dfrac{a-b}{a+b}\cot\dfrac C2=\dfrac{2}{2\sqrt3}\cot30^\circ=\dfrac1{\sqrt3}\cdot\sqrt3=1\implies\dfrac{A-B}{2}=45^\circ\implies A-B=90^\circ$ . Also $A+B=180^\circ-60^\circ=120^\circ$ . Solving: $A=\dfrac{120+90}{2}=105^\circ$ , $B=\dfrac{120-90}{2}=15^\circ$ . For $c$ : $c^2=a^2+b^2-2ab\cos C$ . $a^2=(\sqrt3+1)^2=4+2\sqrt3$ , $b^2=(\sqrt3-1)^2=4-2\sqrt3 …

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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.Solve the triangle in which $a = \sqrt{3}+1$ , $b = \sqrt{3}-1$ and $C = 60^\circ$ .

Concept understanding — Cosine Rule

Cosine Rule

Where it comes from

How it is used

Q.Solve the triangle in which a=3+1a = \sqrt{3}+1a=3​+1, b=3−1b = \sqrt{3}-1b=3​−1 and C=60∘C = 60^\circC=60∘.

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Q.Solve the triangle in which $a = \sqrt{3}+1$ , $b = \sqrt{3}-1$ and $C = 60^\circ$ .

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