Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

Answered by GDVilla last updated on 24/Dec/24

$$3?$$

Answered by A5T last updated on 24/Dec/24

Commented by A5T last updated on 24/Dec/24

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

Commented by cherokeesay last updated on 24/Dec/24

$$thankyousir!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com