Standing on the top of a tower, 100 m high, Swati observes two cars parked on the opposite sides of the tower. If their angles of depression are 45 ° and 30 ° find the distance between the cars.
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Let P be the position of the tower where Swati is standing. Let A and B be the positions of the two cars such that angle of depression at A and B are 30° and 45°.
Let AM= x m , BM = y m
Given : height of the tower (PM)= 100 m
In ∆ AMP
tan 30° = PM / AM ( P/B)
1/√3= 100 / x
x= 100 / √3 …….,...(1)
In ∆ BMP
tan 45° = PM / BM ( P/B)
1= 100 / y
y= 100 ………(2)
From eq 1 & eq 2
AB = x+y
AB = 100/√3+100
AB = 100( 1/√3+ 1)
AB = 100 ( 1/ 1.732 +1)
[ √3= 1.732]
AB = 100 (( 1+1.732)/1.732)
AB =100 ( 2. 732/ 1.732) = 273.2/ 1.732
AB =273200/1732= 157.736
AB = 157.736 m
Hence, the distance between the cars= 157.736 m
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Hope this will help you.....
Let AM= x m , BM = y m
Given : height of the tower (PM)= 100 m
In ∆ AMP
tan 30° = PM / AM ( P/B)
1/√3= 100 / x
x= 100 / √3 …….,...(1)
In ∆ BMP
tan 45° = PM / BM ( P/B)
1= 100 / y
y= 100 ………(2)
From eq 1 & eq 2
AB = x+y
AB = 100/√3+100
AB = 100( 1/√3+ 1)
AB = 100 ( 1/ 1.732 +1)
[ √3= 1.732]
AB = 100 (( 1+1.732)/1.732)
AB =100 ( 2. 732/ 1.732) = 273.2/ 1.732
AB =273200/1732= 157.736
AB = 157.736 m
Hence, the distance between the cars= 157.736 m
==================================================================================
Hope this will help you.....
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