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.
Balaji said:
1 decade ago
Let me explain here, here carefully,
Here in the problem tree is taken as perpendicular i.e. AB and they also informed that the shadow of tree 3 times the tree i.e. 3 AB. Here shadow is nothing but base.
So we have formula that cot thita = cos θ/ sin θ
Cot θ= AC/AB
= 3 * AB/AB
Cot θ= 3
Cot θ= Cot 30
θ = 30
Here in the problem tree is taken as perpendicular i.e. AB and they also informed that the shadow of tree 3 times the tree i.e. 3 AB. Here shadow is nothing but base.
So we have formula that cot thita = cos θ/ sin θ
Cot θ= AC/AB
= 3 * AB/AB
Cot θ= 3
Cot θ= Cot 30
θ = 30
Kshyama Sagar said:
9 years ago
Considering the above Fig.
Assume that the angle of elevation made by the Sun in the shadow of tree is < acb= θ and height of tree (AB) is = x (in meter).
According to question,
shadow of the tree is √3 times the height of the tree.
shadow of a tree(CA) will be = √3x.
tan θ = AB/AC = x/√3x.
tan θ =1/√3.
tan θ = tan 1/√3.
tan θ = tan30°.
θ = 30° (Require answer).
Assume that the angle of elevation made by the Sun in the shadow of tree is < acb= θ and height of tree (AB) is = x (in meter).
According to question,
shadow of the tree is √3 times the height of the tree.
shadow of a tree(CA) will be = √3x.
tan θ = AB/AC = x/√3x.
tan θ =1/√3.
tan θ = tan 1/√3.
tan θ = tan30°.
θ = 30° (Require answer).
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)
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)
Jeslin said:
7 years ago
Let tree AB=x.
Shadow AC= √3 times height of the tree.
so, AC= √3x.
Here, the opposite and adjacent sides are involved so, we are using tan.
Tan theta = opp/adj =AB/AC =x/√3x =1/√3.
Theta = tan inverseof( 1/√3).
Theta=30°.
(we know that tan 30° =1/√3).
Hope it helps.
Shadow AC= √3 times height of the tree.
so, AC= √3x.
Here, the opposite and adjacent sides are involved so, we are using tan.
Tan theta = opp/adj =AB/AC =x/√3x =1/√3.
Theta = tan inverseof( 1/√3).
Theta=30°.
(we know that tan 30° =1/√3).
Hope it helps.
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)
Anchal said:
1 decade ago
We can also solve like this...let AB=x, then AC=3x, Now
AB/AC=tanθ
then x/3=tanθ
After crossing x from x..we get
1/3=tanθ
so tan 30=1/3
so θ=30...Ans..
AB/AC=tanθ
then x/3=tanθ
After crossing x from x..we get
1/3=tanθ
so tan 30=1/3
so θ=30...Ans..
Lavanya said:
10 years ago
Guys they said shadow of the tree is root 3 times of its height that is if we take height as "x" then base is "root 3x".
If we do by using Tan then:
Tan theta = Height/Base.
= x/root 3x.
By cancelling we will get 1/root 3.
Theta = tan-1(1/root 3).
i.e theta = 30 that's it.
If we do by using Tan then:
Tan theta = Height/Base.
= x/root 3x.
By cancelling we will get 1/root 3.
Theta = tan-1(1/root 3).
i.e theta = 30 that's it.
Sandeep sai kumar said:
7 years ago
We consider;
ab=x then,
ca=√3x,
So tanθ=opposite/adjacent.
Tanθ=ab/ac.
Tanθ=x/√3x,
Tanθ=1/√3,
θ=tanθ 1(1/√3),
θ=tan (√3),
θ=30°.
ab=x then,
ca=√3x,
So tanθ=opposite/adjacent.
Tanθ=ab/ac.
Tanθ=x/√3x,
Tanθ=1/√3,
θ=tanθ 1(1/√3),
θ=tan (√3),
θ=30°.
Swapnali wadhavne said:
7 years ago
Here shadow is √3 times the tree height.
Therefore AC=√3AB.
NOW tanθ =AB/AC,
tanθ =AB/√3AB,
tanθ=1/√3,
θ=tan^-1((1/√3)),
θ=30°.
Therefore AC=√3AB.
NOW tanθ =AB/AC,
tanθ =AB/√3AB,
tanθ=1/√3,
θ=tan^-1((1/√3)),
θ=30°.
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