The three paths intersect in pairs at the points A, B, and C. Measure and record the angle formed at each intersection. Also measure and record the lengths of the straight paths, or sides. Capture the image, and paste or insert it in the space below.

Respuesta :

Answer:

Just draw 3 non parallel lines. These paths will form a triangle with vertices A, B, C. So you know that the angles have to add up to 180.  

Step-by-step explanation: I HAD A DIFFICULT TIME DOING THIS ON PLATO I'LL COME BACK WITH AN UPDATE TO SEE IF ITS CORRECT. ( I HOPE I HELP FOR NOW)

The dimensions of length can be measured using a ruler, and the angle measurement can be made with the use of a protractor

The dimensions (calculated) of the figure are:

Angles formed at each intersection;

Angle A is approximately 97.125°

Angle B is approximately  53.6°

Angle C is approximately  26.6°

Lengths of the straight sides;

Side AB is approximately  3.6

Side BC is 8

Side AC is approximately  6.7

Please find attached the screen grab of the image

The reason the above values are correct is as follows:

Question: The coordinates of the points A, B, and C, obtained from a similar question online are; A(3, 4), B(1, 1), and C(9, 1)

Please find attached the diagram of the tree paths that intersect  at the points A, B, and C

The required parameter:

To measure and record the angle formed at each intersection

The angles can be measured using a protractor, however, by calculation, we have;

The length of AB = √((3 - 1)² + (4 - 1)²) = √13 ≈ 3.6

The length of BC = √((9 - 1)² + (1 - 1)²) = 8

The length of AC = √((9 - 3)² + (1 - 4)²) = √45 = 3·√5 ≈ 6.7

By cosine rule, we have;

[tex]cos(A) = \dfrac{8^2 - \left( \left(\sqrt{13} \right)^2 + 3 \cdot \left(\sqrt{5} \right)^2 \right)}{-2 \times \sqrt{13} \times 3 \cdot \left(\sqrt{5} \right) } = \dfrac{-\sqrt{65} }{65}[/tex]

[tex]A = arcos\left( \dfrac{-\sqrt{65} }{65} \right) \approx 97.125 ^{\circ}[/tex]

[tex]cos(B) = \dfrac{ \left(3 \cdot \left(\sqrt{5} \right)\right)^2 - \left( \left(\sqrt{13} \right)^2 +8^2 \right)}{-2 \times \sqrt{13} \times8 \right) }[/tex]

[tex]B = arcos\left(\dfrac{2 \times \left(-\sqrt{65} \right)}{65} \right) \approx 56.3[/tex]

[tex]cos(C) = \dfrac{ 13 - \left( \left(3 \cdot\sqrt{5} \right)^2 +8^2 \right)}{-2 \times 3 \cdot \sqrt{5} \times 8 \right) } = 2 \times\dfrac{\sqrt{5} }{5}[/tex]

[tex]C=arcos \left( 2 \times\dfrac{\sqrt{5} }{5}\right) \approx26.6^{\circ}[/tex]

Therefore, the angles formed at each intersection are;

Angle A ≈ 97.125°

Angle B ≈ 53.6°

Angle C ≈ 26.6°

The lengths of the straight side are;

Side AB ≈ 3.6

Side BC = 8

Side AC ≈ 6.7

Please find attached the image of the triangle ABC created with MS Excel

Learn more about finding the dimensions of triangle given the coordinates here:

https://brainly.com/question/13937524

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