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.
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)
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°.
(12)
Ravi said:
5 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)
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)
Siddharth said:
3 weeks 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°
(1)
Idiotshere said:
1 decade 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:
6 years ago
AB/AC=1/√3.
TANθ=30θ.
TANθ=30θ.
(1)
Lory said:
5 years ago
I can't understand the question yet.
(1)
Vaibhav said:
5 years ago
Thank you all for explaining.
(1)
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