Part 1
Sine is one of the ratio of the sides of a right triangle. The sine is opposite over hypotenuse.
Cosine is another one of the ratios of the sides of a right triangle. This cosine is adjacent over hypotenuse.
Tangent is another one of the ratios of the sides of a right triangle. The tangent is opposite over adjacent.
ArcSine is the inverse function for Sine. sin−1(x)
ArcCosine is the inverse function for Cosine. cos−1(x)
ArcTangent is inverse function for Tangent. tan−1(x)
Law of Sines is the relationship between the sides and angles of oblique triangles. The ratio of the length of the side of a triangle to the sine of the angle opposite of that side is the same for all sides and angles of a triangle.
sin (A) / a = sin (B) / b = sin (C) / c
Law of Cosines is for calculating one side of a triangle when the angle opposite and other sides are known. It relates the three sides of a triangle to an angle.
You can prove the Pythagorean Theorem by getting a right triangle and making a square with each side length. If you get the area of the two smallest squares, and add it together, you get the area of the big triangle. So that means that a²+b²=c². This can be used to find an area of a triangle.
You can use the Pythagorean Theorem to get the Distance Formula by first placing the right triangle on a coordinate plane. You fill in the equation with the coordinates then take the square root of it. That is the distance formula.
To Use the Distance Formula to find the equation of a circle in a coordinate plan, you need to first use the Distance Formula to find the equation of a circle. Then you substitute it with the coordinates of the center of the circle and the point on the circle. Square each side. That is the equation of a Circle.
A unit circle is a circle that is on the center (0,0) and has a radius of 1.
To find points on a unit circle, first substitute the known coordinate in the unit circle equation. Square that coordinate and subtract the value from each side. Take the square root of each side and that will give you the missing coordinate.
To use symmetry to find the remaining point on a unit circle, you can flip the line and connect the coordinates to make a triangle. You can use the previous equation to find the coordinates.
To use a unit circle to define sine and cosine you first get the coordinate on the circle and draw it perpendicular to the x axis. Then that makes a triangle and you can label sine and cosine. This helps you figure out where the opposite and adjacent are and angle theta.
The tangent of an angle is the length of an opposite side divided by the length of the adjacent side.
You can use similarity and proportions to find trigonometric functions by understanding that the longest side of a right triangle is the hypotenuse, the side on the other side of that is adjacent and the remaining one is opposite.
sinθ=opposite/hypotenuse
cosθ=adjacent/hypotenuse
tan=opposite/adjacent
Arc Sine, Arc Cosine, and Arc Tangent on a unit circle with its positive counterpart would look like a reflection or opposite of the coordinate.
When we did the Mount Everest problem, we first labeled the sides and used the Law of Sines to figure out what the missing sides and heights of smaller triangles were.
The Law of Sines is a/Sin A=b/Sin B=c/Sin C. A, B, C, are angles and a, b, c, are sides of the triangle.
The Law of Cosines is
a2=b2+c2−2bccosAa2=b2+c2−2bccosA, b2=a2+c2−2accosBb2=a2+c2−2accosB, c2=a2+b2−2abcosCc2=a2+b2−2abcosC
Cosine is another one of the ratios of the sides of a right triangle. This cosine is adjacent over hypotenuse.
Tangent is another one of the ratios of the sides of a right triangle. The tangent is opposite over adjacent.
ArcSine is the inverse function for Sine. sin−1(x)
ArcCosine is the inverse function for Cosine. cos−1(x)
ArcTangent is inverse function for Tangent. tan−1(x)
Law of Sines is the relationship between the sides and angles of oblique triangles. The ratio of the length of the side of a triangle to the sine of the angle opposite of that side is the same for all sides and angles of a triangle.
sin (A) / a = sin (B) / b = sin (C) / c
Law of Cosines is for calculating one side of a triangle when the angle opposite and other sides are known. It relates the three sides of a triangle to an angle.
You can prove the Pythagorean Theorem by getting a right triangle and making a square with each side length. If you get the area of the two smallest squares, and add it together, you get the area of the big triangle. So that means that a²+b²=c². This can be used to find an area of a triangle.
You can use the Pythagorean Theorem to get the Distance Formula by first placing the right triangle on a coordinate plane. You fill in the equation with the coordinates then take the square root of it. That is the distance formula.
To Use the Distance Formula to find the equation of a circle in a coordinate plan, you need to first use the Distance Formula to find the equation of a circle. Then you substitute it with the coordinates of the center of the circle and the point on the circle. Square each side. That is the equation of a Circle.
A unit circle is a circle that is on the center (0,0) and has a radius of 1.
To find points on a unit circle, first substitute the known coordinate in the unit circle equation. Square that coordinate and subtract the value from each side. Take the square root of each side and that will give you the missing coordinate.
To use symmetry to find the remaining point on a unit circle, you can flip the line and connect the coordinates to make a triangle. You can use the previous equation to find the coordinates.
To use a unit circle to define sine and cosine you first get the coordinate on the circle and draw it perpendicular to the x axis. Then that makes a triangle and you can label sine and cosine. This helps you figure out where the opposite and adjacent are and angle theta.
