23. Find the height of a tree, if it casts a shadow 15 m long on the level of ground, when the
angle of elevation of the sun is 45
please fast
Answers
Answer:
✪SOLUTION✪
To solve these type of Questions we have to assume something.
From the attached figure, we have assumed that:
OX is ground level.
AB is height of the tower.
BC is casted shadow by tower AB
∠ABC=90°
∠ACB=45° (Angle of elevation)
We have to find height of the tower i.e AB.
Since:
∠ABC=90°, therefore, ∆ABC is right angled at B
Now:
In right ∆ABC, we have:-
AB= perpendicular (p)
BC= base (b)
AC= hypotenuse (h)
As we know:
= > tanθ = \frac {p}{b} where \:θ = {45}^{0}=>tanθ=
b
p
whereθ=45
0
= > tan {45}^{0} = \frac{p}{15}=>tan45
0
=
15
p
☞tan45°=1
= > 1 = \frac{p}{15}=>1=
15
p
= > p = 15=>p=15
Hence perpendicular (p) which is height of the tower is 15cm.
Another shortcut method:
If angle of elevation is 45° then the value of perpendicular and base are same.
Here, length of shadow is 15cm and angle of elevation is 45°, therefore, height will also be 15cm.
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