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Firdavs [7]
3 years ago
14

An airplane pilot can see the top of a traffic control tower at a 20 degree angle of depression. the airplane is 5,000 feet away

from the tower horizontally. how far above the tower is the airplane? round to the nearest hundredth.
Mathematics
2 answers:
daser333 [38]3 years ago
8 0

The given question describes a right triangle with with one of the angles as 20 degrees and the side adjacent to the angle 20 degrees is of length 5,000 feet. We are looking for the length of the side opposite the angle 20 degrees.

Let the required length be x, then

\tan{20^o}=\cfrac{opp}{hyp}=\cfrac{x}{5,000}\\ \\ \Rightarrow x=5,000\tan{20^o}=1,819.85

Therefore, the height of the airplane above the tower is 1,819.85 feet.

natali 33 [55]3 years ago
7 0

Answer:

y \approx 1819.851\,ft

Step-by-step explanation:

The vertical distance between the airplane and the traffic control tower is obtained by means of the following trigonometric relation:

y = x \cdot \tan \theta

y = (5000\,ft)\cdot \tan 20^{\textdegree}

y \approx 1819.851\,ft

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In triangle PQR, m∠P = 83°, PQ = 7.6, and PR = 8.6. What is m∠R to the nearest degree? A. 45° B. 55° C. 35° D. 41°
Mila [183]

Answer:

The unknown measurement of angle R to the nearest degree is 41°

Step-by-step explanation:

As you can see, we made a diagram form the given information in the problem. We are trying to find the measurement of angle R which is our unknown angle represented by the question mark.

in order to solve this problem, we are going to be using tangent. Tangent uses the opposite side and the adjacent side from the given or from the unknown angle measurement.

So, our equation will look like this.

tanx=\frac{7.6}{8.6}

Since, we do not know the measurement of the unknown angle, then are going to use the inverse of tangent.

x=\frac{7.6}{8.6}(tan^-^1)

Now, we solve. You can use a calculator to do these calculations.

The unknown measurement of the unknown angle to the nearest degree is 41° which is answer choice D.

3 0
3 years ago
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Amanda has an envelope that is 2 1/4 inches tall. The envelope is 4 1/3 times as long as it is tall. How long is her envelope?
Sidana [21]

Answer:

its 2 1/2 its width

Step-by-step explanation:

4 0
2 years ago
I don't get it and I have tried it multiple ways
jasenka [17]

9514 1404 393

Answer:

  9350 m²

Step-by-step explanation:

There are several ways you can go at this. Here are 3 of them:

  1. along an extension of the lower right vertical edge, divide the figure into a rectangle and a right triangle.
  2. between the upper left corner and the inside corner at lower right draw a diagonal line to divide the figure into a trapezoid and a triangle.
  3. subtract the area of the trapezoid at the right from the area of the bounding rectangle

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The relevant formulas for area are ...

  rectangle: A = LW . . . . . . . . . . . length times width

  trapezoid: A = (1/2)(b1 +b2)h . . . b1, b2 are the bases, h is the height

  triangle: A = (1/2)bh . . . . . . . . . . . b is the base, h is the height

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Here's what those computations look like:

1. rectangle: 60m wide by 80 m high = 4800 m².

  right triangle: base 190-60 = 130 m; height = 80-10 = 70 m.

     area = (1/2)(130 m)(70 m) = 4550 m²

  figure area = 4800 m² +4550 m² = 9350 m² . . . . area of figure

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2. The trapezoid has bases of 80 m and 10 m, and a height of 60 m. Its area is ...

  A = (1/2)(80 m +10 m)(60 m) = 2700 m²

The triangle has a base of 190 m and a height of 80-10 = 70 m. Its area is ...

  A = (1/2)(190 m)(70 m) = 6650 m²

Then the total area is 2700 m² +6650 m² = 9350 m² . . . total area

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3. The bounding rectangle is 190 m by 80 m, so its area is ...

  A = LW = (190 m)(80 m) = 15200 m²

The (negative) trapezoid at right has bases of 10 m and 80 m, and a width of 190-60 = 130 m. Its area is ...

  A = (1/2)(10 m +80 m)(130 m) = 5850 m²

The area of the figure is the difference between these:

  figure area = 15200 m² -5850 m² = 9350 m² . . . figure area

7 0
3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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adelina 88 [10]

OPTIONS:

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C. The upper quartile of the trail mix data is equal to the maximum value of the cracker data.

D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

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Step-by-step explanation:

With the given information about how the box plot looks like, let's examine each option to see if they are true or not.

<u><em>Option A: "The interquartile range of the trail mix data is greater than the range of the cracker data."</em></u>

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Option A is NOT TRUE.

<u><em>Option B: "The value 70 is an outlier in the trail mix data.</em></u>"

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<em><u>Option C: "The upper quartile of the trail mix data is equal to the maximum value of the cracker data."</u></em>

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Max value of cracker data = 100

This statement is NOT TRUE.

<em><u>Option D: "The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers."</u></em>

The greater the range value, the greater the variation. Thus,

Range value of the trail mix data = 115 - 70 = 45

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This is statement is correct because trail mix data have a greater range value, hence, it has a greater variation in the number of calories.

6 0
3 years ago
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