Sunday, March 16, 2014

BQ #1: Unit P: Concept 2: Law of Sines SSA and Concept 4: Area of An Oblique Triangle

2. Law of Sines


In an SSA triangle, our three angles are not carved in stone as AAS or ASA triangles where even though we had two angles, in reality we had all three because all one had to do was add the two and subtract it from 180 to get the third angle's values, meaning there was no ambiguity there.
With SSA we only know one angle's value! (and two sides) There is a possibility however that with only this information that there may be no triangle at all! We always start the problem assuming there is two possible triangles, until you "hit a wall" which will lead to only one possible triangle or no possible triangle. A "wall" can either be if SinA=1.1 or 2.8, something that is not possible. Remember from our unit circle trig functions, we know that sin and cos are not compatible and do not work with numbers greater than one, (-1 less than or equal to sin theta less than or equal to 1), or if the angles add up to be something that is greater than 180.
In this triangle, we are solving for angle B and C, and side b. We have information for both angle A and side a, our magic pair, and we have information for side C but not side c, which is our bridge. We set our magic pair and bridge in proportion to one another to solve for the angle of C and use inverse sin to find it. But because we are assuming we have two triangles, and we are using sin (remember that with sin, we have positive answers in the first and second quadrant but our calculator only gives us the positive answer in the first quadrant), we use the reference angle of the first quadrant and subtract it from 180 to find the measurement of angle C'. In this case the angle of C' of 171 added with angle A of 155 was much bigger than 180 meaning only one triangle is possible. Had it been an acute angle like angle C of 9, there  could have been a possibility of two triangles.

4. Area Formulas

In a normal triangle where the area could be found with the formula of A = 1/2bh, we know that the perpendicular height of the triangle is h and b is the base. While the oblique triangles (where all sides are different lengths) area is one half of the product of two sides and the sine of their included angle <- very important to realize, it must be in between the two sides. 
We know that sinC=h/a from concept 1, so once you mulitply 'a' on both sides h (height) = asinC, and plugging 'h' into that A=1/2bh formula we really have A=1/2b(asinC)

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