Question Number 191602 by ajfour last updated on 26/Apr/23
Commented by ajfour last updated on 26/Apr/23
$${Load}\:{is}\:{to}\:{be}\:{taken}\:{from}\:{A}\:{to}\:{B}. \\ $$$${If}\:{s}={h}\:,\:{find}\:{h}\:\:{in}\:{terms}\:{of}\:{a},{b}. \\ $$
Answered by mr W last updated on 27/Apr/23
$$\sqrt{\left({h}+{b}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} }−\left({h}+{b}\right)=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${h}^{\mathrm{2}} −\mathrm{2}{h}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\left.{h}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right.}\right) \\ $$
Commented by ajfour last updated on 27/Apr/23
$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+{b}+{h}=\sqrt{\left({b}+{h}\right)^{\mathrm{2}} +{s}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}\left({b}+{h}\right)\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={h}^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} −\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{h}−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${h}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 27/Apr/23
$${yes}\:{sir},\:{thanks}. \\ $$