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As part of summer camp, Henry goes on a treasure hunt. He starts at the base of a tree and walks 180 feet due north. He then turns and walks 80 feet due east. He turns again and walks 80 feet due south . How far is Henry from the treel Round your answer to the nearest foot.

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Answer:128.06 ft

Step-by-step explanation:

First Henry walks 180 ft North

Then 80 feet East and after that 80 ft south

So he is at a direction of North east from the starting Point

Vertically he is at a distance of 100 ft from starting point

and Horizontally at a distance of 80 ft from starting point

So, net distance from starting point is

[tex]d=\sqrt{100^2+80^2}[/tex]

[tex]d=128.06\ ft[/tex]

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      Henry is 128 feet far from the tree.

Given in the question,

  • Henry starts from summer camp (point O) and walks 180 feet due North.
  • He turns right and walks 80 feet due East (from A to B).
  • Then he turns again and walks 80 feet due South (from B to C).

From the figure attached,

AB = CD = 80 feet

AO = AD + OD

180 = 80 + OD

OD = 100 feet

To get the distance from the last location of Henry and base of the tree, apply Pythagoras Theorem in ΔODC.

OC² = OD² + CD²

OC² = (100)² + (80)²

OC = [tex]\sqrt{10000+6400}[/tex]

OC = 128.06 feet

OC ≈ 128 feet

     Therefore, Henry is 128 feet far from the tree.

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