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An h = 135-ft tower is located on the side of a mountain that is inclined 32° to the horizontal. A guy wire is to be attached to the top of the tower and anchored at a point 55 ft downhill from the base of the tower. Find the shortest length of wire needed. (Round your answer to the nearest foot.)

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lublana

Answer:

171 foot

Step-by-step explanation:

We are given that

Height of tower=135 ft

AC=55 ft

Let x be the length of shortest length of wire.

The angle of inclination of mountain with horizontal =[tex]32^{\circ}[/tex]

The angle between sides AB and BC=[tex]90+32=122^{\circ}[/tex]

Exterior angle is equal to sum interior angles of a triangle on the opposite sides .

We have to find the value of x.

We know that law of cosine

[tex]c=\sqrt{a^2+b^2-2ab cos\theta}[/tex]

Substitute the values in the given formula we get

[tex]x=\sqrt{(135)^2+(55)^2-2(135)(55)cos 122}[/tex]

[tex]x=170.64 ft[/tex]

Hence, the length of wire needed=171 foot

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akposevictor

By applying the Law of Cosines, the shortest distance of the needed wire, to the nearest foot, is: 171 ft.

Recall:

  • Law of Cosines is given as: [tex]\mathbf{e^2 = d^2 + f^2 - 2(d)(f)(CosE)}[/tex]

The given scenario has been sketched in the diagram showing the tower that is located on the side of a mountain alongside other information that are provided.

The shortest distance is represented as x.

To find x, we would apply the law of Cosines,  [tex]\mathbf{e^2 = d^2 + f^2 - 2(d)(f)(CosE)}[/tex].

  • Where:

[tex]E = 90 + 32 = 122^{\circ}[/tex]

d = 55 ft

f = 135 ft

e = x

  • Substitute

[tex]x^2 = 55^2 + 135^2 - 2(55)(135)(Cos122)\\\\x^2 = 21,250 - (-7,869.3)\\\\x^2 = 21,250 + 7,869.3\\\\x^2 = 29,119.3\\\\[/tex]

  • Square both sides

[tex]\sqrt{x^2} = \sqrt{29,119.3} \\\\\mathbf{x = 171 $ ft}[/tex]

Therefore, by applying the Law of Cosines, the shortest distance of the needed wire, to the nearest foot, is: 171 ft.

Learn more here:

https://brainly.com/question/11399406

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