BQ #1- Unit P: Law of Cosines and Area of an Oblique Triangle

3. Law of Cosines- We need Law of Cosines because it can solve for any angle and side of any triangle. As long as we have the right information (three sides or a side angle side), we can easily solve for any value as it is derived from the distance formula. Since we know that on a triangle, angle C is at (b,0) and angle B is at (cCosA, cSinA) since b is the distance from angle B to between angle A and angle C (Since you draw a perpendicular line down angle B) we can derive the (b,0). Also, since c is the distance from angle C to between angle A and angle B, we can also derive the latter point.

                      https://www.youtube.com/watch?v=lPP-pABvwdA

In the above video, we see the method to deriving the Law of Cosines. We see our desired formula of a^2=b^2+c^2-2bccosA. However, how do we derive that formula? First of all, we draw a triangle (as shown in the video, not a right triangle) and label the corners appropriately as A,B, and C. We then label their opposing sides as a,b, and c. We then draw a perpendicular line down angle C as demonstrated in the video to form two right triangles. We now have to cut the c side into two different values. Label the left triangle as "x", and since that is x, then the other side's value must be c-x because c is the total side value while x is one portion of it. Now that we know all the side values (as variables) we can use the Pythagorean Theorem. For the left triangle, we have x^2+h^2=b^2 (remember that h is the height we made from cutting the triangle in 2). For the right triangle we have (c-x)^2+h^2=a^. We have to foil this, so after we do that, we get c^2-2cx+x^2+h^2=a^2
            Where do we go from this?  Since x^2+h^2 is present in both the left and right triangles, and since the left triangle says that it equals b^2, then we can substitute that in for the right triangle formula to make c^2-2cx+b^2=a^2. We can move b^2 to the front and have b^2+c^2-2cx=a^2. As we reverse the equation to make it more like our original, desired formula, we have a^2=b^2+c^2-2cx. The x value in that formula does not correspond to what we want, so how do we get it? We get it by using our first triangle and taking the cos of angle A to get cosA=x/b. We solve for x and get x=bcosA. We substitute that in for our formula and get the desired formula, thus deriving the Law of Cosines.

4. Area of an Oblique Triangle- As we already know in an ABC labeled triangle, the sin of angle C is h/a because like the last problem, we draw a perpendicular line down the middle angle to make 2 right triangles. Since sinC=h/a, then that also means that h=asinC when we solve for h. Now when we substitute our common area equation for h, we get A=1/2b(asinC), since t'he regular area equation is A=1/2bh. This relates to the area equation because we use both base and height as it does except that we go a step beyond by using our laws of sin to derive our formula.


       https://www.youtube.com/watch?v=gMnBMJwpyc0

In the above video, we see how an oblique triangle is derived. This formula is very simple. It is A=1/2bcsinA. Of course, there are variations, so it can also be 1/2acsinB or 1/2 absinC. When you know the SAS value of the triangle, then you can easily plug in the corresponding values and then plug into your calculator the equation. For example, in the video, you merely plug in 1/2 x 12 x 8 x sin(135) to get your answer.

References:
http://www.mathsisfun.com/algebra/trig-cosine-law.html
https://www.youtube.com/watch?v=gMnBMJwpyc0
https://www.youtube.com/watch?v=lPP-pABvwdA