This unit started off with the review and proof of the Pythagorean Theorem. The first proof we did was called the “Proof by Rugs”. In this worksheet we had to explain why two rugs that were shaded differently had the same total area shaded and thus made the game of throwing a dart into the shaded area, fair. When we started, right away I noticed that each diagram had 4 triangles that my group and I figured had to be congruent because they each had a side length, A, and a side length, B. Since the 4 triangles and big squares or “rugs” were congruent, I concluded that the shaded areas were the same despite their different designs. After we used this worksheet to review and prove the Pythagorean Theorem, we then used our new understanding of the theorem to derive the Distance Formula.
To find the distance between two points, you’d have to create a triangle using the point where the two points meet up. Then you substitute in for the proper variables into the equation D = (x₂-x₁)²+(y₂-y₁)² and then you solve it to get your distance. Before going in depth about the Distance formula, I instinctively knew that if you used Pythagorean’s Theorem, a² + b² = c², on the triangle created by the two points then you should be able to get the hypotenuse created by the distance of the points but I never understood exactly what was going on. Using the Distance Formula allowed me to see that subtracting x₂ from the x₁ gave you the length across minus the distance from the Y-axis and subtracting y₂ from y₁ gave you the height minus the distance from the X axis. Then when you finished solving the equation you got the distance between the two points. Similarly, we used the Pythagorean Theorem to solve for any point on a circle with a radius of 1. Since the radius is 1 no matter where you measure across, whatever point you decide to solve for must equal to 1. Anything less than one meant it was inside the circle, anything equal to one meant it was on the circle and satisfied the equation, while if it was more than one that meant it was outside the circle. This special circle is called the unit circle, meaning it’s center is located at the origin and it has a radius of 1.
Keeping this circle in mind, solving for distances when given certain angles and a point on the circle was easy once we figured out the pattern. I first solved for the side lengths when the angle was 45°. Since the triangle had two congruent angles and sides it meant that x₁ actually equaled y₁ and when you put that into the Pythagorean Theorem and solve it you get both side lengths being 22. Doing the same thing with a 30° angle, I realized that if you reflect the triangle over the X-axis, you end up making an equilateral triangle which means all sides and angles are equal. Knowing that the side length of the entire equilateral triangle is one because the radius is one, when thinking about our smaller triangle which is half of the equilateral triangle, you could assume that the y₁ length is 12. If you substitute that into the equation then you get 32 for the x₁ length. Solving for a 60° angle was similar to solving for the 30° angle because the interior and exterior angles were just switched so the sides were reversed and put on the other side. After we figured out these angles we used symmetry across the axes to figure out what they’d be in different quadrants. Each coordinate was the same except it had a negative sign in front of it depending on which quadrant it was in.
At this point we used the unit circle to integrate cosine and sine. Cosine being the relationship between the angle ϴ and the corresponding x-coordinate and sine being the relationship between the angle ϴ and the corresponding y-coordinate. An important thing I learned during this section was SOHCAHTOA. It was an easy trick to remember that sine equals the opposite side divided by the hypotenuse, cosine equals the adjacent side divided by the hypotenuse, and tangent equals opposite divided by adjacent. Tangent, another trigonometry term, is when a line is drawn perpendicular to the radial line of a unit circle. This opens up to the tangent function which is when you use the tangent line to make a proportional relationship between the bigger triangle that’s created with said tangent to the original triangle. If you set up a proportion that sets the proportional radial line to the original radial line and one that sets the height to the length, because the two triangles are similar, you could simplify it to eventually get tanϴ = sinϴ/cosϴ. All of these trigonometry functions have their own opposite function which return them to the angles they were solved for. These are called the arc-functions; arcSine, arcCosine, and arcTangent.
If we didn’t know one of the side lengths, we would use the Law of Cosines which states c²=a² + b²-2ab(cosϴ). As long as you know two sides and the opposite angle (SAS), you could solve for the missing side length. The Law of Sines states that Asin a = Bsin b= Csin c. Used in problems like the Mount Everest problem, the Law of Sines helps you find missing sides as long as you know two angles and one side (AAS)
To find the distance between two points, you’d have to create a triangle using the point where the two points meet up. Then you substitute in for the proper variables into the equation D = (x₂-x₁)²+(y₂-y₁)² and then you solve it to get your distance. Before going in depth about the Distance formula, I instinctively knew that if you used Pythagorean’s Theorem, a² + b² = c², on the triangle created by the two points then you should be able to get the hypotenuse created by the distance of the points but I never understood exactly what was going on. Using the Distance Formula allowed me to see that subtracting x₂ from the x₁ gave you the length across minus the distance from the Y-axis and subtracting y₂ from y₁ gave you the height minus the distance from the X axis. Then when you finished solving the equation you got the distance between the two points. Similarly, we used the Pythagorean Theorem to solve for any point on a circle with a radius of 1. Since the radius is 1 no matter where you measure across, whatever point you decide to solve for must equal to 1. Anything less than one meant it was inside the circle, anything equal to one meant it was on the circle and satisfied the equation, while if it was more than one that meant it was outside the circle. This special circle is called the unit circle, meaning it’s center is located at the origin and it has a radius of 1.
Keeping this circle in mind, solving for distances when given certain angles and a point on the circle was easy once we figured out the pattern. I first solved for the side lengths when the angle was 45°. Since the triangle had two congruent angles and sides it meant that x₁ actually equaled y₁ and when you put that into the Pythagorean Theorem and solve it you get both side lengths being 22. Doing the same thing with a 30° angle, I realized that if you reflect the triangle over the X-axis, you end up making an equilateral triangle which means all sides and angles are equal. Knowing that the side length of the entire equilateral triangle is one because the radius is one, when thinking about our smaller triangle which is half of the equilateral triangle, you could assume that the y₁ length is 12. If you substitute that into the equation then you get 32 for the x₁ length. Solving for a 60° angle was similar to solving for the 30° angle because the interior and exterior angles were just switched so the sides were reversed and put on the other side. After we figured out these angles we used symmetry across the axes to figure out what they’d be in different quadrants. Each coordinate was the same except it had a negative sign in front of it depending on which quadrant it was in.
At this point we used the unit circle to integrate cosine and sine. Cosine being the relationship between the angle ϴ and the corresponding x-coordinate and sine being the relationship between the angle ϴ and the corresponding y-coordinate. An important thing I learned during this section was SOHCAHTOA. It was an easy trick to remember that sine equals the opposite side divided by the hypotenuse, cosine equals the adjacent side divided by the hypotenuse, and tangent equals opposite divided by adjacent. Tangent, another trigonometry term, is when a line is drawn perpendicular to the radial line of a unit circle. This opens up to the tangent function which is when you use the tangent line to make a proportional relationship between the bigger triangle that’s created with said tangent to the original triangle. If you set up a proportion that sets the proportional radial line to the original radial line and one that sets the height to the length, because the two triangles are similar, you could simplify it to eventually get tanϴ = sinϴ/cosϴ. All of these trigonometry functions have their own opposite function which return them to the angles they were solved for. These are called the arc-functions; arcSine, arcCosine, and arcTangent.
If we didn’t know one of the side lengths, we would use the Law of Cosines which states c²=a² + b²-2ab(cosϴ). As long as you know two sides and the opposite angle (SAS), you could solve for the missing side length. The Law of Sines states that Asin a = Bsin b= Csin c. Used in problems like the Mount Everest problem, the Law of Sines helps you find missing sides as long as you know two angles and one side (AAS)