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Question Number 67430 by TawaTawa last updated on 27/Aug/19

Commented by MJS last updated on 27/Aug/19

$coordinate method$ $turn the triangle \to C A is the base$ $side length = s$ $s = 8 + x \Rightarrow x = s - 8$ $C = (\begin{matrix} - \frac{s}{2} \\ 0 \end{matrix}) A = (\begin{matrix} \frac{s}{2} \\ 0 \end{matrix}) B = (\begin{matrix} 0 \\ \frac{\sqrt{3} s}{2} \end{matrix})$ $D = (\begin{matrix} \frac{s}{2} - 3 \\ 0 \end{matrix}) E = (\begin{matrix} 5 - \frac{s}{2} \\ 0 \end{matrix})$ $line B D$ $l_{1} : y = - \frac{\sqrt{3} s}{s - 6} x + \frac{\sqrt{3} s}{2} \Rightarrow \tan θ_{1} = - \frac{\sqrt{3} s}{s - 6}$ $line B E$ $l_{2} : y = \frac{\sqrt{3} s}{s - 10} x + \frac{\sqrt{3} s}{2} \Rightarrow \tan θ_{2} = \frac{\sqrt{3} s}{s - 10}$ $θ_{2} - θ_{1} = \pm 30 °$ $\arctan \frac{\sqrt{3} s}{s - 10} + \arctan \frac{\sqrt{3} s}{s - 6} = \pm 30 °$ $[\tan (\arctan α + \arctan β) = \frac{α + β}{1 - α β}]$ $- \frac{\sqrt{3} (s^{2} - 8 s)}{s^{2} + 8 s - 30} = \pm \frac{\sqrt{3}}{3}$ $\Rightarrow s = 15$ $\Rightarrow x = 7$
Commented by TawaTawa last updated on 27/Aug/19

$God bless you sir . I appreciate .$
	Answered by Kunal12588 last updated on 27/Aug/19

	$s = 8 + x$ $a = \sqrt{b^{2} + c^{2} - 2 b c c o s A}$ $B E = \sqrt{5^{2} + {(x + 8)}^{2} - 10 (x + 8) c o s 60 °}$ $D B = \sqrt{3^{2} + {(x + 8)}^{2} - 6 (x + 8) c o s 60 °}$ $x = D E = \sqrt{{B E}^{2} + {D B}^{2} - 2 (B E) (D B) c o s 30 °}$
	Commented by TawaTawa last updated on 27/Aug/19

	$God bless you sir . I appreciate .$
	Answered by mr W last updated on 27/Aug/19

	$l e t ∠ A B D = α, ∠ C B E = β$ $α + β + 30 ° = 60 °$ $\Rightarrow α + β = 30 °$ $s i d e l e n g t h o f e q u i l a t e r a l = 3 + x + 5 = 8 + x$ $\frac{\sin α}{A D} = \frac{\sin (α + 60 °)}{A B}$ $\frac{\sin α}{3} = \frac{\sin (α + 60 °)}{8 + x} = \frac{\sin α + \sqrt{3} \cos α}{2 (8 + x)}$ $\frac{2 (8 + x)}{3} = 1 + \sqrt{3} \cot α$ $\Rightarrow \cot α = \frac{13 + 2 x}{3 \sqrt{3}} \Rightarrow \tan α = \frac{3 \sqrt{3}}{2 x + 13}$ $\frac{\sin β}{E C} = \frac{\sin (β + 60 °)}{B C}$ $\frac{\sin β}{5} = \frac{\sin (β + 60 °)}{8 + x} = \frac{\sin β + \sqrt{3} \cos β}{2 (8 + x)}$ $\frac{2 (8 + x)}{5} = 1 + \sqrt{3} \cot β$ $\Rightarrow \cot β = \frac{11 + 2 x}{5 \sqrt{3}} \Rightarrow \tan β = \frac{5 \sqrt{3}}{2 x + 11}$ $\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β} = \tan 30 ° = \frac{1}{\sqrt{3}}$ $\Rightarrow \frac{\frac{3 \sqrt{3}}{2 x + 13} + \frac{5 \sqrt{3}}{2 x + 11}}{1 - \frac{3 \sqrt{3}}{2 x + 13} \times \frac{5 \sqrt{3}}{2 x + 11}} = \frac{1}{\sqrt{3}}$ $\Rightarrow \frac{3 (2 x + 11) + 5 (2 x + 13)}{(2 x + 13) (2 x + 11) - 45} = \frac{1}{3}$ $\Rightarrow \frac{16 x + 98}{4 x^{2} + 48 x + 98} = \frac{1}{3}$ $\Rightarrow x^{2} - 49 = 0$ $\Rightarrow x = 7$
	Commented by TawaTawa last updated on 27/Aug/19

	$God bless you sir . I appreciate .$
	Commented by TawaTawa last updated on 27/Aug/19

	$Sir, how did you get α + 60$
	Commented by mr W last updated on 27/Aug/19

	$\sin ∠ A D B = \sin (180 ° - α - ∠ A)$ $= \sin (α + ∠ A) = \sin (α + 60 °)$
	Commented by TawaTawa last updated on 27/Aug/19

	$I appreciate your time sir .$
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