Question:
In a \(\Delta ABC\), \(\frac{a}{b}=2+\sqrt{3}\), and \(\angle C=60^\circ\). Then the ordered pair \((\angle A,\angle B)\) is equal to:
In a \(\Delta ABC\), \(\frac{a}{b}=2+\sqrt{3}\), and \(\angle C=60^\circ\). Then the ordered pair \((\angle A,\angle B)\) is equal to:
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Use \(A+B=120^\circ\) and Napier's analogy (or the sine rule) with \(a/b=2+\sqrt3\).
Updated On: Nov 11, 2026
- \((105^\circ,\,15^\circ)\)
- \((15^\circ,\,105^\circ)\)
- \((45^\circ,\,75^\circ)\)
- \((75^\circ,\,45^\circ)\)
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The Correct Option is A
Solution and Explanation
\(A+B=180^\circ-60^\circ=120^\circ\).
By Napier's analogy \(\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac C2\).
\(\frac{a-b}{a+b}=\frac{(2+\sqrt3)-1}{(2+\sqrt3)+1}=\frac{1+\sqrt3}{3+\sqrt3}=\frac{1}{\sqrt3}\), and \(\cot30^\circ=\sqrt3\).
So \(\tan\frac{A-B}2=1\Rightarrow A-B=90^\circ\).
With \(A+B=120^\circ\): \(A=105^\circ,\ B=15^\circ\).
Answer: (1) \((105^\circ,\,15^\circ)\)
By Napier's analogy \(\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac C2\).
\(\frac{a-b}{a+b}=\frac{(2+\sqrt3)-1}{(2+\sqrt3)+1}=\frac{1+\sqrt3}{3+\sqrt3}=\frac{1}{\sqrt3}\), and \(\cot30^\circ=\sqrt3\).
So \(\tan\frac{A-B}2=1\Rightarrow A-B=90^\circ\).
With \(A+B=120^\circ\): \(A=105^\circ,\ B=15^\circ\).
Answer: (1) \((105^\circ,\,15^\circ)\)
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