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Triss [41]
3 years ago
8

In which quadrant does the terminal side of the angle 387° lie?

Mathematics
2 answers:
gulaghasi [49]3 years ago
7 0

Answer:

Quadrant I

Step-by-step explanation:

To find in which quadrant an angle is, you have to locate it between 0 and 360.

Since your angle is larger than 360 degrees, we subtract 360 degrees from it.... to get 27 degrees.

27 degrees and 387 degrees lie on the same spot, since they are exactly one turn around from each other.

Now, angles between 0 degrees and 90 degrees are in quadrant I

angles between 90 and 180 degrees are in quadrant II

Angles between 180 and 270 degrees are in quadrant III

Angles between 270 and 360 degrees are in quadrant IV

So, an angle of 27 degrees would land in Quadrant I.... so is 27 + 360x, where x is any integer representing a number of turns.

Shalnov [3]3 years ago
4 0
<h2>Hello!</h2>

The answer is:

The terminal side of the angle 387° lies at the Quadrant I.

<h2>Why?</h2>

We know that there are four quadrants on the plane, those quadrants goes from 0° to 360°, however, since the quadrant is going only from 0° to 360°, if we want to know where an angle greater than 360° is located, we need to take only the excess, so:

We are asked to find where the angle 387° is located, we can write the angle by the following way:

387\°=360\°+27\°\\

So, we have that there are 27° more than 360°, if we want to find where the 367° we only need to locate where the excess (27°) angle is located.

Also, we know that:

Quadrant I: 0° to 90°

Quadrant II: 90° to 180°

Quadrant III: 180° to 270°

Quadrant IV: 270° to 360°

We have that 27° is between 0° and 90°, so, it's located on the first quadrant.

Hence, we have that the angle 367° is located at the Quadrant I.

Have a nice day!

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If you look at the second graph and were to add the percentages, they add far higher than 100%. Why is that the case?
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The length of a rectangle is 3 more than 3 times its width. The perimeter of the rectangle is 174 inches. What is the length of
saveliy_v [14]

Answer:

<em>l = w + 3cm</em>

<em>l = w + 3cmp = 2l + 2w = 58cm</em>

<em>l = w + 3cmp = 2l + 2w = 58cm </em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13 </em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13 Plug back in:</em>

<em>l = w + 3cmp = 2l + 2w = 58cm Solve by substitution:2l + 2w = 58 ⇒ 2(w + 3) + 2w = 58⇒ 2w + 6 + 2w = 4w + 6 = 58⇒ 4w = 52 ⇒ w = 13 Plug back in:l = (13cm) + 3cm = 16cm</em>

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I hope this helps you.

5 0
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