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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

$$$$$$$$$$\mathrm{Taking}\mathrm{it}\mathrm{at}\mathrm{Right}\mathrm{angle}$$$$\mathrm{L}=\sqrt{({40}^2-{10}^2)}$$$$\mathrm{L}=\sqrt{1500}$$$$\mathrm{L}=38.7$$$$?=2\mathrm{L}$$$$=2(38.7)$$$$=77.45$$

Commented by TawaTawa1 last updated on 27/Mar/20

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

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

$$\mathrm{answer}=2\mathrm{L}=2\times \sqrt{{40}^2-{10}^2}$$$$=20\sqrt{15}\mathrm{m}$$

Commented by TawaTawa1 last updated on 27/Mar/20

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

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

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

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

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

Commented by TawaTawa1 last updated on 27/Mar/20

$$19$$

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

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

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

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

Answered by MJS last updated on 26/Mar/20

$$\mathrm{width}\mathrm{between}\mathrm{the}\mathrm{poles}=w$$$$\mathrm{length}\mathrm{of}\mathrm{cable}=l=80$$$$\mathrm{height}\mathrm{of}\mathrm{hanging}\mathrm{cable}=h=10$$$$w=2h(\frac{l^2}{4h^2}-1){\tanh }^{-1}\frac{2h}{l}\approx 76.62m$$

Commented by TawaTawa1 last updated on 27/Mar/20

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

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