Saved Bookmarks
| 1. |
A person observed the angle of elevation of the top of a tower as 30 |
| Answer» Let height of the tower be DC = h m and BC = x m AC = (50 + x) mIn\xa0{tex}\\triangle {/tex}DBC,\xa0{tex}\\frac { h } { x } = \\tan 60 ^ { \\circ } = \\sqrt { 3 }{/tex}\xa0{tex}\\Rightarrow \\quad h = \\sqrt { 3 } x{/tex} ..(i)\xa0In\xa0{tex}\\triangle {/tex}DAC,\xa0{tex}\\frac { h } { x + 50 } = \\tan 30 ^ { \\circ } = \\frac { 1 } { \\sqrt { 3 } }{/tex},\xa0{tex}\\Rightarrow \\quad \\sqrt { 3 } h = x + 50{/tex}\xa0...(ii)Substituting the value of h from (i) in (ii), we get3x = x + 50or, 3x - x = 50\xa0{tex}\\Rightarrow {/tex}\xa02x = 50{tex}\\Rightarrow {/tex} x = 25 m{tex}h = 25 \\sqrt { 3 } = 25 \\times 1 \\cdot 732 \\mathrm { m }{/tex}\xa0= 43.3 mHence, Height of tower = 43.3 m. | |