Saved Bookmarks
| 1. |
From a point 100 m above the lake |
| Answer» Let AB be the surface of the lake and P be the point of observation such that {tex}AP = 100 m{/tex}. Let C be the position of the helicopter and C\' be its reflection in the lake.Then,\xa0{tex}CB = C\'B.{/tex}Let PM be perpendicular from P on CB.Then, {tex}\\angle{/tex}{tex}CPM = 30\xa0^\\circ{/tex}and {tex}\\angle{/tex}C\'PM =\xa0{tex}60^\\circ{/tex}Then,\xa0{tex}CM = h, \\ CB = h + 100.{/tex}In right {tex}\\triangle{/tex}CMP{tex}\\tan 30^{\\circ}=\\frac{\\mathrm{C} \\mathrm{M}}{\\mathrm{PM}} \\Rightarrow{\\frac{1}{ \\sqrt{3}}}=\\frac{h}{\\mathrm{PM}}{/tex}{tex}\\Rightarrow PM =\\sqrt3 h{/tex}...(i)In right {tex}\\triangle{/tex}PMC\'{tex}\\tan 60^{\\circ}=\\frac{\\mathrm{C}^{\\prime} \\mathrm{M}}{\\mathrm{PM}} \\Rightarrow \\sqrt{3}=\\frac{\\mathrm{C}^{\\prime} \\mathrm{B}+\\mathrm{BM}}{\\mathrm{PM}}{/tex}{tex}\\Rightarrow \\sqrt{3}=\\frac{h+100+100}{P M}{/tex}{tex}\\Rightarrow P M=\\frac{h+200}{\\sqrt{3}}{/tex}\xa0...(ii)From (i) and (ii), we get\xa0{tex} \\sqrt{3} h=\\frac{h+200}{\\sqrt{3}}{/tex}{tex} 3h = h + 200\xa0\\\\ 2h = 200\\\\ h = 100{/tex}Now,{tex}CB = CM + MB\\\\ = h + 100\\\\ = 100 + 100\\\\ = 200\\\\{/tex}\xa0Hence, the height of the helicopter from the surface of the lake = 200 m | |