Question Number 222730 by fantastic last updated on 06/Jul/25
Commented by fantastic last updated on 06/Jul/25
$${find}\:{x}\:{in}\:{terms}\:{of}\: \\ $$$$\alpha,\beta,\gamma, \\ $$
Answered by mr W last updated on 06/Jul/25
$${x}=\sqrt{\frac{\left(\alpha+\beta\right)^{\mathrm{2}} +\gamma^{\mathrm{2}} }{\mathrm{2}}} \\ $$
Answered by gabthemathguy25 last updated on 06/Jul/25
$$\mathrm{Diagonal}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:=\:{x}\sqrt{\mathrm{2}} \\ $$$$\:{x}\centerdot\frac{\mathrm{1}}{\mathrm{sin}\left(\alpha\right)} \\ $$$${x}\sqrt{\mathrm{2}}=\frac{{x}}{\mathrm{sin}\left(\alpha\right)}+\frac{{x}}{\mathrm{sin}\left(\beta\right)}+\frac{{x}}{\mathrm{sin}\left(\gamma\right)} \\ $$$${x}\sqrt{\mathrm{2}}={x}\left(\frac{\mathrm{1}}{{sin}\left(\alpha\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\beta\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\gamma\right)}\right) \\ $$$$\sqrt{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{sin}\left(\alpha\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\beta\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\gamma\right)} \\ $$$$\left.{x}=\frac{\sqrt{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{sin}\left(\alpha\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\beta\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\gamma\right)}}\::\right) \\ $$
Commented by fantastic last updated on 06/Jul/25
$${those}\:{are}\:{representing}\:{length}\:{not}\:{angle} \\ $$