The angles of depression of the top and bottom of a 50m high building from the top of a tower are 45 and 60 degree respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use root 3 = 1.73)



Let AB be the tower of height h metres and CD be the building of height 50 m.Draw CLAB.Now, LB = CD = 50 m; AL = h-50 mAlso, LC = BD Let ACL = 45° and ADB = 60°In ABD,tan 60° = ABBD3 = hBDBD = h3    ......1In ALC, tan 45° = ALLC1 = h-50BDBD = h - 50    ......2From 1 and 2, we get     h - 50 = h3h - h3 = 503h - h = 503h3 - 1 = 503h = 5033-1 = 5033-1×3+13+1 = 150 + 5033-1 = 150+5032 = 75+253 = 75 + 25×1.73 = 118.25 mNow, from 2, we getBD = 118.25 - 50 = 68.25 mSo, height of the tower, h = 118.25 mhorizontal distance between tower and building = 68.25 m

  • 145
What are you looking for?