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Ghella [55]
3 years ago
5

The sand pit for the long jump has a width of 2.75 meters and a length of 9.54 meters. Just in case it rains, the principal want

s to cover the sand pit with a piece of plastic the night before the event. How many square meters of plastic will the principal need to cover the sand pit?
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
7 0

Answer:

The principal will need 26.24 m² of plastic to cover the sand pit.

Step-by-step explanation:

In order to calculate the amount of plastic, we need to find out the area of the sandpit. The shape of the sandpit is a rectangle, so we can use the formula:

A=W×H

where,

W is the width of the sandpit

L is the length of the sandpit

Therefore,

A=2.75 m × 9.54 m = 26.24 m²

The amount of plastic needed is 26.24 m²

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3 years ago
two points on the same side of a tree are 65 feet apart. the angles of elevation of the top the tree are 21 19 from one point an
zloy xaker [14]
The question is find the height of the tree, given that at two points 65 feet apart on the same side of the tree and in line with it, the angles of elevaton of the top of the tree are  21° 19' and 16°20'.

1) Convert the angles to decimal form:

19' * 1°/60' = 0.32° => 21° 19' = 21.32°

20' * 1°/60' = 0.33° => 16° 20' = 16.33°

2) Deduce the trigonometric ratios from the verbal information.

You can form a triangle with

- horizontal leg  x + 65 feet
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- vertical leg height of the tree, h

=> trigonometric ratio: tan (16.33) = h /( x + 65) => h = (x+65) * tan(16.33)

You can form a second triangle with:

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=> trigonometric ratio: tan(21.32) = h / x  => h = x * tan(21.32)

Now equal the two expressions for h:

(x+65)*tan(16.33) = x*tan(21.32)

=> x*tan(16.33) + 65*tan(16.33) = x*tan(21.32)
=> x*tan(21.32) - x*tan(16.33) = 65*tan(16.33)
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=> h = 195.73 * tan(21.32) = 76.39 feet.

Answer: 76.39 feet
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3 years ago
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