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
