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Question Number 29896 by ajfour last updated on 13/Feb/18

Commented by ajfour last updated on 13/Feb/18

$A t l e a s t i f α = β = γ = θ (s a y) .$
	Answered by mrW2 last updated on 13/Feb/18
	$take B as origin and BC as x−axis. B(0,0) C(a,0) A((a/2),(((√3)a)/2)) Eqn. of BE: y=tan α x Eqn. of CF: y=−tan ((π/3)−β) (x−a) Eqn. of AD: y=(((√3)a)/2)+tan ((π/3)+γ) (x−(a/2)) y_D =tan α x_D =(((√3)a)/2)+tan ((π/3)+γ) (x_D −(a/2)) [tan α−tan ((π/3)+γ)] x_D =(a/2)[(√3)−tan ((π/3)+γ)] ⇒x_D =((a[(√3)−tan ((π/3)+γ)])/(tan α−tan ((π/3)+γ))) ⇒y_D =((a[(√3)−tan ((π/3)+γ)]tan α)/(tan α−tan ((π/3)+γ))) y_E =tan α x_E =−tan ((π/3)−β) (x_E −a) (tan α+tan ((π/3)−β)] x_E =a tan ((π/3)−β) ⇒x_E =((a tan ((π/3)−β))/(tan α+tan ((π/3)−β))) ⇒y_E =((a tan ((π/3)−β) tan α)/(tan α+tan ((π/3)−β))) y_F =−tan ((π/3)−β) (x_F −a)=(((√3)a)/2)+tan ((π/3)+γ) (x_F −(a/2)) [tan ((π/3)+γ)+tan ((π/3)−β)] x_F =(a/2)[2tan ((π/3)−β)+tan ((π/3)+γ)−(√3)] ⇒x_F =((a[2tan ((π/3)−β)+tan ((π/3)+γ)−(√3)])/(2[tan ((π/3)+γ)+tan ((π/3)−β)])) ⇒y_F =((a[tan ((π/3)+γ)+(√3)]tan ((π/3)−β))/(2[tan ((π/3)+γ)+tan ((π/3)−β)])) A_(Blue) =(1/2)[x_D (y_E −y_F )+x_E (y_F −y_D )+x_F (y_D −y_E )] A_(Blue) =(a^2 /2)((((√3)−tan ((π/3)+γ))/(tan α−tan ((π/3)+γ)))×{((tan ((π/3)−β)tan α)/(tan α+tan ((π/3)−β)))−(([tan ((π/3)+γ)+(√3)]tan ((π/3)−β))/(2[tan ((π/3)+γ)+tan ((π/3)−β)]))}+((tan ((π/3)−β))/(tan α+tan ((π/3)−β)))×{(([tan ((π/3)+γ)+(√3)]tan ((π/3)−β))/(2[tan ((π/3)+γ)+tan ((π/3)−β)]))−(([(√3)−tan ((π/3)+γ)]tan α)/(tan α−tan ((π/3)+γ)))}+(([2tan ((π/3)−β)+tan ((π/3)+γ)−(√3)]tan α)/(2[tan ((π/3)+γ)+tan ((π/3)−β)]))×{(((√3)−tan ((π/3)+γ))/(tan α−tan ((π/3)+γ)))−((tan ((π/3)−β))/(tan α+tan ((π/3)−β)))})$
	$t a k e B a s o r i g i n a n d B C a s x - a x i s .$ $B (0, 0)$ $C (a, 0)$ $A (\frac{a}{2}, \frac{\sqrt{3} a}{2})$ $E q n . o f B E :$ $y = \tan α x$ $E q n . o f C F :$ $y = - \tan (\frac{π}{3} - β) (x - a)$ $E q n . o f A D :$ $y = \frac{\sqrt{3} a}{2} + \tan (\frac{π}{3} + γ) (x - \frac{a}{2})$ $y_{D} = \tan α x_{D} = \frac{\sqrt{3} a}{2} + \tan (\frac{π}{3} + γ) (x_{D} - \frac{a}{2})$ $[\tan α - \tan (\frac{π}{3} + γ)] x_{D} = \frac{a}{2} [\sqrt{3} - \tan (\frac{π}{3} + γ)]$ $\Rightarrow x_{D} = \frac{a [\sqrt{3} - \tan (\frac{π}{3} + γ)]}{\tan α - \tan (\frac{π}{3} + γ)}$ $\Rightarrow y_{D} = \frac{a [\sqrt{3} - \tan (\frac{π}{3} + γ)] \tan α}{\tan α - \tan (\frac{π}{3} + γ)}$ $y_{E} = \tan α x_{E} = - \tan (\frac{π}{3} - β) (x_{E} - a)$ $(\tan α + \tan (\frac{π}{3} - β)] x_{E} = a \tan (\frac{π}{3} - β)$ $\Rightarrow x_{E} = \frac{a \tan (\frac{π}{3} - β)}{\tan α + \tan (\frac{π}{3} - β)}$ $\Rightarrow y_{E} = \frac{a \tan (\frac{π}{3} - β) \tan α}{\tan α + \tan (\frac{π}{3} - β)}$ $y_{F} = - \tan (\frac{π}{3} - β) (x_{F} - a) = \frac{\sqrt{3} a}{2} + \tan (\frac{π}{3} + γ) (x_{F} - \frac{a}{2})$ $[\tan (\frac{π}{3} + γ) + \tan (\frac{π}{3} - β)] x_{F} = \frac{a}{2} [2 \tan (\frac{π}{3} - β) + \tan (\frac{π}{3} + γ) - \sqrt{3}]$ $\Rightarrow x_{F} = \frac{a [2 \tan (\frac{π}{3} - β) + \tan (\frac{π}{3} + γ) - \sqrt{3}]}{2 [\tan (\frac{π}{3} + γ) + \tan (\frac{π}{3} - β)]}$ $\Rightarrow y_{F} = \frac{a [\tan (\frac{π}{3} + γ) + \sqrt{3}] \tan (\frac{π}{3} - β)}{2 [\tan (\frac{π}{3} + γ) + \tan (\frac{π}{3} - β)]}$ $A_{B l u e} = \frac{1}{2} [x_{D} (y_{E} - y_{F}) + x_{E} (y_{F} - y_{D}) + x_{F} (y_{D} - y_{E})]$ $A_{B l u e} = \frac{a^{2}}{2} (\frac{\sqrt{3} - \tan (\frac{π}{3} + γ)}{\tan α - \tan (\frac{π}{3} + γ)} \times {\frac{\tan (\frac{π}{3} - β) \tan α}{\tan α + \tan (\frac{π}{3} - β)} - \frac{[\tan (\frac{π}{3} + γ) + \sqrt{3}] \tan (\frac{π}{3} - β)}{2 [\tan (\frac{π}{3} + γ) + \tan (\frac{π}{3} - β)]}} + \frac{\tan (\frac{π}{3} - β)}{\tan α + \tan (\frac{π}{3} - β)} \times {\frac{[\tan (\frac{π}{3} + γ) + \sqrt{3}] \tan (\frac{π}{3} - β)}{2 [\tan (\frac{π}{3} + γ) + \tan (\frac{π}{3} - β)]} - \frac{[\sqrt{3} - \tan (\frac{π}{3} + γ)] \tan α}{\tan α - \tan (\frac{π}{3} + γ)}} + \frac{[2 \tan (\frac{π}{3} - β) + \tan (\frac{π}{3} + γ) - \sqrt{3}] \tan α}{2 [\tan (\frac{π}{3} + γ) + \tan (\frac{π}{3} - β)]} \times {\frac{\sqrt{3} - \tan (\frac{π}{3} + γ)}{\tan α - \tan (\frac{π}{3} + γ)} - \frac{\tan (\frac{π}{3} - β)}{\tan α + \tan (\frac{π}{3} - β)}})$
	Commented by ajfour last updated on 14/Feb/18

