Given angle A = 60 ∘ , c = √ 3 − 1 , b = √ 3 + 1 . Solve the triangle Options: A . &nbsp a = √ 6 , B = 15 ∘ , C = 100 ∘ B . &nbsp a = √ 6 , B = 15 ∘ , C = 105 ∘ C . &nbsp a = √ 12 , B = 15 ∘ , C = 105 ∘ D . &nbsp a = √ 6 , B = 105 ∘ , C = 15 ∘

Answer: Option D : D A = 60 ∘ ⇒ B + C = 120 ∘ W e k n o w , t a n B − C 2 = b − c b + c c o t A 2 ⇒ t a n B − C 2 = ( √ 3 + 1 ) − ( √ 3 − 1 ) ( √ 3 + 1 ) + ( √ 3 − 1 ) c o t ( 60 2 ) = 2 2 √ 3 ( √ 3 ) = 1 ∴ B − C 2 = 45 ∘ ⇒ B − C = 90 ∘ B + C = 120 ∘ ∴ B = 105 ∘ a n d C = 15 ∘ , F r o m s i n e r u l e , a = b s i n A s i n B = √ 3 + 1 ( s i n 60 s i n 105 ) s i n 105 = s i n ( 60 ∘ + 45 ∘ ) = √ 3 2 ( 1 √ 2 ) + 1 2 ( 1 √ 2 ) = √ 3 + 1 2 √ 2 a = ( √ 3 + 1 ) ( √ 3 ) ( 2 ) ( √ 2 ) ( √ 3 + 1 ) 2 = √ 6 ∴ a = √ 6 , B = 105 ∘ , C = 15 ∘ Option d is correct.

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

Discussion Forum : Solution Of Triangles

Question -

Given angle $A = 60^{\circ}, c = \sqrt{3} - 1, b = \sqrt{3} + 1$ . Solve the triangle

Options:

A .

a = \sqrt{6}, B = 15^{\circ}, C = 100^{\circ}

B .

a = \sqrt{6}, B = 15^{\circ}, C = 105^{\circ}

C .

a = \sqrt{12}, B = 15^{\circ}, C = 105^{\circ}

D .

a = \sqrt{6}, B = 105^{\circ}, C = 15^{\circ}

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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