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The tallest tower in a city 100mt high and a multistoryed hotel at the city center is 20mt high. The angle of elevation of the top of the tower of the hotel is 3°36' . A building H meter high is situated on the road connecting the tower with city center at a distance of 1Km from the tower. Find the value of H .if the top of the hotel, top of the building and top of the tower in a straight line .Also find distance of the tower from the city center?
Asked by abhitanshj | 11 Sep, 2016, 02:13: PM
answered-by-expert Expert Answer
begin mathsize 16px style In space the space above space problem comma space the space value space of space the space angle space of space elevation space is space in space degrees. The space value space of space the space angle space of space elevation space should space be space given space in space the space question comma space unless space it space is space straight a space known space angle space like space 30 degree comma space 45 degree comma space 60 degree comma space etc. straight I space have space straight a space similar space question comma space where space instead space of space 3 degree 36 apostrophe comma space the space value space of space tanθ equals space 0.0629 space is space given space in space the space question. Kindly space check space the space question. Refer space to space the space figure space above. AC space is space the space tower. FD space is space the space building space whose space height space is space apostrophe straight H apostrophe. GH space is space the space hotel space in space the space city space centre space whose space height space is space 20 space mt. Angle space of space elevation space of space apostrophe straight A apostrophe space from space apostrophe straight G apostrophe space is space straight theta. angle FGE equals angle AFB In space triangle space AFB comma space tanθ equals AB over BF rightwards double arrow 0.0629 equals AB over CD rightwards double arrow AB equals 0.0629 cross times 1000 equals 62.9 space straight m BC equals AC minus AB equals 100 minus 62.9 equals 37.1 In space the space figure space FD equals BC equals 37.1 space left parenthesis height space of space the space building right parenthesis.. left parenthesis 1 right parenthesis So comma space straight H equals 37.1 space straight m. In space triangle space FEG comma space FE equals FD minus ED equals FD minus GH equals 37.1 minus 20 equals 17.1 space straight m In space triangle FEG comma space tanθ equals FE over EG rightwards double arrow EG equals fraction numerator 17.1 over denominator 0.0629 end fraction equals 271.86 space straight m CH equals CD plus DH equals 1000 plus 271.86 equals 1271.86 space straight m Hence comma space distance space of space the space tower space from space the space city space centre space is space 1271.86 space straight m end style
Answered by Rebecca Fernandes | 11 Sep, 2016, 07:43: PM

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