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Question Number 187608 by ajfour last updated on 19/Feb/23

	Answered by a.lgnaoui last updated on 19/Feb/23

	$\tan α = \frac{X}{c} = \frac{X + 1}{2 X + c} \Rightarrow$ $(2 X + c) X = c (X + 1)$ ${2 X}^{2} + c X = c X + c$ ${2 X}^{2} = c$ $X = \sqrt{\frac{c}{2}}$
	Commented by a.lgnaoui last updated on 19/Feb/23

	Commented by ajfour last updated on 19/Feb/23
	how verical is 2x ?
	Commented by a.lgnaoui last updated on 19/Feb/23

	$O H w i t h O H + C = 2 X + C$
	Commented by a.lgnaoui last updated on 19/Feb/23

	Commented by a.lgnaoui last updated on 19/Feb/23

	$i f t h e q u a d r i l a t e r e i s n o t$ $s q u a r t t h a t i s a n a u t h e r e$ $v a l u e$ $y o u r s o l u t i o n i s g e n e r a l$ $t h a n k s .$
	Commented by mr W last updated on 19/Feb/23

	$y o u j u s t i g n o r e t h e s e m i c i r c l e i n$ $o r d e r t o b e a b l e t o s o l v e t h e p r o b l e m,$ $b u t t h i s i s n o t a r e a l s o l u t i o n, b e c a u s e$ $y o u h a v e c h a n g e d t h e q u e s t i o n .$
	Commented by mr W last updated on 19/Feb/23

	Commented by a.lgnaoui last updated on 19/Feb/23

	$h c a n t a k e v a l u e 2 x .$
	Commented by mr W last updated on 20/Feb/23

	${i t}^{'} s t h e n n o t t h e a s k e d q u e s t i o n,$ $t h e r e f o r e n o n - s e n s e .$
	Commented by ajfour last updated on 20/Feb/23
	yeah if the company agrees it can give me the car for 5$.
	Answered by ajfour last updated on 19/Feb/23

	$\tan α = \frac{x}{c} = \frac{1}{h} = \frac{h}{2 x + 1}$ $\Rightarrow h^{2} = 2 x + 1 = \frac{c^{2}}{x^{2}}$ $\Rightarrow 2 x^{3} + x^{2} = c^{2}$ $o r x \sqrt{2 x + 1} = c$
	Answered by mr W last updated on 19/Feb/23

	$h^{2} = 1 \times (2 x + 1) \Rightarrow h = \sqrt{2 x + 1}$ $\frac{c}{x} = \frac{h}{1}$ $\Rightarrow x \sqrt{2 x + 1} = c$ $x^{2} (2 x + 1) - c^{2} = 0$ $\frac{1}{x^{3}} - \frac{1}{c^{2} x} - \frac{2}{c^{2}} = 0$ $\Rightarrow \frac{1}{x} = \sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 - \frac{1}{27 c^{2}}} + 1)} - \sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 - \frac{1}{27 c^{2}}} - 1)}$ $\Rightarrow x = \frac{1}{\sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 - \frac{1}{27 c^{2}}} + 1)} - \sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 - \frac{1}{27 c^{2}}} - 1)}}$
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