Given angle A=60∘, c=√3−1, b=√3+1. Solve t

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Question

Given angle

A = 60^{\circ}, c = \sqrt{3} - 1, b = \sqrt{3} + 1

. Solve the triangle

Options:

A . a=√6, B=15∘, C=100∘

B . a=√6, B=15∘, C=105∘

C . a=√12, B=15∘, C=105∘

D . a=√6, B=105∘, C=15∘

Answer: Option D
:
D

A = 60^{\circ} \Rightarrow B + C = 120^{\circ} W e k n o w, t a n \frac{B - C}{2} = \frac{b - c}{b + c} c o t \frac{A}{2} \Rightarrow t a n \frac{B - C}{2} = \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{(\sqrt{3} + 1) + (\sqrt{3} - 1)} c o t (\frac{60}{2}) = \frac{2}{2 \sqrt{3}} (\sqrt{3}) = 1 ∴ \frac{B - C}{2} = 45^{\circ} \Rightarrow B - C = 90^{\circ} B + C = 120^{\circ} ∴ B = 105^{\circ} a n d C = 15^{\circ}, F r o m s i n e r u l e, a = \frac{b s i n A}{s i n B} = \sqrt{3} + 1 (\frac{s i n 60}{s i n 105}) s i n 105 = s i n (60^{\circ} + 45^{\circ}) = \frac{\sqrt{3}}{2} (\frac{1}{\sqrt{2}}) + \frac{1}{2} (\frac{1}{\sqrt{2}}) = \frac{\sqrt{3} + 1}{2 \sqrt{2}} a = \frac{(\sqrt{3} + 1) (\sqrt{3}) (2) (\sqrt{2})}{(\sqrt{3} + 1) 2} = \sqrt{6} ∴ a = \sqrt{6}, B = 105^{\circ}, C = 15^{\circ}

Option d is correct.

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