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yulyashka [42]
3 years ago
10

Find the measure of the exterior angle shown​

Mathematics
1 answer:
RUDIKE [14]3 years ago
5 0

Answer:

• exterior angle of a triangle is equal to sum of opposite interior angle

(6x-7) = (103-x) + 2x

6x-7 = 103 - x + 2x

6x-7 = 103 + x

6x - 7 - x -103 = 0

6x - x - 110 = 0

5x - 110 = 0

5x = 110

x = 110/5

x = 22

as exterior angle is (6x-7)

= 6 × 22 - 7

= 125

-> Exterior angle is 125°

pls do mark as brainliest. . . ☺

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Step-by-step explanation:

A diagram can help you understand the question, and can give you a clue as to how to find the answer. A diagram is attached. The problem can be described as finding the sum of two vectors whose magnitude and direction are known.

__

<h3>understanding the direction</h3>

In navigation problems, direction angles are specified a couple of different ways. A <em>bearing</em> is usually an angle in the range [0°, 360°), <em>measured clockwise from north</em>. In land surveying and some other applications, a bearing may be specified as an angle east or west of a north-south line. In this problem we are given the bearing of the second leg of the walk as ...

  N 35° E . . . . . . . 35° east of north

Occasionally, a non-standard bearing will be given in terms of an angle north or south of an east-west line. The same bearing could be specified as E 55° N, for example.

<h3>the two vectors</h3>

A vector is a mathematical object that has both magnitude and direction. It is sometimes expressed as an ordered pair: (magnitude; direction angle). It can also be expressed using some other notations;

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In the latter case, "cis" is an abbreviation for the sum cos(θ)+i·sin(θ), where θ is the direction angle.

Sometimes a semicolon is used in the polar coordinate ordered pair to distinguish the coordinates from (x, y) rectangular coordinates.

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The first leg of the walk is 3 meters due north. The angle from north is 0°, and the magnitude of the distance is 3 meters. We can express this vector in any of the ways described above. One convenient way is 3∠0°.

The second leg of the walk is 2.5 meters on a bearing 35° clockwise from north. This leg can be described by the vector 2.5∠35°.

<h3>vector sum</h3>

The final position is the sum of these two changes in position:

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Some calculators can compute this sum directly. The result from one such calculator is shown in the second attachment:

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This tells you the magnitude of the distance from the original position is about 5.25 meters. (This value is also shown in the first attachment.)

__

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  a² = b² +o² -2bo·cos(A) . . . . where A is the internal angle at A, 145°

Using the values we know, this becomes ...

  a² = 3² +2.5² -2(3)(2.5)cos(145°) ≈ 27.5373

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The distance from the original position is about 5.25 meters.

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The vector sum can also be calculated in terms of rectangular coordinates. Position A has rectangular coordinates (0, 3). The change in coordinates from A to B can be represented as 2.5(sin(35°), cos(35°)) ≈ (1.434, 2.048). Then the coordinates of B are ...

  (0, 3) +(1.434, 2.048) = (1.434, 5.048)

The distance can be found using the Pythagorean theorem:

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