Dterminer la valeur de x


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Question Number 181404 by a.lgnaoui last updated on 24/Nov/22

$D t e r m i n e r l a v a l e u r d e x$
Commented by mr W last updated on 24/Nov/22

$g e n e r a l l y t h e r e a r e t w o p o s s i b l e$ $s o l u t i o n s .$
Commented by mr W last updated on 24/Nov/22

Commented by a.lgnaoui last updated on 24/Nov/22

$l e s s e g m e n t s A P B P P C s o n t d e t e r m i n e$ $p a r d e s m e s u r e s (n o n p a s d e s$ $m e d i a n e s n i b i s s e c t r i c e s)$ $c e l a a p p a r a i t c l a i r e d a n s (f i g)$
Commented by a.lgnaoui last updated on 24/Nov/22

Commented by Acem last updated on 24/Nov/22

$S a l u t m o n s i o e u r, {M e}^{'} d i a n e s o u b i s s e c t r i c e s$ $d^{'} a n g l e s ?$
	Answered by mr W last updated on 24/Nov/22

	Commented by mr W last updated on 24/Nov/22

	$\cos α = \frac{x^{2} + c^{2} - a^{2}}{2 c x}$ $\cos β = \frac{x^{2} + c^{2} - b^{2}}{2 c x} = \sin α$ ${(x^{2} + c^{2} - a^{2})}^{2} + {(x^{2} + c^{2} - b^{2})}^{2} = 4 c^{2} x^{2}$ $x^{4} - (a^{2} + b^{2}) x^{2} + \frac{a^{4} + b^{4}}{2} - (a^{2} + b^{2} - c^{2}) c^{2} = 0$ $x^{2} = \frac{1}{2} [a^{2} + b^{2} \pm \sqrt{4 (a^{2} + b^{2} - c^{2}) c^{2} - {(a^{2} - b^{2})}^{2}}]$ $\Rightarrow x = \sqrt{\frac{a^{2} + b^{2} \pm \sqrt{4 (a^{2} + b^{2} - c^{2}) c^{2} - {(a^{2} - b^{2})}^{2}}}{2}}$ $w i t h a = 15, b = 12, c = 9 :$ $x = \sqrt{\frac{15^{2} + 12^{2} \pm \sqrt{4 (15^{2} + 12^{2} - 9^{2}) 9^{2} - {(15^{2} - 12^{2})}^{2}}}{2}}$ $= \frac{3 \sqrt{82 \pm 6 \sqrt{119}}}{2} \approx 18.214 o r 6.102$
	Commented by a.lgnaoui last updated on 24/Nov/22

	$e x a c t l y t h a n k y o u P r o f$
	Answered by HeferH last updated on 26/Nov/22

	Commented by HeferH last updated on 26/Nov/22

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