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Question Number 181201 by HeferH last updated on 22/Nov/22

	Answered by ARUNG_Brandon_MBU last updated on 23/Nov/22

	$\tan θ = \frac{3}{4}$ $\tan 2 θ = \frac{\frac{3}{4} + \frac{3}{4}}{1 - \frac{3}{4} \cdot \frac{3}{4}} = \frac{24}{7} = \frac{A C}{A E} = \frac{A C}{4}$ $\Rightarrow A C = \frac{96}{7}$ $area (A B C) = \frac{1}{2} \times A B \times A C$ $= \frac{1}{2} \times 9 \times \frac{96}{7} = \frac{432}{7}$
	Commented by ARUNG_Brandon_MBU last updated on 23/Nov/22
	OK
	Commented by HeferH last updated on 23/Nov/22

	$I t h i n k t h a t y o u a r e a s u m i n g A D = 3, w h i c h$ $i s n o t n e c e s s a r i l y t r u e .$
	Answered by mr W last updated on 23/Nov/22

	$A D = 4 \tan α$ $A D + D C = 4 \tan 2 α$ $D C = 4 (\tan 2 α - \tan α) = D B$ $D B = \sqrt{9^{2} + {(4 \tan α)}^{2}}$ $4 (\tan 2 α - \tan α) = \sqrt{9^{2} + {(4 \tan α)}^{2}}$ $4 \tan α (\frac{2}{1 - \tan^{2} α} - 1) = \sqrt{81 + 16 \tan^{2} α}$ $16 \tan^{2} α {(\frac{1 + \tan^{2} α}{1 - \tan^{2} α})}^{2} = 81 + 16 \tan^{2} α$ $l e t t = \tan^{2} α$ $16 t {(\frac{1 + t}{1 - t})}^{2} = 81 + 16 t$ $17 t^{2} - 162 t + 81 = 0$ $t = \frac{81 - 72}{17} = \frac{9}{17} = \tan^{2} α$ $\Rightarrow \tan α = \frac{3}{\sqrt{17}}$ $A C = 4 \times \frac{2 \times \frac{3}{\sqrt{17}}}{1 - {(\frac{3}{\sqrt{17}})}^{2}} = 3 \sqrt{17}$ $Δ A B C = \frac{9}{2} \times 3 \sqrt{17} = \frac{27 \sqrt{17}}{2} ✓$
	Answered by HeferH last updated on 23/Nov/22

	Commented by HeferH last updated on 23/Nov/22

	$Y e s, I u s e d t h e b i s e c t o r t h e o r e m$
	Commented by HeferH last updated on 23/Nov/22

	$I^{'} l l s h a r e m y s o l u t i o n :)$ $t h a n k y o u a l l f o r a n s w e r i n g$
	Commented by Acem last updated on 23/Nov/22

	$@ H e f e r h, S i r, y o u r m e t h o d d e p e n d s o n$ $t h e c i r c u m c i r c l e ? o r o n s o m e t h i n g e l s e ?$
	Commented by HeferH last updated on 23/Nov/22

	Commented by HeferH last updated on 23/Nov/22

	$A c l e a n e r s o l u t i o n (c o m p a r e d t o w h a t i d i d)$
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