Question
Mon February 18, 2013 By: Sahil
 

The angles of depression of the top and bottom of a 20m high building from the top of a multistoreyed building are 30degrees and 45degrees respectively. Find the height of the multistoreyed building. [ Take root3= 1.732]

Expert Reply
Mon February 18, 2013
Answer : Given: The angles of depression of the top and bottom of a 20m high building from the top of a multistoreyed building are 30degrees and 45degrees respectively.
 
To Find : the height of the multistoreyed building. 
let the height of the multistoreyed building be AE , and the height of the building be CD i.e 20 m (given)
 
Now in triangle ABC
angle ACB = 30 deg (vertically opposite angle are equal)
similarily in triangle ADE
angle ADE = 45 degree
 
Now in triangle ABC
=> AB/BC = tan 30 degree
=> AB/BC = 1/31/2 = 1/1.732........eq 1      { tan 30 = 1/31/2 , and 31/2 = 1.732 given }
 
 
In triangle ADE
=> AE/ED = tan 45
=> AE /ED = 1       { tan 45 =1 }
=> AE/ BC = 1              { since ED = BC as EDCD is a rectangle and opposite sides of a rectangle are equal}
 
Dividing the above eq by eq 1 we get
=> AE /AB = 1.732
=> (AB + BE )/ AB = 1.732      {as AE = AB + BE}
=> AB + BE = 1.732 * AB
=> BE = (1.732-1) AB 
=> BE = 0.732 AB 
=> CD = 0.732 AB   { since BE = CD as opposite side of rectangle BEDC}
=> 20 / 0.732 = AB 
=> AB = 27.32 m ( approx)
 
therefore total height of multistoreyed buliding is = AB + BE = AB + CD = 27.32+20 = 47.32 m Answer
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