The law of cosines. Topics in trigonometry

THE LAW OF COSINES

WE USE THE LAW OF COSINES to solve triangles that are not right-angled. In particular, when we know two sides of a triangle and their included angle, then the Law of Cosines enables us to find the third side.

Thus if we know sides a and b and their included angle θ, then the Law of Cosines states:

c² = a² + b² − 2ab cos θ

(The Law of Cosines is a extension of the Pythagorean theorem; because if θ were a right angle, we would have c² = a² + b².)

Example 1. In triangle DEF, side e = 8 cm, f = 10 cm, and the angle at D is 60°. Find side d.

Solution.. We know two sides and their included angle. Therefore, according to the Law of Cosines,

d² = e² + f² − 2ef cos 60°

d² = 8² + 10² − 2· 8· 10· ½, since cos 60° = ½,

d² = 164 − 80

d² = 84.

d = .

Problem 1. In the oblique triangle ABC, find side b if side a = 5 cm, c = cm, and they include and angle of 45°. No Tables.

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b² = a² + c² − 2ac cos 45°

= 5² + ()² − 2· 5· · cos 45°

= 25 + 2 − 10· · ½, since cos 45° = ½,

= 25 + 2 − 10, (· = 2)

= 17.

b = cm.

Problem 2. In the oblique triangle PQR, find side r if side p = 5 in, q = 10 in, and they include and angle of 14°. (Table)

r² = 5² + 10² − 2· 5· 10 cos 14°

= 25 + 100 − 100(.970), from the Table.

= 125 − 97

= 28.

r = in.

Example 2. In Example 1, we found that d = , which is approximately 9.17.

Use the Law of Sines to complete the solution of triangle DEF. That is, find angles E and F.

Solution. To find angle F, we have this version of

Unknown
Known

sin F sin D	=	f d

sin F sin 60°	=	10 9.17

sin F	=	(.866)	10 9.17	from the Table,

		.944	with the aid of a calculator.

Therefore, on inspecting the Table for the angle whose sine is closest to .944,

Angle F 71°.

And therefore

Angle E	=	180° − (71° + 60°)

	=	180° − 131°

	=	49°.

And so using the Laws of Sines and Cosines, we have completely solved the triangle.

Proof of the Law of Cosines

Let ABC be a triangle with sides a, b, c. We will show

c² = a² + b² − 2ab cos C.

Draw BD perpendicular to CA, separating triangle ABC into the two right triangles BDC, BDA. BD is the height h of triangle ABC.

Call CD x. Then DA is the whole b minus the segment x: b − x.

Also, since

x
a

= cos C,

then

x = a cos C . . . . . . . (1)

Now, in the right triangle BDC, according to the Pythagorean theorem,

h² + x² = a²,

so that

h² = a² − x². . . . . . (2)

In the right triangle BDA,

c² = h² + (b − x)²

c² = h² + b² − 2bx + x².

(The square of a binomial)

For h², let us substitute line (2):

c² = a² − x² + b² − 2bx + x²

c² = a² + b² − 2bx.

Finally, for x, let us substitute line (1):

c² = a² + b² − 2b· a cos C.

That is,

c² = a² + b² − 2ab cos C.

This is what we wanted to prove.

In the same way, we could prove that

a² = b² + c² − 2bc cos A

and

b² = a² + c² − 2ac cos B.

This is the Law of Cosines.

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