A vertical pole of length 6cm casts a shadow 4m long on the ground .At the same time another tower casts a shadow 30m long .find the height of the tower.
Answers
Answer:
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Step-by-step explanation:
Hint:We will be using the concepts of height and distance to solve the problem. We will be using trigonometry to further simplify the solution. We will be using the fact that the angle of inclination of the sun is the same for all objects at a particular time. We will start by finding the ratio
tan
θ=
p
B
tanθ=pB
, where p is perpendicular and B is base, from triangle ABC and then from triangle DEF. Equating these two ratios, we will be able to find the height of the tower.
Complete step-by-step solution:
Now we have been given a vertical pole of 6m which casts a shadow of 4m on the ground and at the same time, a tower casts a shadow 28m long.
We have let the angle of inclination be
θ
θ
in
ΔABC
ΔABC
. We know that
tan
θ=
p
B
tanθ=pB
.
Where P is perpendicular and B is base. Therefore,
tan
θ=
A B
B C
=
6
4
tanθ=ABBC=64
tan
θ=
3
2
tanθ=32
…………………… (1)
Now, we will apply
tan
θ
tanθ
in
ΔDEF
ΔDEF
. We know that
tan
θ=
p
B
tanθ=pB
tan
θ=
D E
E F
⇒tan
θ=
DE
28
tanθ=DEEF⇒tanθ=DE28
Now, we will substitute the value of
tan
θ
tanθ
from (1)
3
2
=
DE
28
32=DE28
On cross multiply we have
3×28=2×DE
⇒DE=
3×28
2
⇒DE=3×14
⇒DE=42m
3×28=2×DE⇒DE=3×282⇒DE=3×14⇒DE=42m
Therefore, the height of tower DE=42.
Note: To solve these types of questions it is important to note the fact that the sun has the same angle of inclination for all objects at a particular time also one must know some important trigonometric formulae like
tan
θ=
Perpendicular
Hypotenuse
tanθ=PerpendicularHypotenuse
. Students can also try using the Pythagoras theorem and finding the perpendicular from triangle ABC and proceed or they might try to compute the angle. But, this is not required here, since the angle for both the cases would be the same. So, they can directly apply the tangent ratio, equate them, and get the answer.
Answer:
Height of pole=AB=6 m
Length of shadow of pole =BC=4 m
Length of shadow of tower=EF=28 m
In △ABC and △DEF
∠B=∠E=90
∘
both 90
∘
as both are vertical to ground
∠C=∠F (same elevation in both the cases as both shadows are cast at the same time)
∴△ABC∼△DEF by AA similarity criterion
We know that if two triangles are similar, ratio of their sides are in proportion
So,
DE
AB
=
EF
BC
⇒
DE
6
=
28
4
⇒DE=6×7=42 m
Hence the height of the tower is 42 m