Aptitude - Height and Distance - Discussion
Discussion Forum : Height and Distance - General Questions (Q.No. 6)
6.
The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:
Answer: Option
Explanation:
Let AB be the tree and AC be its shadow.
Let 
ACB = 
.
| Then, | AC | = | 3  cot = 3 |
| AB |
= 30°.
Discussion:
62 comments Page 1 of 7.
Jabbar jr said:
2 years ago
The length of the shadow fo the tree is √3.
Here we want length of the tree only, So;
tanθ=opp/adj.
tanθ = x/√3.
θ=x/√3 => θ=1/√3.
So, the answer is θ = 30°.
Here we want length of the tree only, So;
tanθ=opp/adj.
tanθ = x/√3.
θ=x/√3 => θ=1/√3.
So, the answer is θ = 30°.
(13)
Shaik Nayab said:
3 years ago
Consider <ABC,
Height of the tree taken as = h.
Given, length of the shadow of a tree √3 times the height of the tree = √3 h.
I have taken, tan θ = BC/AC.
= h/√3 h.
=> numerator and denominator " h" canceled;
Remaining = 1/√3.
So, tan θ = 1 /√3 (Given data on both sides, opposite and adjacent sides, so I can take tan).
Then, θ = 30°.
Height of the tree taken as = h.
Given, length of the shadow of a tree √3 times the height of the tree = √3 h.
I have taken, tan θ = BC/AC.
= h/√3 h.
=> numerator and denominator " h" canceled;
Remaining = 1/√3.
So, tan θ = 1 /√3 (Given data on both sides, opposite and adjacent sides, so I can take tan).
Then, θ = 30°.
(13)
Siddharth said:
8 months ago
The length of the tree = AB, shadow of tree = CB.
Assume AB = x then CB = √3x.
Now , tanθ = AB / CB = x/√3x = 1/√3.
we know tan30° = 1/√3.
So, θ = 30°
Assume AB = x then CB = √3x.
Now , tanθ = AB / CB = x/√3x = 1/√3.
we know tan30° = 1/√3.
So, θ = 30°
(12)
Ravi said:
6 years ago
Here we solve the problem by using the angle of depression. Let β the angle of depression made by the sun on the tree.
Then angle β is equal to the angle of BCA.
Given the length of shadow is √3 times the height of tree i.e., AC=√3AB.
tanβ = AB/AC = AB/√3AB = 1/√3
tanβ = tan30° =>β = 30°.
Then angle β is equal to the angle of BCA.
Given the length of shadow is √3 times the height of tree i.e., AC=√3AB.
tanβ = AB/AC = AB/√3AB = 1/√3
tanβ = tan30° =>β = 30°.
(3)
Nicholas Hatontola said:
4 years ago
Why are we considering BC as the length of the shadow instead of AC? Isn't the shadow supposed to be the base?
Please explain me.
Please explain me.
(2)
Aditya said:
5 years ago
Why cot is taken here. Why we cannot take tan? Please explain me.
(2)
Aditi said:
5 years ago
Here we are taking base AC as a shadow of height then we can solve it like;
Given-
Shadow of a tree is √3 times the height of the tree.
Height = √3AB.
3AB/AC.
3 = AC/AB.
tan = 1/cot.
tan = Perpendicular/Base so the inverse of tan is cot.
Cot = AC/AB (Base/Perpendicular).
Cot 3 = 30°.
Given-
Shadow of a tree is √3 times the height of the tree.
Height = √3AB.
3AB/AC.
3 = AC/AB.
tan = 1/cot.
tan = Perpendicular/Base so the inverse of tan is cot.
Cot = AC/AB (Base/Perpendicular).
Cot 3 = 30°.
(2)
Idiotshere said:
2 decades ago
Help me please!.
I'm not getting here why we are saying ACB why don't BCA ?
And another is that why AC/AB why don't AB/AC = tan60 ?
I'm not getting here why we are saying ACB why don't BCA ?
And another is that why AC/AB why don't AB/AC = tan60 ?
(1)
Aravind said:
7 years ago
AB/AC=1/√3.
TANθ=30θ.
TANθ=30θ.
(1)
Lory said:
6 years ago
I can't understand the question yet.
(1)
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