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

i. Law of Sines - We need the Law of Sines in order to find missing angle(s) and missing side(s) of a triangle that is not a right triangle. We can use trig functions to derive the Law of Sines.

Law of Sines: In the triangle ABC, SinA/a = SinB/b = SinC/c

When using the law of sines, we only use two out of the three.

You have a non-right triangle and draw a perpendicular line straight down the triangle from the angle B. That will be labeled "h" for height. Because we drew that perpendicular line, we now have two right triangles. We can now apply our trig skills.

To find the Law of Sines from the two triangles, we must use Sin. So using angle A, we have SinA = h/c (opposite over hypotenuse) and we have SinC = h/a (also opposite over hypotenuse). We then multiply each by the denominator to get rid of the ratio. We would then have cSinA=h and aSinC=h. Now, because both equal "h", that would also mean that they equal each other. So we set cSinA = aSinCWe and would then divide each by ac and that would cross out the a in aSinC and the c in cSinA. We would then be left with SinA/a = SinC/c.

iv. Area of an Oblique Triangle - We use this when we have SAS.

The area formula for a right triangle is A=1/2bh. The area of an oblique triangle is derived through also sin. For example, in this picture we will use angle C in the triangle that has a perpendicular line draw through it.

We have to find the "h" height. We know that sinC = h/a and that aSinC = h. So since we do not know what "h" is, we can substitute "h" for aSinC. Therefore, we would then get A=1/2baSinC. This relates to the original formula we are familiar with because it uses the same format of 1/2bh, but instead of being stuck with not know two variables (A and h), we can use our knowledge of the Law of Sines to see that h would equal aSinC and that we substitute aSinC (and the answer we have for aSinc) for "h".