The law of sines

The law of sines are used to solve the following two cases:

1. When the measure of two sides and the angle opposite one of them is known, and
2. When the measure of two angles and one side is known.

Example 1

Given: A = 33°, a = 9.4 and c = 14.3
Solve the triangle.

Solution

We are given:

sides angles a = 9.4 A = 33° b = ? B = ? c = 14.3 C = ?

Using the law of sines:


B = 180° - 33° - 55.95° = 91.05°

We can now find the length of the third side.

Now we have:

sides angles a = 9.4 A = 33° b = 17.26 B = 91.05° c = 14.3 C = 55.96°

Example 2

Given: A = 82.17°, B = 64.43° and c = 9.12
Solve the triangle.

Solution

We are given:

sides angles a = ? A = 82.17° b = ? B = 64.43° c = 9.12 C = ?

We can find angle C:

C = 180° - 82.17° - 64.43° = 33.40°

Now use the law of sines:

Now find side b using the law of sines:

Now we have:

sides angles a = 16.41 A = 82.17° b = 14.94 B = 64.43° c = 9.12 C = 33.40°

Example 3

Given: a = 20, b = 24, and A = 55.4°
Solve the triangle.

Solution

We are given:

sides angles a = 20 A = 55.4° b = 24 B = ? c = ? C = ?

Since sin is positive in the 1st and 2nd quadrants,
then B could also be 180° - 81.03° = 98.97°.

Now we have:

sides angles a = 20 A = 55.4° b = 24 B = 81.03° c = ? C = 43.57°		sides angles a = 20 A = 55.4° b = 24 B = 98.97° c = ? C = 25.63°
C = 180° - 55.4° - 81.03°    = 43.57°		C = 180° - 55.4° - 98.97°    = 25.63°

We now need to use the law of sines with C = 43.57° and C = 25.63°:

C sin 55.40° = 20 sin 43.57°		C sin 55.40° = 20 sin 25.63°

Finally we have:

sides angles a = 20 A = 55.4° b = 24 B = 81.03° c = 16.75 C = 43.57°

sides angles a = 20 A = 55.4° b = 24 B = 98.97° c = 10.51 C = 25.63°

Note:

If sin is larger than 1, then the triangle is impossible.