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Question Number 214964 by cherokeesay last updated on 24/Dec/24

Answered by GDVilla last updated on 24/Dec/24

3?

$$\mathrm{3}? \\ $$

Answered by A5T last updated on 24/Dec/24

Commented by A5T last updated on 24/Dec/24

x=y; y^2 =z(z+z)=2z^2 ⇒z^2 =(y^2 /2)  (2z)^2 =x^2 +(2y)^2 −2x(2y)cosθ  ⇒4z^2 =2y^2 =y^2 +4y^2 −4y^2 cosθ⇒cosθ=(3/4)  ⇒x^2 +y^2 −2xycosθ=1^2 +1^2 −2×1^2 cos(180−θ)  ⇒2x^2 =((2+(2×(3/4)))/(1−(3/4)))=14⇒x=(√7)

$${x}={y};\:{y}^{\mathrm{2}} ={z}\left({z}+{z}\right)=\mathrm{2}{z}^{\mathrm{2}} \Rightarrow{z}^{\mathrm{2}} =\frac{{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{2}{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\mathrm{2}{y}\right)^{\mathrm{2}} −\mathrm{2}{x}\left(\mathrm{2}{y}\right){cos}\theta \\ $$$$\Rightarrow\mathrm{4}{z}^{\mathrm{2}} =\mathrm{2}{y}^{\mathrm{2}} ={y}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} {cos}\theta\Rightarrow{cos}\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xycos}\theta=\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}^{\mathrm{2}} {cos}\left(\mathrm{180}−\theta\right) \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} =\frac{\mathrm{2}+\left(\mathrm{2}×\frac{\mathrm{3}}{\mathrm{4}}\right)}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}}=\mathrm{14}\Rightarrow{x}=\sqrt{\mathrm{7}} \\ $$

Commented by cherokeesay last updated on 24/Dec/24

thank you sir !

$${thank}\:{you}\:{sir}\:! \\ $$

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