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: Sine rule:
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ gives $\dfrac{\sin A}{\sin B} = \dfrac{\sqrt3+1}{\sqrt3-1} = 2+\sqrt3$.

Step 2: Test option A:
$\sin105^\circ = \dfrac{\sqrt6+\sqrt2}{4}$ and $\sin15^\circ = \dfrac{\sqrt6-\sqrt2}{4}$. Their ratio is $\dfrac{\sqrt3+1}{\sqrt3-1} = 2+\sqrt3$. It matches.

Step 3: Confirm angle sum:
$105^\circ + 15^\circ + 60^\circ = 180^\circ$, so option (A) is right. The other pairs give different sine ratios.

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