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The angles of depression of the top and the bottom of a building 50 metres high as observed from the top of a tower are and  respectively. Find the height of the tower and also the horizontal distance between the building and the tower.
Asked by Topperlearning User | 03 Nov, 2017, 12:52: PM
answered-by-expert Expert Answer

 

      begin mathsize 12px style Let space AB thin space be space the space building equals 50 CE space be space the space tower equals 50 plus straight y Let space BC equals AD equals straight x equals the space distance space between space them tan 30 to the power of 0 equals DE over AD rightwards double arrow fraction numerator 1 over denominator square root of 3 end fraction equals straight y over straight x end style

      begin mathsize 12px style In space right space increment EBC comma tan 60 degree space equals space CE over BC square root of 3 equals fraction numerator 50 space plus space straight y over denominator straight x end fraction square root of 3 straight x space equals space 50 space plus space straight y square root of 3 open parentheses square root of 3 straight y close parentheses space equals space 50 space plus space straight y........ from space left parenthesis straight i right parenthesis 3 straight y space minus space straight y space equals space 50 2 straight y space equals space 50 straight y space equals space 25 Height space of space the space tower space CE space equals space 50 space plus space straight y equals 50 space plus space 25 equals space 75 space cm Now space taking space the space value space of space straight y space in space left parenthesis straight i right parenthesis comma space we space get straight x space equals space square root of 3 cross times 25 straight x space equals space 1.73 cross times 25 straight x space equals space 43.25 Required space horizontal space distance space straight x space equals space 43.25 space straight m end style

          

Answered by | 03 Nov, 2017, 02:52: PM

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