MATH 1113 PreCalculus

Day #33, Law of Cosines

The "Generalized Pythagorean Theorem"

I think of the Law of Cosines as a Pythagorean Theorem with a correction factor.  The actual Pythagorean Theorem: c² = a² + b² only applies in the case that the triangle is a right triangle, with side "c" being opposite the right angle (c is the hypotenuse.)

The Law of Cosines states the amazing fact that this relationship "almost" holds if the angle "C" is "almost" a right angle, and it tells us exactly how far off the relationship is, in terms of the sides "a" and "b" and the angle "C":

The Law of Cosines:

For ANY triangle ABC, with sides a, b, and c, and opposite angles A, B, and C:

c² = a² + b² - 2*a*b*cos(C)
a² = b² + c² - 2*b*c*cos(A)
b² = a² + c² - 2*a*c*cos(B)

So, the Pythagorean Theorem is off by a correction factor of "-2*a*b*cos(C)."  Notice that, since cos(90^o) = 0, in the case of a right triangle, this reduces to the usual Pythagorean Theorem.

Here's a Geogebra Applet that demonstrates the Law of Cosines:

Applet

It looks sort of complicated, but if you think of it as a correction factor added to the Pythagorean Theorem that you're used to, it's not hard to remember it and rearrange it for whatever situation applies.

Using the Law of Cosines to handle the other two cases.

Basically, if you know two sides and an angle, and it isn't the ambiguous case (so you lack any ratio of side to angle for the Law of Sines), you can use the Law of Cosines to find the third side.  (This is the S-A-S case):

Example (S-A-S):

Suppose that side b has length 8.4, side c has length 11.6, and the angle, A, in between them is 73 degrees.  Solve the triangle.

Here's a Geogebra Applet that lets you play with the S-A-S situation.

Applet

Notice that you must first find the missing side by using the Law of Cosines. The formula is in the picture, above.

Once you've found side "a," you might be tempted to use the Law of Sines to find the next angle, since it's easier.  This is OK, AS LONG AS YOU USE IT TO FIND THE SMALLER ANGLE (the one opposite the shorter side.)  Since the sine of an angle and its supplement are the same, you want to be sure that you're looking for an acute angle.  To avoid this problem, use the Law of Cosines again to find the missing second angle, by using the Inverse Cosine function.

So, here, we solved for angle "B:"

b² = a² + c² - 2*a*c*cos(B);

cos(B) = (b² - (a² + c²))/(-2*a*c) OR (a² + c² - b² )/(2*a*c)

So, cos(B) = 0.7513, or B = 41.3^o.

You can always find the third angle by subtraction.

Angle C = 180 - (41.3 + 73) = 65.7^o.

Notice that, in this case, for any angle 0<A<180, and any lengths b and c, there is ALWAYS a unique triangle.

Page Last Modified: 28 March, 2007

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