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General Sherman, a tree located in Sequoia National Park, stands 275 feet tall. To see the top of the tree,
Carlos looks up at a 19° angle of elevation. If Carlos is 6 feet tall, how far is he from the base of the tree to the
nearest foot?

Sagot :

facundo3141592

By using trigonometric relations, we will see that the distance between Carlos and the tree is:

D = 781.23 ft

How far is Carlos for the base of the tree?

We can model this with a right triangle. We know that Carlos is 6ft tall, and the height of the tree is 275ft.

Then one of the cathetus of the triangle is equal to:

275ft - 6ft = 269ft.

That cathetus is the opposite cathetus to the angle of 19° (the elevation angle). And the adjacent cathetus is the distance between Carlos and the tree.

Then we can use the relation:

tan(a) = (opposite cathetus)/(adjacent cathetus).

Where:

  • a = 19°
  • opposite cathetus = 269ft
  • adjacent cathetus = D.

Replacing that, we get:

tan(19°) = 269ft/D

Solving for D, we get:

D = 269ft/tan(19°) = 781.2ft

So we conclude that Carlos is at 781.23 ft of the tree.

If you want to learn more about right triangles:

https://brainly.com/question/2217700

#SPJ1

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