Example 1 :
A boy standing on the ground, spots a balloon moving with the wind in a horizontal line at a constant height . The angle of elevation of the balloon from the boy at an instant is 60°. After 2 minutes, from the same point of observation, the angle of elevation reduces to 30°. If the speed of wind is 29√3 m/min. then, find the height of the balloon from the ground level.
Solution :
Distance covered by the balloon = BC
BC = Time x Speed ==> 2 x 29 √3 ==> 58√3 m
AB = x then AC = x + 58√3
In triangle DAC :
∠DAC = 30°
tan θ = opposite side/Adjacent side
tan 30° = DC/AC
1/√3 = DC/(x + 58√3)
DC = (x + 58√3)/√3 ----(1)
In triangle EAB :
∠EAB = 60°
tan θ = opposite side/Adjacent side
tan 60° = EB/AB
√3 = EB/x
x√3 = EB
EB = √3x ---->(2)
Since EB = DC
(1) = (2)
(x + 58√3)/√3 = √3x
x + 58√3 = 3x
3x - x = 58√3
2x = 58√3
x = 58√3/2 ==> 29√3 m
Height of the balloon from ground level EB = √3 x
= 29 √3 (√3)
= 29(3) ==> 87 m
Hence height of the balloon from ground level = 87 m.
Example 2 :
A straight highway leads to the foot of a tower . A man standing on the top of the tower spots a van at an angle of depression of 30°. The van is approaching the tower with a uniform speed. After 6 minutes, the angle of depression of the van is found to be 60°. How many more minutes will it take for the van to reach the tower?
Solution :
From the given information, we can draw a rough diagram
Distance covered by the van to reach D from C = 6 minutes
time taken = x
Distance between D and C = 6x
In triangle ACB
∠ACB = 30°
tan θ = opposite side/Adjacent side
tan 30° = AB/BC
1/√3 = AB/(BD+DC)
1/√3 = AB/(BD+6x)
(BD+6x)/√3 = AB ----(1)
In triangle ABD
∠ABD = 60°
tan θ = opposite side/Adjacent side
tan 60° = AB/BD
√3 = AB/BD ==> AB = BD√3 -----(2)
(1) = (2)
(BD+6x)/√3 = BD√3
BD + 6x = BD(3)
3BD - BD = 6x
2BD = 6x
BD = 6x /2 = 3x
Here 3 represents number of minutes covered by the van and x stands for time taken.
Hence 3 more minutes will it take for the van to reach the tower.
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