The tangent of an angle is the length of an opposite side divided by the length of the adjacent side.
You can use similarity and proportions to find trigonometric functions by understanding that the longest side of a right triangle is the hypotenuse, the side on the other side of that is adjacent and the remaining one is opposite.
sinθ=opposite/hypotenuse
cosθ=adjacent/hypotenuse
tan=opposite/adjacent
Arc Sine, Arc Cosine, and Arc Tangent on a unit circle with its positive counterpart would look like a reflection or opposite of the coordinate.
When we did the Mount Everest problem, we first labeled the sides and used the Law of Sines to figure out what the missing sides and heights of smaller triangles were.
The Law of Sines is a/Sin A=b/Sin B=c/Sin C. A, B, C, are angles and a, b, c, are sides of the triangle.
The Law of Cosines is
a2=b2+c2−2bccosAa2=b2+c2−2bccosA, b2=a2+c2−2accosBb2=a2+c2−2accosB, c2=a2+b2−2abcosCc2=a2+b2−2abcosC
Part 2
Trigonometry
For the Trigonometry portion I chose to measure my math notebook because I knew I could measure the triangles that can be found in it, if I cut it into two. I used trigonometry to find the hypotenuse of the triangles and their angles.
First, I started looking for the hypotenuse and used the Pythagorean Theorem, a²+b²=c²
The length of the sides are 10.5 ianches, 8 inches, so i put it in the equation which made 18.5
To find angle a, I divided 10.5 by 18.5 to get 0.5675
I wrote got the sin of it, sin⁻¹(0.5675)=34.58 °.
For angle b, I did the same, 8 divided by 18.5 which is 0.4324.
Then, I did cos⁻¹(0.4324)=64.37°
For the Trigonometry portion I chose to measure my math notebook because I knew I could measure the triangles that can be found in it, if I cut it into two. I used trigonometry to find the hypotenuse of the triangles and their angles.
First, I started looking for the hypotenuse and used the Pythagorean Theorem, a²+b²=c²
The length of the sides are 10.5 ianches, 8 inches, so i put it in the equation which made 18.5
To find angle a, I divided 10.5 by 18.5 to get 0.5675
I wrote got the sin of it, sin⁻¹(0.5675)=34.58 °.
For angle b, I did the same, 8 divided by 18.5 which is 0.4324.
Then, I did cos⁻¹(0.4324)=64.37°
Area
For the Area portion, I chose to measure my math notebook because since I realized I could find the area in two different ways to prove my answer is correct using the information from the previous slides. I used the area of a triangle and area of a rectangle to find the area of my notebook.
First I looked at the cut triangle I made before and drew it out.
I labeled all of the sides.
I used the equation for finding the area of a triangle which is a=b*h*½
I replaced it with the side lengths which makes it a=8*10.5*½
The answer is 42.
I multiply 42 by 2 because there are two triangles and get 84.
Another way to get the answer is using the equation to find the area of a rectangle, l*w=a
I replaced the variables with the side lengths, 10.5*8=84
For the Area portion, I chose to measure my math notebook because since I realized I could find the area in two different ways to prove my answer is correct using the information from the previous slides. I used the area of a triangle and area of a rectangle to find the area of my notebook.
First I looked at the cut triangle I made before and drew it out.
I labeled all of the sides.
I used the equation for finding the area of a triangle which is a=b*h*½
I replaced it with the side lengths which makes it a=8*10.5*½
The answer is 42.
I multiply 42 by 2 because there are two triangles and get 84.
Another way to get the answer is using the equation to find the area of a rectangle, l*w=a
I replaced the variables with the side lengths, 10.5*8=84
Volume
For the Volume portion, I chose to measure a chocolate box that is shaped as hexagonal prism because it isn’t a shape you see normally during everyday life. I used the equation to find the volume of a hexagonal prism to find the volume of my chocolate box.
First I drew my object and labeled the length of the base and height.
Then I wrote out the equation to find the volume of a hexagonal prism, which is v=3√3/2*a²*h.
When I replaced the variables with my side lengths, which is 3√3/2*a²h=32.15
For the Volume portion, I chose to measure a chocolate box that is shaped as hexagonal prism because it isn’t a shape you see normally during everyday life. I used the equation to find the volume of a hexagonal prism to find the volume of my chocolate box.
First I drew my object and labeled the length of the base and height.
Then I wrote out the equation to find the volume of a hexagonal prism, which is v=3√3/2*a²*h.
When I replaced the variables with my side lengths, which is 3√3/2*a²h=32.15
Reflection
I used the habit Stay Organized when writing out my diagrams neatly.
I used the habit Be Confident and Persistent for when I made mistakes with my calculations.
If I were to do this project again I would find more difficult objects to measure and do the project.
I used the habit Stay Organized when writing out my diagrams neatly.
I used the habit Be Confident and Persistent for when I made mistakes with my calculations.
If I were to do this project again I would find more difficult objects to measure and do the project.