Question:
In \( \Delta ABC \), if \( A+B=120^{\circ} \), \( a=\sqrt{3}+1 \) and \( b=\sqrt{3}-1 \), then \( A:B= \)
In \( \Delta ABC \), if \( A+B=120^{\circ} \), \( a=\sqrt{3}+1 \) and \( b=\sqrt{3}-1 \), then \( A:B= \)
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Whenever a problem involves asymmetric radical side variations like \( \sqrt{3} \pm 1 \), it strongly points to angle values containing component combinations of \( 45^{\circ} \) and \( 60^{\circ} \). Using Napier's formula is the fastest way to extract their clean difference.
Updated On: Jun 8, 2026
- \( 9:7 \)
- \( 7:1 \)
- \( 5:3 \)
- \( 3:1 \)
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The Correct Option is B
Solution and Explanation
Concept:
According to Napier’s Analogy (Law of Tangents) for any triangle \( ABC \):
\[
tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} cot\left(\frac{C}{2}\right)
\]
Since \( A+B=120^{\circ} \), the third interior angle must be \( C = 180^{\circ} - 120^{\circ} = 60^{\circ} \).
Step 1: Calculating the coefficient terms from side lengths.
\[ a - b = (\sqrt{3}+1) - (\sqrt{3}-1) = 2 \] \[ a + b = (\sqrt{3}+1) + (\sqrt{3}-1) = 2\sqrt{3} \] \[ \frac{a-b}{a+b} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \]
Step 2: Substituting values into Napier's formula.
Since \( C = 60^{\circ} \), we have \( \frac{C}{2} = 30^{\circ} \), which gives \( cot(30^{\circ}) = \sqrt{3} \): \[ tan\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1 \] \[ \frac{A-B}{2} = 45^{\circ} \implies A - B = 90^{\circ} \quad \cdots (1) \]
Step 3: Solving the linear system of equations for angles \( A \) and \( B \).
We are given \( A + B = 120^{\circ} \cdots (2) \). Adding equations (1) and (2): \[ 2A = 210^{\circ} \implies A = 105^{\circ} \] Substituting back to find \( B \): \[ B = 120^{\circ} - 105^{\circ} = 15^{\circ} \]
Step 4: Finding the final ratio.
\[ A : B = 105^{\circ} : 15^{\circ} = \frac{105}{15} = 7 : 1 \]
Step 1: Calculating the coefficient terms from side lengths.
\[ a - b = (\sqrt{3}+1) - (\sqrt{3}-1) = 2 \] \[ a + b = (\sqrt{3}+1) + (\sqrt{3}-1) = 2\sqrt{3} \] \[ \frac{a-b}{a+b} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \]
Step 2: Substituting values into Napier's formula.
Since \( C = 60^{\circ} \), we have \( \frac{C}{2} = 30^{\circ} \), which gives \( cot(30^{\circ}) = \sqrt{3} \): \[ tan\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1 \] \[ \frac{A-B}{2} = 45^{\circ} \implies A - B = 90^{\circ} \quad \cdots (1) \]
Step 3: Solving the linear system of equations for angles \( A \) and \( B \).
We are given \( A + B = 120^{\circ} \cdots (2) \). Adding equations (1) and (2): \[ 2A = 210^{\circ} \implies A = 105^{\circ} \] Substituting back to find \( B \): \[ B = 120^{\circ} - 105^{\circ} = 15^{\circ} \]
Step 4: Finding the final ratio.
\[ A : B = 105^{\circ} : 15^{\circ} = \frac{105}{15} = 7 : 1 \]
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