Question-191602

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

(√((h+b)^2 +h^2 ))−(h+b)=(√(a^2 +b^2 )) h^2 −2h(√(a^2 +b^2 ))−(a^2 +b^2 +2b(√(a^2 +b^2 )))=0 h=(√(a^2 +b^2 ))+(√(2(a^2 +b^2 +b(√(a^2 +b^2 )))))

$$\sqrt{\left({h}+{b}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} }−\left({h}+{b}\right)=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${h}^{\mathrm{2}} −\mathrm{2}{h}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\left.{h}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right.}\right) \\ $$

Commented by ajfour last updated on 27/Apr/23

(√(a^2 +b^2 ))+b+h=(√((b+h)^2 +s^2 )) ⇒ a^2 +b^2 +2(b+h)(√(a^2 +b^2 ))=h^2 h^2 −2(√(a^2 +b^2 ))h−(a^2 +b^2 )−2b(√(a^2 +b^2 ))=0 ⇒ h=(√(a^2 +b^2 ))+(√(2(a^2 +b^2 )+2b(√(a^2 +b^2 ))))

$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+{b}+{h}=\sqrt{\left({b}+{h}\right)^{\mathrm{2}} +{s}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}\left({b}+{h}\right)\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={h}^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} −\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{h}−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${h}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$

Commented by ajfour last updated on 27/Apr/23

$${yes}\:{sir},\:{thanks}. \\ $$

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