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Question Number 86040 by TawaTawa1 last updated on 26/Mar/20

Answered by sakeefhasan05@gmail.com last updated on 26/Mar/20

Commented by Serlea last updated on 26/Mar/20

Taking it at Right angle L=(√((40^2 −10^2 ))) L=(√(1500)) L=38.7 ?=2L =2(38.7) =77.45

$$ \\ $$$$ \\ $$$$\mathrm{Taking}\:\mathrm{it}\:\mathrm{at}\:\:\mathrm{Right}\:\mathrm{angle} \\ $$$$\mathrm{L}=\sqrt{\left(\mathrm{40}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \right)} \\ $$$$\mathrm{L}=\sqrt{\mathrm{1500}} \\ $$$$\mathrm{L}=\mathrm{38}.\mathrm{7} \\ $$$$?=\mathrm{2L} \\ $$$$=\mathrm{2}\left(\mathrm{38}.\mathrm{7}\right) \\ $$$$=\mathrm{77}.\mathrm{45} \\ $$

Commented by TawaTawa1 last updated on 27/Mar/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by sakeefhasan05@gmail.com last updated on 26/Mar/20

answer=2L=2×(√(40^2 −10^2 )) =20(√(15)) m

$$\mathrm{answer}=\mathrm{2L}=\mathrm{2}×\sqrt{\mathrm{40}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} } \\ $$$$=\mathrm{20}\sqrt{\mathrm{15}}\:\mathrm{m} \\ $$

Commented by TawaTawa1 last updated on 27/Mar/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by sakeefhasan05@gmail.com last updated on 27/Mar/20

thank u very much how old r u

$$\mathrm{thank}\:\mathrm{u}\:\mathrm{very}\:\mathrm{much}\:\mathrm{how}\:\mathrm{old}\:\mathrm{r}\:\mathrm{u} \\ $$

Commented by sakeefhasan05@gmail.com last updated on 27/Mar/20

thank u very much how old r u

$$\mathrm{thank}\:\mathrm{u}\:\mathrm{very}\:\mathrm{much}\:\mathrm{how}\:\mathrm{old}\:\mathrm{r}\:\mathrm{u} \\ $$

Commented by TawaTawa1 last updated on 27/Mar/20

19

$$\mathrm{19} \\ $$

Commented by sakeefhasan05@gmail.com last updated on 27/Mar/20

I am 20 from Sri Lanka where r u from?

$$\mathrm{I}\:\mathrm{am}\:\mathrm{20} \\ $$$$\mathrm{from}\:\mathrm{Sri}\:\mathrm{Lanka}\:\mathrm{where}\:\mathrm{r}\:\mathrm{u}\:\mathrm{from}? \\ $$

Commented by sakeefhasan05@gmail.com last updated on 27/Mar/20

I am 20 from Sri Lanka where r u from?

$$\mathrm{I}\:\mathrm{am}\:\mathrm{20} \\ $$$$\mathrm{from}\:\mathrm{Sri}\:\mathrm{Lanka}\:\mathrm{where}\:\mathrm{r}\:\mathrm{u}\:\mathrm{from}? \\ $$

Answered by MJS last updated on 26/Mar/20

width between the poles = w length of cable =l=80 height of hanging cable = h=10 w=2h((l^2 /(4h^2 ))−1)tanh^(−1) ((2h)/l) ≈76.62m

$$\mathrm{width}\:\mathrm{between}\:\mathrm{the}\:\mathrm{poles}\:=\:{w} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{cable}\:={l}=\mathrm{80} \\ $$$$\mathrm{height}\:\mathrm{of}\:\mathrm{hanging}\:\mathrm{cable}\:=\:{h}=\mathrm{10} \\ $$$${w}=\mathrm{2}{h}\left(\frac{{l}^{\mathrm{2}} }{\mathrm{4}{h}^{\mathrm{2}} }−\mathrm{1}\right)\mathrm{tanh}^{−\mathrm{1}} \:\frac{\mathrm{2}{h}}{{l}}\:\approx\mathrm{76}.\mathrm{62}{m} \\ $$

Commented by TawaTawa1 last updated on 27/Mar/20

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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