Question:
With the usual notation, in \(\Delta\)ABC, if \(\angle A+\angle B=120^\circ\), \(a=\sqrt3+1\), \(b=\sqrt3-1\) then the ratio \(\angle A:\angle B\), is :
With the usual notation, in \(\Delta\)ABC, if \(\angle A+\angle B=120^\circ\), \(a=\sqrt3+1\), \(b=\sqrt3-1\) then the ratio \(\angle A:\angle B\), is :
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Since \(\angle C=60^\circ\), use the Law of Tangents: \(\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{C}{2}\).
Updated On: Nov 5, 2026
- 9 : 7
- 7 : 1
- 5 : 3
- 3 : 1
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The Correct Option is B
Solution and Explanation
\(\angle C=180^\circ-120^\circ=60^\circ\).
Law of tangents: \(\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{C}{2}\).
\(a-b=2\), \(a+b=2\sqrt3\), so \(\frac{a-b}{a+b}=\frac{1}{\sqrt3}\); \(\cot 30^\circ=\sqrt3\).
\(\tan\frac{A-B}{2}=\frac{1}{\sqrt3}\cdot\sqrt3=1\), so \(\frac{A-B}{2}=45^\circ\) and \(A-B=90^\circ\).
With \(A+B=120^\circ\): \(A=105^\circ\), \(B=15^\circ\).
\(A:B=105:15=7:1\).
Answer: (2) 7 : 1
Law of tangents: \(\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{C}{2}\).
\(a-b=2\), \(a+b=2\sqrt3\), so \(\frac{a-b}{a+b}=\frac{1}{\sqrt3}\); \(\cot 30^\circ=\sqrt3\).
\(\tan\frac{A-B}{2}=\frac{1}{\sqrt3}\cdot\sqrt3=1\), so \(\frac{A-B}{2}=45^\circ\) and \(A-B=90^\circ\).
With \(A+B=120^\circ\): \(A=105^\circ\), \(B=15^\circ\).
\(A:B=105:15=7:1\).
Answer: (2) 7 : 1
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