The Law of Cosines: Videos & Practice Problems

To solve for a side in a triangle when given certain sides and angles, the Law of Sines is often a go-to method. However, there are situations where this approach is not applicable, particularly when you have two sides but lack the angles opposite those sides, or when you have an angle but not its corresponding side. In such cases, the Law of Cosines becomes essential.

The Law of Cosines provides a way to relate the lengths of the sides of a triangle to the cosine of one of its angles. The general formula for the Law of Cosines is expressed as:

\[c^2 = a^2 + b^2 - 2ab \cdot \cos(C)\]

In this equation, \(c\) is the side opposite angle \(C\), while \(a\) and \(b\) are the other two sides. This formula allows you to calculate the length of side \(c\) when you know the lengths of sides \(a\) and \(b\) and the measure of angle \(C\).

For example, if you have \(a = 3\), \(b = 2\), and \(C = 60^\circ\), you can substitute these values into the Law of Cosines:

\[c^2 = 3^2 + 2^2 - 2 \cdot 3 \cdot 2 \cdot \cos(60^\circ)\]

Calculating this gives:

\[c^2 = 9 + 4 - 12 \cdot \frac{1}{2}\]

\[c^2 = 13 - 6 = 7\]

Taking the square root results in:

\[c = \sqrt{7}\]

This method not only allows for the calculation of missing sides but can also be used to find missing angles if all three sides are known. It's important to note that the Law of Cosines encompasses the Pythagorean theorem as a special case, where the angle is \(90^\circ\). Understanding these relationships and formulas is crucial for solving various triangle problems effectively.

In solving triangles, the law of cosines is essential for determining unknown sides and angles, particularly in SAS (Side-Angle-Side) and SSS (Side-Side-Side) configurations. When given two sides and the included angle in an SAS triangle, the law of cosines allows us to find the third side. The formula used is:

\[c^2 = a^2 + b^2 - 2ab \cdot \cos(C)\]

In this context, if we have a triangle where side a is 4, side b is 3, and angle C is 60 degrees, we can substitute these values into the formula. This results in:

\[c^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \cos(60^\circ)\]

Calculating this gives:

\[c^2 = 16 + 9 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2}\]

\[c^2 = 25 - 12 = 13\]

Thus, c is equal to the square root of 13, or approximately 3.6, although it is often preferable to leave it in radical form to avoid rounding errors.

Next, to find one of the remaining angles, we can again apply the law of cosines. For angle A, the formula is:

\[A^2 = b^2 + c^2 - 2bc \cdot \cos(A)\]

Substituting the known values, we have:

\[4^2 = 3^2 + 13 - 2 \cdot 3 \cdot \sqrt{13} \cdot \cos(A)\]

Rearranging and simplifying leads to:

\[16 = 9 + 13 - 6\sqrt{13} \cdot \cos(A)\]

From this, we can isolate cos(A):

\[-6\sqrt{13} \cdot \cos(A) = -6\]

Thus, we find:

\[\cos(A) = \frac{1}{\sqrt{13}}\]

Taking the inverse cosine yields:

\[A \approx 73.9^\circ\]

Since the cosine function only provides one angle in the range of 0 to 180 degrees, we do not need to consider a second possible angle. With angle A determined, we can find the last angle B using the angle sum property of triangles:

\[A + B + C = 180^\circ\]

Substituting the known values gives:

\[73.9 + B + 60 = 180\]

Solving for B results in:

\[B = 180 - 73.9 - 60 = 46.1^\circ\]

In conclusion, the triangle's dimensions are now fully determined: the third side is c = \(\sqrt{13}\), angle A = 73.9^\(\circ\), and angle B = 46.1^\(\circ\). This systematic approach using the law of cosines effectively resolves the unknowns in SAS and SSS triangles.