THE LAW OF COSINES
WE USE THE LAW OF COSINES AND THE LAW OF SINES to solve triangles that are not right-angled. Such triangles are called oblique triangles. The Law of Cosines is used much more widely than the Law of Sines. Specifically, 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.
To see the answer, pass your mouse over the colored area.
b2 = a2 + c2 − 2ac cos 45°
= 52 + ()2 − 2· 5· · cos 45°
= 25 + 2 − 10· · ½, since cos 45° = ½,
= 25 + 2 − 10, (· = 2)
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
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.
Therefore, on inspecting the Table for the angle whose sine is closest to .944,
Angle F 71°.
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.
(The trigonometric functions are defined in terms of a right-angled triangle. Therefore it is only with the aid of right-angled triangles that we can prove anything)
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.
x = a cos C . . . . . . . (1)
Now, in the right triangle BDC, according to the Pythagorean theorem,
h2 + x2 = a2,
h2 = a2 − x2. . . . . . (2)
In the right triangle BDA,
c2 = h2 + (b − x)2
c2 = h2 + b2 − 2bx + x2.
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.
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
b2 = a2 + c2 − 2ac cos B.
This is the Law of Cosines.
Please make a donation to keep TheMathPage online.
Copyright © 2022 Lawrence Spector
Questions or comments?
E-mail: [email protected]