Question Number 198114 by Tawa11 last updated on 10/Oct/23
Commented by mr W last updated on 10/Oct/23
$${it}\:{should}\:{be}\:\mathrm{20}\:{m}\:\cancel{{longer}}\:{shorter}. \\ $$
Answered by mr W last updated on 10/Oct/23
Commented by Tawa11 last updated on 10/Oct/23
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Commented by mr W last updated on 10/Oct/23
$${l}_{\mathrm{1}} =\frac{{h}}{\mathrm{tan}\:\mathrm{30}°} \\ $$$${l}_{\mathrm{2}} =\frac{{h}}{\mathrm{tan}\:\mathrm{60}°} \\ $$$$\Delta{l}={l}_{\mathrm{1}} −{l}_{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{30}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{60}°}\right){h} \\ $$$$\Rightarrow{h}=\frac{\Delta{l}}{\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{30}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{60}°}}=\frac{\mathrm{20}}{\:\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}=\mathrm{10}\sqrt{\mathrm{3}} \\ $$