Sunday, October 16, 2011

Law of Cosines

Theres a triangle with legs 4, 8, 10 labeled ABC. a = 4, b = 8, c = 10. Find all of the angles.
Formula: opposite side^2 = adj. side^2 + other adj. side^2 -2(adj. side)(other adj. side)cos(angle b/w)

It doesn't matter which angle u solve for first but I'm going to solve for A first.

4^2 = 8^2 + 10^2 -2(8)(10)cosA (You don't have an angle so you will eventually have to take the inverse).
(Subtract 8^2 + 10^2)
4^2-8^2-10^2=-2(8)(10)cosA (Divide)
A= cos(inverse) ((4^2-8^2-10^2)
                            (-2(8)(10))) 
=  22.332 (You enter that into your calculator exactly) (Label angle A on your triangle)
You would do the same thing for the next angle (I'm solving for B)
You would get Angle B = 49. 458 and label your triangle
Add the two angles you found and subtract that number from 180 to find Angle C.
Angle C = 108.210

~ Parrish Masters

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