	$U n l i m i t e d t h a n k s S i r .$ $I w i l l t r y s o m e v e c t o r a p p r o a c h .$
	Answered by ajfour last updated on 13/Feb/18

	$A r e a (△ D E F) = \sqrt{3} a^{2} \sin^{2} (\frac{π}{6} - θ) .$
	Answered by ajfour last updated on 14/Feb/18

	$A = \frac{a^{2}}{2} \sin (\frac{π}{3} + α - β) [\frac{\sin (\frac{2 π}{3} + β)}{\sin (\frac{2 π}{3} + β - α)} - \frac{\sin γ}{\sin (\frac{π}{3} + γ - α)}]$ $\times [\frac{\sin (\frac{2 π}{3} + γ)}{\sin (\frac{π}{3} + β - γ)} - \frac{\sin α}{\sin (\frac{2 π}{3} + β - α)}] .$
	Answered by ajfour last updated on 14/Feb/18

	$e q . o f B D E :$ $x = p \cos α; y = p \sin α$ $e q . o f C E F :$ $x = a + q \cos (\frac{2 π}{3} + β); y = q \sin (\frac{2 π}{3} + β)$ $e q . o f A F D :$ $x = \frac{a}{2} + r \cos (\frac{π}{3} + γ)$ $y = \frac{\sqrt{3}}{2} \sin (\frac{π}{3} + γ)$ $p o i n t E (x_{2}, y_{2}) l i e s o n l i n e B D E$ $a n d l i n e C E F, s o$ $p_{2} \cos α = a + q_{1} \cos (\frac{2 π}{3} + β)$ $p_{2} \sin α = q_{1} \sin (\frac{2 π}{3} + β)$ $\Rightarrow p_{2} = \frac{a \sin (\frac{2 π}{3} + β)}{\sin (\frac{2 π}{3} + β - α)}$ $q_{1} = \frac{a \sin α}{\sin (\frac{2 π}{3} + β - α)}$ $p o i n t D (x_{1}, y_{1}) l i e s o n B D E$ $a n d A F D, s o$ $p_{1} \cos α = \frac{a}{2} + r_{2} \cos (\frac{π}{3} + γ)$ $p_{1} \sin α = \frac{a \sqrt{3}}{2} + r_{2} \sin (\frac{π}{3} + γ)$ $\Rightarrow p_{1} = \frac{a \sin γ}{\sin (\frac{π}{3} + γ - α)}$ $p o i n t F (x_{3}, y_{3}) l i e s o n C E F a n d$ $A F D, s o$ $a + q_{2} \cos (\frac{2 π}{3} - β) = r_{1} \cos (\frac{π}{3} + γ)$ $q_{2} \sin (\frac{2 π}{3} + β) = \frac{a \sqrt{3}}{2} + r_{1} \sin (\frac{π}{3} + γ)$ $\Rightarrow q_{2} = \frac{a \sin (\frac{2 π}{3} + γ)}{\sin (\frac{π}{3} + β - γ)}$ $A = \frac{1}{2} (p_{2} - p_{1}) (q_{2} - q_{1}) \sin (\frac{π}{3} + α - β)$ $h e n c e t h e a n s w e r .$
	Answered by mrW2 last updated on 14/Feb/18

