So - it's taken me a little while to prepare a post on the Law of Cosines. Mostly because it took me a little while to wrap my brain around it - but a little bit of lightning struck my brain and I suddenly got it. Here's the gist:
Now - note that there are no right angles here, which makes this triangle different from any of the other ones we've used so far. The law of cosines and the law of sines are used to determine the relationships between the angles and sides of these triangles, and frankly I think the law of cosines is the more useful of the two, but I'll show you both.
Now, for a right triangle we can calculate the relationship between its sides pretty simply, using the Pythagorean theorem. The Law of Cosines, in fact, is just an extension of that which will work for any triangle. Here it is:
It also works for finding the length of any of the three sides, not just c, as long as you replace the angle after "cos" with the angle opposite the side you're trying to find.
So let's give this a shot. Back to our original triangle, let's assign some values.
c = 11
B = 37 degrees.
b^2 = a^2 + c^2 - 2ab cos(B)
b^2 = 8^2 + 11^2 - 2 8 11 cos(37 deg)
b^2 = 64 + 121 - 176 cos(37deg)
b^2 = 185 - 176 cos(37deg)
We'll have to use a calculator to find the cosine of 37 degrees, or look it up. Here it is.
cos (37 degrees) = 0.798...
b^2 = 185 - 176 (0.798...)
b^2 = 185 - 140.559...
b^2 = 44.440...
b = sqrt(44.440...)
b = 6.666...
Wow. I promise I didn't plan that.
I did use a problem that I had the answer to, just to verify that I did it correctly but I didn't look at the solution until I was done. Success! Now we can find the other two angles, if we want. If you're looking for the angle, you just have to reconfigure the equation solving for cos(x).
c^2 = a^2 + b^2 - 2ab cos(C)
c^2 - a^2 - b^2 = -2ab cos(C)
-c^2 + a^2 + b^2 = 2ab cos(C)
(-c^2 + a^2 + b^2) / 2ab = cos (C)
There you have it. Yay algebra! Now, here's the less useful Law of Sines.
The Law of Sines
The Law of SInes is simply a statement of the ratios between the angles and their opposite sides. So, in our same triangle from before:
The Quadratic Surprise
I had forgotten completely about the quadratic formula. Imagine my surprise when I stumbled upon a trig problem that seemed just unsolveable at first because I couldn't isolate the variable.... when I looked up the solution, I saw that you had to rephrase the equation as a quadratic first. I don't remember the details of the problem but a quadratic equation, basically, looks like this:
ax^2 + bx + c = 0
Where a, b, and c are given, and are known as the quadratic coefficients. Then you use the quadratic formula to solve for x:
Until next time...