BQ #1: Unit P Concept 3 and 4: Law of Cosines and Area of an Oblique Triangle

Law of Cosines
1. Why do we need it?

• When dealing with real life examples of things like triangles, we need to realize that not all of these situations are ideal. There will be triangles that are not right triangles but we still need to know what to do with them. This is where the Laws of Cosines comes in handy. With this, we will be able to find missing parts of practically any triangle, may it be a right triangle or a non-right triangle.
• We can only use Law of Cosines when we are given SSS (all three sides) or SAS (side, angle, side).

2. How is it derived from what we already know? 

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

• At first glance, it is obvious that this triangle is not a right triangle. As a result, we drop a perpendicular from angle B to create two right triangles (ABD and BCD). We want to identify the perpendicular as "h" (the height) and the distance between angle A to the right angle (D) as "r". 
• From here, we can begin our derivation. For this, we want to find the distance between angle B and C (side a, basically). If we were to plot this triangle on the graph (beginning at the origin), angle C would be plotted at (b,0) because we move b units on the x-axis and we do not move at all on the y-axis. Angle B would have the ordered pair of (cCosA, cSinA). We are able to get these values with our knowledge of trig functions. Looking at cosA, we know that it is equal to r/c (adjacent/hypotenuse). Now, "r" is the amount of units we go on the x-axis, therefore, we want to change cosA = r/c to r = cCosA. That is why our x-value for Angle B is cCosA. The reason why we have cSinA as our y-value is because sinA = h/c (opposite/hypotenuse), and after we manipulate the equation, we discover that h = cSinA. "H" is the amount of units we go up, therefore, cSinA is our y-value. 
• Now that we have these two points, we can plug them into our distance formula. We use the distance formula because we are trying to find the distance/length between angle B and C. 

• In order to have an easier problem we can work with, we square both sides of the equation to get rid of the square root. Now, we can begin substituting certain values for others in order to shape the distance formula into our Law of Cosines. 

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm
*Keep in mind that these values were plugged in d^2 = (y-y)^2 + (x-x)^2 (just a different arrangement)*

• Because we are looking for the distance between angles B and C, also known as side a, we change the "d" to an "a", allowing us to be more specific. Now we plug in the values. First, we plug in the y-values: cSinA and 0. Because subtracting 0 does nothing, we leave cSinA by itself in the parentheses. In the next set of parentheses, we plug in the x-values: b and cCosA. That is how we get (b - cCosA). 
• We want to expand these values next, so we foil them as we would with polynomials. From expanding (cSinA)^2, we get c^2sin^2A. When we expand (b - cCosA)^2, we get -2bcCosA + c^2cos^2A. 
• It looks overwhelming now, but we can simplify the equation by taking out c^2 and leaving the rest of the problem. At this point, we are left with (sin^2A + cos^2A) within the parentheses. Because of this, we are able to simplify it to 1. 
• The reason why we can simplify sin^2A + cos^2A is because it is like our Pythagorean Theorem. x^2 + y^2 = r^2. If we divide everything by r^2, we get (x^2/r^2) + (x^2/r^2) = 1 (r^2/r^2). We know from our knowledge of trig functions that cosine is x/r and sine is y/r. Therefore, when we square cosine and sine we get (x^2/r^2) and (x^2/r^2). That is why we are able to have (sin^2A + cos^2A) equal 1. 
• Now, because we are only multiplying c^2 by 1, we are left with the final equation: a^2 = c^2 + b^2 - 2bc cosA. Keep in mind that we are able to derive this equation for different sides of the triangle using similar processes. If we were looking for side b or c, we would end up with similar equations. 

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm
*These are the Law of Cosines variations for different sides of the triangle other than side a*

Area of an Oblique Triangle:

1. How is the “area of an oblique” triangle derived?

http://hotmath.com/hotmath_help/topics/law-of-sines/law-of-sines-image015.gif

• We usually rely on our perpendicular bases and heights in order to find areas of triangles. However, oblique triangles do not have perpendicular bases and heights, so we must create them ourselves, either by dropping a perpendicular, or drawing one ourselves.
• Our regular formula is A = 1/2(bh). What we really need to know is our height since 1/2 is a given and "b" is usually already known. We are going to be using our trig functions once more. We will use angle B as an example. Sin B = h/c (h being the perpendicular drawn). Remember that we are looking for "h". As a result, we multiply both sides by c to discover that h = cSinB. Now that we know "h", we can substitute it in the area formula. Now we have A = 1/2(bcSinB). This is our formula for the area of an oblique triangle!



http://wps.prenhall.com/esm_blitzer_algtrig_2/13/3560/911520.cw/index.html

• There are different variations of the formula since we can use different angles (depending on what we are given in the problem).
• We must be given two sides and an angle in between them (SAS) in order to use this.

2. How does it relate to the area formula that you are familiar with?

• We use the original formula (A = 1/2(bh)) as a starting point to find the area formula for an oblique triangle. We still use the same base, however we have to find our own "h" by using trig functions (sine).

Works Cited: 

1. http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

2. http://hotmath.com/hotmath_help/topics/law-of-sines/law-of-sines-image015.gif

3. http://wps.prenhall.com/esm_blitzer_algtrig_2/13/3560/911520.cw/index.html