	$I h a v e c o n s i d e r e d f o l l o w i n g m e t h o d :$ $∠ B E C = π - α - (\frac{π}{3} - β) = π - (\frac{π}{3} + α - β)$ $\frac{B E}{\sin (\frac{π}{3} - β)} = \frac{B C}{\sin [π - (\frac{π}{3} + α - β)]}$ $\Rightarrow B E = \frac{\sin (\frac{π}{3} - β) a}{\sin (\frac{π}{3} - β + α)}$ $A_{Δ B E C} = \frac{a}{2} \times B E \times \sin α$ $\Rightarrow A_{Δ B E C} = \frac{a^{2}}{2} \times \frac{\sin α \sin (\frac{π}{3} - β)}{\sin (\frac{π}{3} - β + α)}$ $\Rightarrow A_{Δ C F A} = \frac{a^{2}}{2} \times \frac{\sin β \sin (\frac{π}{3} - γ)}{\sin (\frac{π}{3} - γ + β)}$ $\Rightarrow A_{Δ A D B} = \frac{a^{2}}{2} \times \frac{\sin γ \sin (\frac{π}{3} - α)}{\sin (\frac{π}{3} - α + γ)}$ $A_{Δ A B C} = \frac{a}{2} \times \frac{\sqrt{3} a}{2} = \frac{\sqrt{3} a^{2}}{4}$ $A_{B l u e} = A_{Δ A B C} - A_{Δ B E C} - A_{Δ C F A} - A_{Δ A D B}$ $\Rightarrow A_{B l u e} = \frac{a^{2}}{2} [\frac{\sqrt{3}}{2} - Σ \frac{\sin α \sin (\frac{π}{3} - β)}{\sin (\frac{π}{3} - β + α)}]$ $\Rightarrow A_{B l u e} = \frac{a^{2}}{2} [\frac{\sqrt{3}}{2} - \frac{\sin α \sin (\frac{π}{3} - β)}{\sin (\frac{π}{3} - β + α)} - \frac{\sin β \sin (\frac{π}{3} - γ)}{\sin (\frac{π}{3} - γ + β)} - \frac{\sin γ \sin (\frac{π}{3} - α)}{\sin (\frac{π}{3} - α + γ)}]$ $i f α = β = γ = θ :$ $\Rightarrow A_{B l u e} = \frac{a^{2}}{2} [\frac{\sqrt{3}}{2} - 3 \frac{\sin θ \sin (\frac{π}{3} - θ)}{\sin (\frac{π}{3})}]$ $= \frac{a^{2}}{2} [\frac{\sqrt{3}}{2} - \sqrt{3} \sin θ (\sqrt{3} \cos θ - \sin θ)]$ $= \frac{a^{2}}{2} [\frac{\sqrt{3}}{2} - \frac{3}{2} \sin 2 θ + \frac{\sqrt{3}}{2} (1 - \cos 2 θ)]$ $= \frac{a^{2}}{2} [\sqrt{3} - \sqrt{3} (\sin \frac{π}{3} \sin 2 θ + \cos \frac{π}{3} \cos 2 θ)]$ $\Rightarrow A_{B l u e} = \frac{\sqrt{3} a^{2}}{2} [1 - \cos (\frac{π}{3} - 2 θ)]$ $o r$ $\Rightarrow A_{B l u e} = \sqrt{3} a^{2} \sin^{2} (\frac{π}{6} - θ)$
	Commented by ajfour last updated on 14/Feb/18

	$M u c h b e t t e r, I n d e e d! S i r .$
	Commented by mrW2 last updated on 15/Feb/18

	$U s i n g t h i s m e t h o d w e c a n a l s o f i n d$ $A_{B l u e} i n a n a r b i t r a r y t r i a n g l e w i t h s i d e$ $l e n g t h s a, b, c :$ $A_{B l u e} = \frac{1}{4} \sqrt{(a + b + c) (a + b - c) (a - b + c) (- a + b + c)} - \frac{1}{2} [\frac{a^{2} \sin α \sin (\frac{π}{3} - β)}{\sin (\frac{π}{3} - β + α)} + \frac{b^{2} \sin β \sin (\frac{π}{3} - γ)}{\sin (\frac{π}{3} - γ + β)} + \frac{c^{2} \sin γ \sin (\frac{π}{3} - α)}{\sin (\frac{π}{3} - α + γ)}]$
	Commented by ajfour last updated on 16/Feb/18

	$T h a n k y o u S i r .$ $(c o n c l u s i o n : A r e a i s e a s i l y o b t a i n e d$ $f r o m a r e a s .)$
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