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 4 of 7.
Shaik sajid said:
8 years ago
Hey answer is cot 30 degrees but we can also say tan 60 degrees.
Because in the options 60 also there.
Because in the options 60 also there.
Sajid said:
8 years ago
Why can't we say tan 60?
Sivaji said:
8 years ago
How to take a cotθ formula?
Aniket said:
8 years ago
We can also take tan instead of cot.
Amul said:
7 years ago
How can say that cot 30°? tan 60 is not compatitable at all I think sin60 or cos60 would be right choice.
Kelzhenry said:
7 years ago
AB/AC = Tanθ.
From the question,
AB/√3AB =Tanθ
1/√3 = Tanθ
θ = 30°.
From the question,
AB/√3AB =Tanθ
1/√3 = Tanθ
θ = 30°.
Biplab Gorai said:
7 years ago
Thanks for the given solution.
Akhila V U said:
7 years ago
Cot θ= AC/AB.
Cot θ =(3 * AB)/AB,
Cot θ= 3.
Therefore :
θ = 30.
Cot θ =(3 * AB)/AB,
Cot θ= 3.
Therefore :
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°.
Dattatray said:
7 years ago
Let x be the height of tree and shadow length √3 x,
Tanθ= x &div √3 x.
θ=tan-1(1 &div √3),
θ=30.
Tanθ= x &div √3 x.
θ=tan-1(1 &div √3),
θ=30.
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