The top and bottom of a tower were seen
to be at angles of depression 30° and 60° from the top of a
hill of height 100 m. Find the height of the tower

• By making A diagram
  AC represents the hill and DE represents the pole
  
  Given that AC = 100 m
  
  angleXAD = angleADB = 30° (∵ AX || BD )
  angleXAE = angleAEC = 60° (∵ AX || CE)
  
  Let DE = h
  
  Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE
  
  tan 60°=ACCE=>3√=100CE=>CE = 1003√--- (1)
  
  tan 30°=ABBD=>13√=100−hBD=>13√=100−h(1003√)
  (∵ BD = CE and Substituted the value of CE from equation 1 )=>(100−h)=13√×1003√=1003=33.33=>h=100−33.33=66.67 m
  
  i.e., the height of the pole = 66.67 m
• 8 years agoHelpfull: Yes(0) No(0)