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Question-197229




Question Number 197229 by sonukgindia last updated on 10/Sep/23
Commented by Sachinkhar last updated on 10/Sep/23
Solve fourier inverse transform of   F(𝛏)=cosat𝛏^2
$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{fourier}}\:\boldsymbol{\mathrm{inverse}}\:\boldsymbol{\mathrm{transform}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\boldsymbol{\mathrm{F}}\left(\boldsymbol{\xi}\right)=\boldsymbol{\mathrm{cosat}\xi}^{\mathrm{2}} \\ $$
Answered by HeferH last updated on 10/Sep/23
? = (((18(√5))^2 )/2) = 810 u^2
$$?\:=\:\frac{\left(\mathrm{18}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{2}}\:=\:\mathrm{810}\:{u}^{\mathrm{2}} \\ $$
Answered by gksingh last updated on 10/Sep/23
Let sides of Ξ” are x,x and y  then x=(√(36^2 +18^2  ))                = 18(√5)  ∴ Area of Ξ”= (1/2) x (18(√5))^2                               =(1/2)Γ—18Γ—18Γ—5                        = 9Γ—90=810
$${Let}\:{sides}\:{of}\:\Delta\:{are}\:{x},{x}\:{and}\:{y} \\ $$$${then}\:{x}=\sqrt{\mathrm{36}^{\mathrm{2}} +\mathrm{18}^{\mathrm{2}} \:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{18}\sqrt{\mathrm{5}} \\ $$$$\therefore\:{Area}\:{of}\:\Delta=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{x}\:\left(\mathrm{18}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}Γ—\mathrm{18}Γ—\mathrm{18}Γ—\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{9}Γ—\mathrm{90}=\mathrm{810} \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

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