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 2 sides and the contained angle, or all three sides.

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:

a2 = b2 + c2 - 2bc cos A

b2 = a2 + c2 - 2ac cos B

c2 = a2 + b2 - 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.

c2 = a2 + b2 - 2ab cos C

c2 = 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 )

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2) A force triangle consists of three forces F1 = 35.07 Kg, F2 = 22.6 Kg, and F3 = 41.72 Kg.
Find the angle between F1 and F2. ( 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 )

. Solutions

1) Solve these triangles ABC:

a) a = 3, b = 5, c = 7  angle C = arccos (- 0.5) = 1200 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,

b2 = 122 + 132 - 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 F1 = 35.07 Kg, F2 = 22.6 Kg, and F3 = 41.72 Kg.
Find the angle between F1 and F2.  angle x = arccos 0.000067 = 89.996° = 90°

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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°

b2 = 10.772 + 13.52 - 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°

a2 = 402 + 502 - 2(40)(50) cos 47° = 1372.007 km

a = 37.04 km

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