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Question Number 42942 by Raj Singh last updated on 05/Sep/18

	Answered by MJS last updated on 05/Sep/18

	$B C = a C A = b A B = c$ $1 . a = 6 β = 50 ° b + c = 8 \Rightarrow c = 8 - b$ $α + γ = 130 ° \Rightarrow γ = 130 ° - α$ $36 = b^{2} + {(8 - b)}^{2} - 2 b (8 - b) \cos α$ $b^{2} = 36 + {(8 - b)}^{2} - 12 (8 - b) \cos 50 °$ ${(8 - b)}^{2} = 36 + b^{2} - 12 b \cos (130 ° - α)$ $\Rightarrow a = 6 b = 4.62 c = 3.38$ $α = 95.93 ° β = 50 ° γ = 34.07 °$ $2 . b = 6 β = 50 ° b + c = 8 \Rightarrow c = 2$ $α + γ = 130 ° \Rightarrow γ = 130 ° - α$ $a^{2} = 36 + 4 - 24 \cos α$ $36 = a^{2} + 4 - 4 a \cos 50 °$ $4 = a^{2} + 36 - 12 a \cos (130 ° - α)$ $\Rightarrow a = 7.09 b = 6 c = 2$ $α = 115.21 ° β = 50 ° γ = 14.79 °$ $3 . c = 6 β = 50 ° b + c = 8 \Rightarrow b = 2$ $α + γ = 130 ° \Rightarrow γ = 130 ° - α$ $a^{2} = 4 + 36 - 24 \cos α$ $4 = a^{2} + 36 - 12 a \cos 50 °$ $36 = a^{2} + 4 - 4 a \cos (130 ° - α)$ $\Rightarrow no real solution$
	Answered by MrW3 last updated on 05/Sep/18

	Commented by MrW3 last updated on 05/Sep/18

	$T h i s i s h o w t o d r a w t h e s e a r c h e d t r i a n g l e .$ $1 . d r a w l i n e B C = 6$ $2 . d r a w l i n e B D = 8 w i t h ∠ B = 50 °$ $3 . c o n n e c t D w i t h C$ $4 . d r a w p e r p e n d i c u l a r b i s e c t o r o f D C w h i c h$ $i n t e r s e c t s w i t h B D a t A$ $5 . c o n n e c t A w i t h C$ $Δ A B C i s t h e s e a r c h e d t r i a n g l e .$ $p r o o f :$ $A C = A D$ $A B + A C = A B + A D = B D = 8$
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