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Q.

The distance between two vertical pillars is 100 m, and the height of one is double of the other. The angles of elevation of their tops at a point on the line joining the foot of the two pillars are 60 ∘  and 30 ∘  respectively. What will be their heights?


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a

h 1 =40 3    m and h 2 =30 3    m

b

h 1 =30 3    m and h 2 =20 3    m

c

h 1 =20 3    m and h 2 =40 3    m

d

h 1 =20 2    m and h 2 =40 3    m 

answer is C.

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Detailed Solution

Given that the distance between towers is 100   m, the angles made by towers are is 60 °  and 30 °   and the height of one is double of the other.
Question ImageThe height of the first tower CD( h 1 )=y  
The height of the second tower AB( h 2 )=2y    The distance between the towers BD=100   m
Let BE=x   therefore DE=100−x  .
We use the identity tanθ= Perpendicular Base   in ΔABE   .
tanθ= AB BE   Substituting the values,
  ⇒tan 60 ° = 2y x ⇒ 3 = 2y x    [∵tan 60 ° = 3 ] ⇒x= 2y 3  
In ΔCDE  ,
tanθ= CD DE   ⇒tan 30 ° = y 100−x ⇒ 1 3 = y 100−x      [∵tan 30 ° = 1 3 ] ⇒100−x= 3 y  
Substituting the value of x in the equation 100 −x= 3 y  ,
⇒100− 2y 3 = 3 y ⇒100= 3 y+ 2y 3 ⇒100= 3y+2y 3 ⇒y= 100 3 5 ⇒y=20 3  m  
Therefore h 1 =20 3 m  .
The height of the second tower,
⇒ h 2 =2y ⇒ h 2 =2 20 3  m ⇒ h 2 =40 3  m  
Thus the heights of both towers will be h 1 =20 3    m and h 2 =40 3    m.
Therefore the correct option is 3.
 
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The distance between two vertical pillars is 100 m, and the height of one is double of the other. The angles of elevation of their tops at a point on the line joining the foot of the two pillars are 60 ∘  and 30 ∘  respectively. What will be their heights?