Law of Cosines for third side length of a triangle, Law of Sines for area of the triangle

If we know the length of two sides of a triangle, call them a and b, and we have some way to find the angle between them, call it gamma, here is how we can use the Law of Cosines and the Law of Sines best. (Here we assume gamma is an angle between 0° and 180°.)

The third side length can be found using

c² = a² + b² - 2abcosgamma

and the area is given by

Area = ½absingamma

We need a and b as  numbers, but we can know gamma a few different ways.


1) gamma is given. Then we go to the calculator and find cosgamma and singamma.

2) cosgamma is given. We can find sine using The Trigonometric Identity. If we are asked for the measure of gamma, use the inverse cosine function.

3) singamma is given and the quadrant for gamma is given or we are told gamma is acute or obtuse.
We can find cosine using The Trigonometric Identity, where cosine will be positive if we are in the first quadrant (gamma is acute) or cosine will be negative if in the second quadrant (gamma is obtuse). If we are asked for the measure of gamma, use the inverse cosine function.


Example 1.

gamma = 135°, a = 6, b = 2.
sin135° = sqrt(2)/2, cos135° = -sqrt(2)/2

c² = 6² + 2² - 2(6)(2)(-sqrt(2)/2) = 40 + 12sqrt(2), so c = sqrt(40 + 12sqrt(2)) or approximately 7.5479 units.

Area =  ½(6)(2)sqrt(2)/2 = 3sqrt(2) or approximately 4.246 square units.


Example 2.

gamma = 45°, a = 6, b = 2.
sin45° = sqrt(2)/2, cos45° = sqrt(2)/2

c² = 6² + 2² - 2(6)(2)(sqrt(2)/2) = 40 - 12sqrt(2), so c = sqrt(40 - 12sqrt(2)) or approximately 4.7989 units.

Area =  ½(6)(2)sqrt(2)/2 = 3sqrt(2) or approximately 4.246 square units.

(Notice the area for the first two examples is the same.)


Example 3.

singamma = .2, gamma in first quadrant, a = 6, b = 2.

We can also write singamma as 1/5, so 1/5² + cos²gamma = 1
1/25 + cos²gamma = 1
cos²gamma = 24/25
cosgamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(2sqrt(6)/5) = 40 - 48sqrt(6)/5, so c = sqrt(40 - 48sqrt(6)/5) or approximately 4.0602 units.

Area =  ½(6)(2).2 = 1.2 square units exactly.





Example 4.

singamma = .2, gamma in second quadrant, a = 6, b = 2.


We can also write singamma as 1/5, so 1/5² + cos²gamma = 1
1/25 + cos²gamma = 1
cos²gamma = 24/25
cosgamma = -sqrt(24/25) = -2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(-2sqrt(6)/5) = 40 + 48sqrt(6)/5, so c = sqrt(40 + 48sqrt(6)/5) or approximately 7.9696 units.

Area =  ½(6)(2).2 = 1.2 square units exactly.
(Notice the area for examples 3 and 4 is the same.)

Example 5.

cosgamma = .2, a = 6, b = 2.

We can also write cosgamma as 1/5, so 1/5² + sin²gamma = 1
1/25 + sin²gamma = 1
sin²gamma = 24/25
singamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(.2) = 40 - 4.8, so c = sqrt(35.2) or approximately 5.9330 units.

Area =  ½(6)(2)2sqrt(6)/5 or approximately 5.8788 square units.


Example 6.

cosgamma = -.2, a = 6, b = 2.


We can also write cosgamma as -1/5, so (-1/5)² + sin²gamma = 1
1/25 + sin²gamma = 1
sin²gamma = 24/25
singamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(-.2) = 40 + 4.8, so c = sqrt(44.8) or approximately 6.6933 units.

Area =  ½(6)(2)2sqrt(6)/5 or approximately 5.8788 square units.

Yet again, the areas of examples 5 and 6 are the same.


Note: this was updated on Oct. 11 to correct errors in the lengths of the third side. Thanks to my student Fred Johnson for pointing out the errors.