Question

In a triangle ABC, with usual notations \(a = \sqrt{3}+1\), \(b = \sqrt{3}-1\) and \(\angle C = 60^{\circ}\) then the values of \(\angle A\) and \(\angle B\) respectively are

• A. \(105^{\circ},15^{\circ}\)
• B. \(100^{\circ},20^{\circ}\)
• C. \(90^{\circ},30^{\circ}\)
• D. \(110^{\circ},10^{\circ}\)

Hint:

Use the tangent rule: tan((A-B)/2) = ((a-b)/(a+b)) cot(C/2).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Napier's analogy: \(\tan\dfrac{A - B}{2} = \dfrac{a - b}{a + b}\cot\dfrac{C}{2}\).

Step 2: Compute:
\(a - b = 2\) and \(a + b = 2\sqrt3\), so \(\dfrac{a-b}{a+b} = \dfrac{1}{\sqrt3}\).
\(\cot 30^\circ = \sqrt3\).
\[ \tan\frac{A-B}{2} = \frac{1}{\sqrt3}\cdot\sqrt3 = 1 \Rightarrow \frac{A-B}{2} = 45^\circ \]
So \(A - B = 90^\circ\).

Step 3: Use the angle sum:
\(A + B = 180^\circ - 60^\circ = 120^\circ\). Solving \(A - B = 90^\circ\) and \(A + B = 120^\circ\):
\[ A = 105^\circ, \quad B = 15^\circ \]

Step 4: Check:
Option (A) satisfies both conditions. Options (B) and (D) have \(A - B = 80^\circ\) and \(100^\circ\), and (C) has \(A - B = 60^\circ\), so none fits.

Final Answer:
A is 105 degrees and B is 15 degrees. \[ \boxed{\text{(A) }105^{\circ},\,15^{\circ}} \]