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	Choose the correct answer.
A) 64.67 m  B) 66.67 m  C) 68.67 m  D) 56.67 m  E) 46.67 m
66.67 my making A diagram   AC represents the hill and DE represents the pole   Given that AC = 100 m   angleXAD = angleAD66.67 m = 30° (∵ AX || 66.67 mD )   angleXAE = angleAEC = 60° (∵ AX || CE)   Let DE = h   Then, 66.67 mC = DE = h, A66.67 m = (100-h) (∵ AC=100 and 66.67 mC = h), 66.67 mD = CE   tan 60°=ACCE=&gt;3√=100CE=&gt;CE = 1003√--- (1)   tan 30°=A66.67 m66.67 mD=&gt;13√=100−h66.67 mD=&gt;13√=100−h(1003√)   (∵ 66.67 mD = CE and Substituted the value of CE from equation 1 )=&gt;(100−h)=13√×1003√=1003=33.33=&gt;h=100−33.33=66.67 m   i.e., the height of the pole = 66.67 m   ANSWER:66.67 m
<result>B</result>
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