Pre-calulus help: Finding height of tree

[tangent]

Hi guys, I have a problem that is confusing me in the way it is worded and hope you guys can give me some clarification on it.

A tree being cut down makes a 70 degree angle with the ground when the tip of the tree is directly above a spot that is 40 feet from the base of the tree. Find the height of the tree.



Here's a sketch of it. Is the 70 degrees referring to angle B or angle A? Is the 40 feet referring to the base (from B to C) or the hypotenuse (from A to B)? Thanks for your help.

 

[Deleted member 4993]

Hi guys, I have a problem that is confusing me in the way it is worded and hope you guys can give me some clarification on it.

A tree being cut down makes a 70 degree angle with the ground when the tip of the tree is directly above a spot that is 40 feet from the base of the tree. Find the height of the tree.

View attachment 3200

Here's a sketch of it. Is the 70 degrees referring to angle B or angle A? Is the 40 feet referring to the base (from B to C) or the hypotenuse (from A to B)? Thanks for your help.

The angle is made with the ground - hence it is angle ABC - aka angle B.

I interpret that to be → BC = 40'

 

[soroban]

Hello, tangent!

Your sketch is completely wrong . . .

A tree being cut down makes a 70 degree angle with the ground
when the tip of the tree is directly above a spot that is 40 feet from the base of the tree.
Find the height of the tree. .??

    A *
      :*
      : *
      :  *
    y :   * x
      :    *
      :     *
      :      *
      :   70[SUP]o[/SUP] *
    C *--------* B
          40

The tree is \(\displaystyle AB;\) its length is \(\displaystyle x = AB.\)

The top of the tree is \(\displaystyle A.\)
It is 40 feet above \(\displaystyle C;\;y = AC.\)
And \(\displaystyle BC = 40\) feet.


What do they mean by "height"?


Is it the length of the tree \(\displaystyle (x)\,?\)

Then: .\(\displaystyle \cos70^o \,=\,\dfrac{40}{x} \quad\Rightarrow\quad x \,=\,\dfrac{40}{\cos70^o}\)


Is it the distance from the top to the ground \(\displaystyle (y)\,?\)

Then: .\(\displaystyle \tan70^o \,=\,\dfrac{y}{40} \quad\Rightarrow\quad y \,=\,40\tan70^o\)

 

[OldMathVermont]

I'm looking at the same problem.

I'm thinking its the same problem anyway. I want to run a pole up top of the tree and beyond by 10 feet. How do I find the top of tallest tree? I believe its basically the same question. I don't understand the previous gentlemen's answer. I'm an older man and in High School for our Algebra class we measured the water tower by its shadow. Thanks