Law of Sines and Cosines

In trigonometry, the Law of Sines and Law of Cosines are used as an alternative to the Pythagorean theorem in cases where the triangle is not a right triangle. Each law applies to different types of triangles with given information.



Law of Sines
The Law of Sines is used in either of three cases, the third of which is the ambiguous case:
 * A side between two given angles is given (ASA, or Angle-Side-Angle)
 * Two angles and an other side are given (AAS, or Angle-Angle-Side)
 * An angle and two other sides are given (SSA, or Side-Side-Angle)

The Law of Sines is as follows:
 * $$\frac{a}{\sin{A}\,}\,=\frac{b}{\sin{B}\,}\,=\frac{c}{\sin{C}\,}\,$$

Example Usage
Many problems will ask that each part of the triangle is solved.

ASA Case
Given than A=55°, c=2.8, and B=27°, solve the triangle. Round to the nearest tenth.

Let's find C first, since this is the easiest thing to do here.
 * $$m\angle{C}\,=180-(55+27)=180-82=98$$

Therefore, C=98°.

Now to find sides a and b.

To find side a, apply the extended proportion. Use it in a way that only side a is a variable. Here, side c and angle C are already known, and so we'll apply it to them.
 * $$\frac{a}{\sin{55}\,}\,=\frac{2.8}{\sin{98}\,}\,$$


 * $$a=\frac{2.8\sin{55}\,}{\sin{98}\,}\,\approx\,2.316166497\approx\,2.3$$

Therefore, a=2.3.

We'll use the same values for side c and angle C to find side b.
 * $$\frac{b}{\sin{27}\,}\,=\frac{2.8}{\sin{98}\,}\,$$


 * $$b=\frac{2.8\sin{27}\,}{\sin{98}\,}\,\approx\,1.283665948\approx\,1.3$$

Therefore, b=1.3. The triangle is solved.

AAS Case
Given that A=23°, C=62°, and a=20, solve the triangle. Round to the nearest tenth.

Let's find B first.
 * $$m\angle{B}\,=180-(23+62)=180-85=95$$

Therefore, B=95°.

Now for side b. Let's use the proportion using side a and angle A.
 * $$\frac{b}{\sin{95}\,}\,=\frac{20}{\sin{23}\,}\,$$


 * $$b=\frac{20\sin{95}\,}{\sin{23}\,}\,\approx\,50.99131477\approx\,51$$

Therefore, b=51.

Now for side c. Let's use the proportion with the same side and angle a and A, respectively.
 * $$\frac{c}{\sin{62}\,}\,=\frac{20}{\sin{23}\,}\,$$


 * $$c=\frac{20\sin{62}\,}{\sin{23}\,}\,\approx\,45.19463787\approx\,45.2$$

Therefore, c=45.2. The triangle is solved.

SSA/Ambiguous Case
This is the trickiest of the three cases. Due to the nature of having two fixed sides and one defined angle, there can either be no solution, 1 solution, or 2 solutions. It depends on if such a triangle under the given parameters can exist.

One of the tests to check if the triangle exists is to look at and compare the lengths of the sides. For any triangle of sides a, b, and c,
 * $$a+b>c$$

Another test that can be used is to see the sum of the angles. For any triangle with angles A,B, and C,
 * $$A+B+C=180$$

That said, let's work on an SSA case. Solve the triangle given b=2, c=5, and B=31°.


 * $$\frac{2}{\sin{31}\,}\,=\frac{5}{\sin{C}\,}\,$$


 * $$\sin{C}\,(\frac{2}{\sin{31}\,}\,)=5$$


 * $$\sin{C}\,=\frac{5\sin{31}\,}{2}\,$$


 * $$C=\arcsin{\frac{5\sin{31}\,}{2}\,}\,\approx\,\arcsin{1.28759518727514}\,$$

This is undefined because the sine of any angle never exceeds 1. Angle C is too large to form a triangle, thus this triangle doesn't exist.

Law of Cosines
The Law of Cosines is used in either of two cases:
 * All three sides are given (SSS, or Side-Side-Side)
 * An angle between two given sides is given (SAS, or Side-Angle-Side)

The Law of Cosines is as follows:
 * $$a=\sqrt{b^{2}+c^{2}-2bc\cos{A}\,}\,$$ (for side a)


 * $$A=\arccos{\frac{b^{2}+c^{2}-a^{2}}{2bc}\,}\,$$ (for angle A)


 * $$b=\sqrt{a^{2}+c^{2}-2ac\cos{B}\,}\,$$ (for side b)


 * $$B=\arccos{\frac{a^{2}+c^{2}-b^{2}}{2ac}\,}\,$$ (for angle B)


 * $$c=\sqrt{a^{2}+b^{2}-2ab\cos{C}\,}\,$$ (for side c)


 * $$C=\arccos{\frac{a^{2}+b^{2}-c^{2}}{2ab}\,}\,$$ (for angle C)

Example Usage
Many problems will ask that each part of the triangle is solved.

SSS Case
Will be completed later

SAS Case
Will be completed later