1. Law of Sines:

Why do we need it?Because most triangles are not right triangles, we must find the sides or angles of a triangle using a different method (Sorry Pythagorean Theorem). In the case that we are given Angle-Angle-Side (AAS) or Angle-Side-Angle (ASA), we can use the law of Sines.

Here is an image of a non-right triangle, labeled A B C for the angles and a b c for the sides.
 http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm
If we drop down a perpendicular line from angle C, we form 2 right triangles. Let's label the line h.

 http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm

Recall that the area of a triangle is Area=1/2bh. In order to find the area, we need to find h, the height. From past concepts with right triangles, we know that Sin A = h/b and that Sin B=h/a. To get h by itself, we will multiply the first equation by b and the second by a. The resulting equations would be bSinA=h and aSinB=h. Because both equations are equal to h, we can use the transitive property to have the equations equal each other: bSinA=aSinB. To simplify, we can divide everything by ab to get SinA/a=SinB/b.

 http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm

We can prove this again by drawing a perpendicular line from Angle A. We now have Sin B=h/c and
Sin C=h/b. We can rewrite both as cSinB=h and bSinC=h. Using the transitive property we have
cSinB = bSinC. By dividing both by bc, we end up with SinB/b and SinC/c. Since SinB/b is equal to SinA/a as formally proved, we can set all of these rations equal to each other. This gives us the law of Sines which is
.

4. Area Formulas:

How is the “area of an oblique” triangle derived?
Recall that the area of a triangle is A=1/2bh. If we need to find the area of a triangle but we don't know what the height is, we can do it using the area of an oblique triangle.
 http://www.compuhigh.com/demo/lesson07_files/oblique.gif

In the image above, we know that SinC is equal to h/a. If we multiply both sides by a we get aSinC=h. If we substitute this value for h into the area of a triangle formula, we get : A=1/2b(aSinC). We can find the area given different information as well. If given side b and c, we need to know angle A, if given side a and c, we need to know Angle B. Similarly to how we solved for the area of the triangle using angle C earlier, we can do the same for angle A and B. Ultimately, we end up with the formula :
 http://wps.prenhall.com/wps/media/objects/551/564474/StudyGuide/7c1h7_1.gif

How does this relate to the area formula you are familiar with? This relates to the area formula because it is used in deriving the are of an oblique triangle. We basically used sine to find the value of the height and substituted the formula into the equation. Ultimately, the area formula is still being used, it just looks different .

References:

 http://wps.prenhall.com/wps/media/objects/551/564474/StudyGuide/7c1h7_1.gif
http://www.compuhigh.com/demo/lesson07_files/oblique.gif
 http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm