TIE - Law of Cosines Lesson

Law of Cosines

Let's consider a triangle in which we know two sides (a and b) and the included angle ( equation image indicator $LaTeX: \angle C$ $\angle C$ ):

triangle ABC with sides a, b, c

We like to think about triangles in terms of right triangles, so let's drawn an altitude and call it h, we will also split up segment b into x and b - x.

triangle ABC with sides a, b, c with height (h) and segments b-x, and x, and angle C denoted

If we write an equation connecting c and h using the Pythagorean Theorem, it would look like this:

triangle ABC with sides a, b, c with height (h) and segments b-x, and x, and angle C denoted with a smaller triangle AbB

equation image indicator $LaTeX: c^2=h^2+\left(b-x\right)^2\\ c^2=h^2+b^2-2xb+x^2$ $c^2=h^2+\left(b-x\right)^2\\ c^2=h^2+b^2-2xb+x^2$

And if we consider the other right triangle in the figure, we can also say:

triangle ABC with sides a, b, c with height (h) and segments b-x, and x, and angle C denoted with smaller triangle BbC

equation image indicator $LaTeX: a^2=x^2+h^2\\ h^2=a^2-x^2$ $a^2=x^2+h^2\\ h^2=a^2-x^2$

Since we do not know h, let's substitute into the equation image indicator equation.

equation image indicator $LaTeX: c^2=\left[a^2-x^2\right]+b^2-2xb+x^2\\ c^2=a^2+b^2-2xb$ $c^2=\left[a^2-x^2\right]+b^2-2xb+x^2\\ c^2=a^2+b^2-2xb$

Now, we still do not know what x is, but we know angle C, so let's come up with an equation that connects those two:

equation image indicator $LaTeX: \cos C=\frac{x}{a}\\ a\cos C=x$ $\cos C=\frac{x}{a}\\ a\cos C=x$

And we can substitute for x

equation image indicator $LaTeX: c^2=a^2+b^2-2\left[a\cos C\right]b\\ c^2=a^2+b^2-2ab\cos C$ $c^2=a^2+b^2-2\left[a\cos C\right]b\\ c^2=a^2+b^2-2ab\cos C$

We've just proven the Law of Cosines! This works when given any two sides and their included angle and could potentially be in any of the following forms:

LAW OF COSINES: Let a, b, and c be the side lengths opposite angles A, B, and C. Then equation image indicator

$LaTeX: a^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C$ $a^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C$

When given two sides and an included angle of a triangle (SAS), use Law of Cosines first to find the third side, and the use Law of Sines to find the smallest of the two remaining angles (the smallest angle is opposite the shortest side). Watch this video to see how to use the Law of Cosines:

If the three sides of a triangle are given, first use the Law of Cosines to find one of the angles. It is usually best to find the largest angle first, the one opposite the longest side. Then, use the Law of Sines to find the second angle. Finally, subtract these angle measures from 180° to find the third angle. Watch the video to see an example.

Summary of Methods - Based on the Information Given

SSS: If the three sides of a triangle are given, first use the Law of Cosines to find one of the angles. It is best to find the largest angle first, (the largest angle is opposite the longest side). Then, use the Law of Sines to find the second angle. Finally, subtract these angle measures from 180° to find the third angle. It is best to find the largest angle first since once you know that angle, the other two angles in the triangle must be acute.

SAS: When given two sides and an included angle of a triangle, use Law of Cosines first to find the third side and then use Law of Sines to find the smallest of the two remaining angles (the smallest angle is opposite the shortest side). Finally, subtract the two angles from 180° to find the last angle.

ASA: If two angles and the included side of a triangle are known, first subtract these angle measures from 180° to find the third angle. Next, use the Law of Sines to find the lengths of the two missing sides.

AAS: If two angles and a nonincluded side of a triangle are known, first subtract these angle measures from 180° to find the third angle. Next, use the Law of Sines to find the lengths of the two missing sides.

SSA: This is the Ambiguous Case. If two sides and a nonincluded angle of a triangle are known, there could be two triangles, one triangle, or no triangle.

Try these problems to see if you've got it:

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