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Question Number 198114 by Tawa11 last updated on 10/Oct/23

Commented by mr W last updated on 10/Oct/23

it should be 20 m longer shorter.

$${it}\:{should}\:{be}\:\mathrm{20}\:{m}\:\cancel{{longer}}\:{shorter}. \\ $$

Answered by mr W last updated on 10/Oct/23

Commented by Tawa11 last updated on 10/Oct/23

Thanks sir. God bless you.

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

Commented by mr W last updated on 10/Oct/23

l_1 =(h/(tan 30°))  l_2 =(h/(tan 60°))  Δl=l_1 −l_2 =((1/(tan 30°))−(1/(tan 60°)))h  ⇒h=((Δl)/((1/(tan 30°))−(1/(tan 60°))))=((20)/( (√3)−(1/( (√3)))))=10(√3)

$${l}_{\mathrm{1}} =\frac{{h}}{\mathrm{tan}\:\mathrm{30}°} \\ $$$${l}_{\mathrm{2}} =\frac{{h}}{\mathrm{tan}\:\mathrm{60}°} \\ $$$$\Delta{l}={l}_{\mathrm{1}} −{l}_{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{30}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{60}°}\right){h} \\ $$$$\Rightarrow{h}=\frac{\Delta{l}}{\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{30}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{60}°}}=\frac{\mathrm{20}}{\:\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}=\mathrm{10}\sqrt{\mathrm{3}} \\ $$

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