Trigonometry Lesson-3 Oblique Triangles: Law of Cosines |

**The Law of Cosines**

Another way of solving oblique triangles is with the Law of Cosines. This method is used **when the given information doesn't include a side opposite a given angle**. We use the Law of Cosines when we are given

The Law of Cosines has 3 statements -- each of which can be used to solve for an unknown side or an unknown angle. They are:

**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**

**Note:** the **side** on the left of the equation **corresponds to the angle** on the right.

The statements of the Law of Cosines can also be used to solve for angles by isolating the

cos A, cos B or cos C term of the equation like this:

, , or

**Examples**

**Example 1:** given 2 sides and the **contained** angle.

We have 2 sides and the contained angle, we use cosine law to find **c**.

**c ^{2} = a^{2} + b^{2} - 2ab cos C**

c^{2} = 4^{ 2} + 5^{ 2} - 2 (4) (5) cos 60° = 16 + 25 - 40 (.5) = 21

** = 4.58 cm**

We find angle B with the Law of Sines:

sin B = sin B = 0.945443

so **angle B = **arcsin 0.945443** =** **70.99**°

**angle A = **180° - (60° + 70.99° )** = 49.01**°

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Now let's do an example where we're given all three sides but no angles.

**Example 2:** given 3 sides

In such a case we should **always solve for the largest**

** angle first**, since it is the only one that might be obtuse.

Solving for **angle C** using

, so **angle C **= arccos (- 0.25)

**angle C = 104.48**°

The second unknown angle can be found using the Law of Sines.

sin B = 0.726176

therefore **angle B = **arcsin 0.726176 , so **angle B = 46.57°**

**angle A = **180° - (104.48° + 46.57° ) = **28.95 ^{ }**°.

Now get a pencil, an eraser and a note book, copy the questions,

do the practice exercise(s), then check your work with the solutions.

If you get stuck, review the examples in the lesson, then try again.

**Practice**

1) Solve the following triangles ABC:

a) a = 3, b = 5, c = 7

( *solution* )

b) a = 12, c = 13, **angle B = **120°

( *solution* )

.

2) A force triangle consists of three forces F_{1} = 35.07 Kg, F_{2} = 22.6 Kg, and F_{3} = 41.72 Kg.

Find the angle between F_{1} and F_{2}. ( *solution* )

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3) A hill is inclined 5° to the horizontal. A 13.85 m pole stands at the top of the hill. What length of rope will reach from the top of the pole to a point 10.77 m downhill from the base of the pole?

( *solution* )

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4) Two ships leave harbor at the same time. The first sails on a course that is 15° west of due north at 25 km/h. The second ship sails on a course that is 32° east of due north at 20 km/h. After 2 hours, how far apart are the 2 ships? ( *solution* )

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**Solutions**

1) Solve these triangles ABC:

a) a = 3, b = 5, c = 7

**angle C = **arccos (- 0.5) = **120 ^{0}**

sin B = 0.61859, so angle B **=** **38.21**°

**angle A** **=** 180° - (120° + 38.21°) = **21.79**°

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b) a = 12, c = 13, **angle B = **120°

*b*^{ 2} = a^{ 2} + c^{ 2} - 2ac cos B,

*b*^{2} = 12^{2} + 13^{2} - 2(12)(13) cos 120°

*b*^{ 2} = 469

sin C = 0.519775

**angle C =** arcsin 0.519775 = **31.32°**

**angle A =** 180° - (120° + 31.32°) = **28.68**°

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2) A force triangle consists of three forces F_{1} = 35.07 Kg, F_{2} = 22.6 Kg, and F_{3} = 41.72 Kg.

Find the angle between F_{1} and F_{2}.

**angle x** = arccos 0.000067 =

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3) A hill is inclined 5° to the horizontal. A 13.85 m pole stands at the top of the hill. What length

of rope will reach from the top of the pole to a point 10.77 m downhill from the base of the pole?

**angle B** = 90° + 5° = 95°

*b*^{2} = 10.77^{2} + 13.5^{2} - 2(10.77)(13.85) cos 95° = 333.81

** b** = = 18.27 m

The rope must be **18**.**27** m long.

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4) Two ships leave harbor at the same time. The first sails on a course that is 15° west of due north at 25 km/h. The second ship sails on a course that is 32° east of due north at 20 km/h. After 2 hours, how far apart are the 2 ships?

**angle A** = 15° + 32° = 47°

*a*^{2} = 40^{2} + 50^{2} - 2(40)(50) cos 47° = 1372.007 km

** a** =

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