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Question Number 210473 by Ismoiljon_008 last updated on 10/Aug/24

	Answered by mrdiane last updated on 11/Aug/24
	$on a { ((7^2 =x^2 +5^2 −10xcos(θ))),((7^2 =x^2 +(x+5)^2 −2x(x+5)cos(Π−θ))) :} ⇔ { ((7^2 =x^2 +5^2 −10xcos(θ) (1))),((7^2 =x^2 +(x+5)^2 +2x(x+5)cos(θ) (2))) :} (1)=(2) on a 2(x+5)cos(θ)+x+10=−10cos(θ) ⇔cos(θ)=((−x−10)/(2x+20)) =((−1)/2) pour car x≥0 ⇔x^2 −24= dans (1) on a x^2 +5x−24=0 ⇔x=3 ou x=−8 comme x≥0 donc x=3 x^2 +5x−24=0$
	$o n a {\begin{cases} 7^{2} = x^{2} + 5^{2} - 10 x c o s (θ) \\ 7^{2} = x^{2} + {(x + 5)}^{2} - 2 x (x + 5) c o s (Π - θ) \end{cases}$ $\Leftrightarrow {\begin{cases} 7^{2} = x^{2} + 5^{2} - 10 x c o s (θ) (1) \\ 7^{2} = x^{2} + {(x + 5)}^{2} + 2 x (x + 5) c o s (θ) (2) \end{cases}$ $(1) = (2) o n a 2 (x + 5) c o s (θ) + x + 10 = - 10 c o s (θ)$ $\Leftrightarrow c o s (θ) = \frac{- x - 10}{2 x + 20} = \frac{- 1}{2} p o u r c a r x ⩾ 0$ $\Leftrightarrow x^{2} - 24 =$ $d a n s (1) o n a x^{2} + 5 x - 24 = 0 \Leftrightarrow x = 3 o u x = - 8 c o m m e x ⩾ 0$ $d o n c x = 3$ $x^{2} + 5 x - 24 = 0$
	Answered by efronzo1 last updated on 11/Aug/24

	$\cos θ = \frac{x^{2} + {(x + 5)}^{2} - 49}{2 x (x + 5)}$ $\cos (180 ° - θ) = \frac{25 + x^{2} - 49}{2.5 . x}$ $\Rightarrow - \cos θ = \frac{x^{2} - 24}{10 x}$ $\Rightarrow - \frac{{2 x}^{2} + 10 x - 24}{2 x (x + 5)} = \frac{x^{2} - 24}{10 x}$ $\Rightarrow - ({10 x}^{2} + 50 x - 120) = (x^{2} - 24) (x + 5)$ $\Rightarrow - {10 x}^{2} - 50 x + 120 = x^{3} + {5 x}^{2} - 24 x - 120$ $\Rightarrow x^{3} + {15 x}^{2} + 26 x - 240 = 0$ $\Rightarrow (x - 3) (x + 8) (x + 10) = 0$ $∴ x = 3$
	Answered by mr W last updated on 10/Aug/24

	Commented by mr W last updated on 10/Aug/24

	$\underset{―}{M e t h o d I}$ $x (y^{2} + 7^{2}) = 2 x (5^{2} + x^{2})$ $\Rightarrow y^{2} = 2 x^{2} + 1$ $5 \times 7^{2} + x \times x^{2} = (5 + x) (y^{2} + 5 x)$ $\Rightarrow 245 + x^{3} = (5 + x) (2 x^{2} + 1 + 5 x)$ $\Rightarrow x^{3} + 15 x^{2} + 26 x - 240 = 0$ $\Rightarrow (x - 3) (x^{2} + 18 x + 80) = 0$ $\Rightarrow x = 3$
	Answered by mr W last updated on 10/Aug/24

	Commented by mr W last updated on 10/Aug/24

	$\underset{―}{M e t h o d I I}$ $A E = x + 2 \times 5$ $C E = \frac{A E}{2} = \frac{x}{2} + 5$ $C F = \frac{x}{2} + 5 - 5 = \frac{x}{2}$ ${D C}^{2} = 7^{2} - {(\frac{x}{2} + 5)}^{2} = x^{2} - {(\frac{x}{2})}^{2}$ $x^{2} + 5 x - 24 = 0$ $(x - 3) (x + 8) = 0$ $\Rightarrow x = 3$
	Commented by Ismoiljon_008 last updated on 10/Aug/24

	$T h a n k y o u v e r y m u c h$
	Answered by mr W last updated on 11/Aug/24

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