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Euclid was the first to use a formula close to the modern law of cosines, citing two different formulas, one for acute triangles, one for obtuse triangles, in his Book 2 of Elements. However, his formulas are incomplete, as neither the negative numbers nor the cosine function had been thought of in Euclid's time. In the 10th century, a Persian mathematician and astronomer named al-Battani used and expnaded upon Euclid's formulas to calculate the distances between stars. Finally, in the 1400's, another Persian mathematician and astronomer named Jamshīd al-Kāshī again expanded on the law of cosines, bringing it to a very similar version of what it is today.
The following link is a Mathematica demonstration that proves the law of sines by showing that the ratio of the sine of the angle to the length of its opposite side is always equal for all three sides. To use the demonstration, move the points around with your to change the length of the side or the angle of the triangle. Notice that the ratio may change, but it is always the same as each of the other sines and opposite sides. This helps students learn the law of sines better by illustrating the effects of changing the side length or angle measures of a triangle.
http://demonstrations.wolfram.com/TheLawOfSines/
Here again is a Mathematica demonstration. It is similar to the law of sines demonstration, and it also proves the law of cosines by showing that no matter what the angle measures or side lengths are changed to, the resultant calculations always equal the expected value of a^2. The operation instructions of this demonstration are the same as for the law of sines demonstration. This helps students learn the law of cosines better because it shows the effects of changing the side lengths and angle measures of a triangle, and how even if those measures are changed, the formula will always hold true.
http://demonstrations.wolfram.com/TheLawOfCosines/
So, now that we have proven the law of sines and observed it algebraically, let's apply it to a real world situation. Suppose a tree grows out of the ground at an angle of 86 degrees to the ground pointing west. From the base of the tree, you walk 57 feet west, and from that point measure the angle of elevation to the top of the tree to be 44 degrees. What is the height of the tree?
Solution:
For this problem, we have two angle measures, and we know that the three angles in a triangle add up to 180 degrees, so the third angle can easily be found by subtracting 180-44-86=50 degrees.Now, we have the angle opposite x and the another angle measure with opposite side. Using those values, we can apply the law of sines, so 57/sin(50)=x/sin(44). Now, we solve for x, so x=(57*sin(44))/(sin(50)). Finally, with a bit of calculator work, we can determine that the tree is 51.688 feet tall.
Attached is a video of me explaining the law of sines to a student and then helping them solve this problem.
Here is a video of me teaching the law of sines to a student.
Finally, now that we have derived the law of cosines, let's use it in a real-life situation. Suppose there is a pond in your backyard and you want to know how wide it is at its widest point. However, since you can't walk across the water, you walk from the point on the eastern-most bank of the pond in a southwestern direction (NOT due southwest) 41 m, then turn 29 degrees towards the north. After that, you walk 18min a straight line to the point on the western-most bank of the pond. Find the width of the pond.
Solution:
In this problem, we are given two sides and the angle between them, and we need to find the third side. Therefore, if the unknown side is x, we can plug all of the information to the law of cosines. The equation would read x^2=18^2+41^2-41*18*cos(29). This simplifies to x^2=714.061, which again simplifies to x=26.721 m.
Here is a video of me teaching the law of cosines to a student
To help illustrate the law of sines and the law of cosines, I used two Mathematica demonstrations found on the Mathematica website. The law of sines and the law of cosines are two properties of trigonometry that are easily proven with the trigonometric properties of a right triangle, but in those proofs, only variables are used. I chose to use these demonstrations because all of the angles and all of the sides of the triangles are measured, and those numbers are then plugged into the formula that is associated with either the law of sines or the law of cosines. This is useful because it shows that no matter what triangle one makes, the formulas always hold true.
Learning can be enhanced by using these tools because they provide additional proof to the geometrical proofs already shown. However, instead of assigning variables to the proof, real numbers are used, and the concept of using the formulas for use in all triangles becomes more realistic. In addition, the triangles in the demonstrations are well-labeled, so if the student was confused about which value is which, this demonstration can provide some guidance.
My three learning objectives were that after going through my presentation, they were able to prove the law of sines and the law of cosines, identify triangles that can be solved using those laws, and use the laws to solve real-life situations. Alex Stratton, to whom I showed the proofs and gave problems to solve afterwards, showed that after listening to the information that is presented in my project, he was able to understand the proofs by using right triangle trigonometry. Also, when I gave him problems to solve using the law of sines and the law of cosines, Alex was able to apply the correct laws and after a bit of work, was able to use those laws to solve the problems, which illustrated real-life situation, like finding the height of a tree or the distance across a pond. Therefore, my learning objectives were met, as Alex was able to understand the proofs of the law of sines and the law of cosines, and he was able to apply those laws to triangles in order to solve real-life situations in which those formulas are needed.
In order to fully understand the law of sines and the law of cosines, one must first figure out where the formulas came from, know what exactly they mean, and be able to prove them. Therefore, the proofs included in this project were a very good first step to understanding how the law of sines and the law of cosines work. Additionally, technology such as the Mathematica demonstrations is a simple yet effective ways of showing the laws in action to help further the understanding of the laws. However, the best way to understand this concept is to simply learn the proofs and formulas, and do many problems, as new situations arise and students learn by experiencing these circumstances on their own, illustrated by the problems I created.
After reading this, students should be able to:
1. prove the law of sines and the law of cosines
2. identify triangles that can be solved using the law of sines and the law of cosines and solve them
3. apply the law of sines and the law of cosines to real-life situations
The law of sines can be derived by dropping an altitude from one corner to its opposite side. Call this altitude h1. Using right triangle trigonometry, find the values of h1 in terms of angle A and angle C. By doing this, we get the formulas h1=asin(C) and h1=csin(A). Then, since these formulas both equal h1, they can be set equal to each other. Using simple division, part of the law of sines can be derived as sin(A)/a=sin(C)/c. Using the same method but with a different altitude, the whole law of sines can be derived, and it states that sin(A)/a=sin(B)/b=sin(C)/c.
The law of cosines can also be proved with some simple algebra. Again, drop an altitude from angle C (you can use any angle, but angle C works well with this picture), and call it h. Also, on side c, label from A to the point where h intersects side c as x, and label the remainder of that side c-x. Now, using the Pythagorean theorem, derive the formula for both sides, and the formula for the cosine of angle A. You should get x^2+h^2=b^2, cosA=x/b (which we will change to x=bcos(A)), and (c-x)^2+h^2=a^2 (which we will simplify as a^2=c^2+(x^2+h^2)-2cx. Then, by substituting the first formula into (x^2+h^2) and the second formula in for x, we derive the formula a^2=c^2+b^2-2bc(cos(A)), which is one of the equations for the law of cosines. The other two (b^2=a^2+c^2-2ac(cos(B)) and c^2=a^2+b^2-2ab(cos(C))) can be derived using different altitudes at the beginning.
http://www-history.mcs.st-andrews.ac.uk/Biographies/Al-Jayyani.html