Trigonometry

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7

THE LAW OF COSINES

Proof of 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.

The the Law of Cosines

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.

The the Law of Cosines

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)

The Law of Cosines

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.

The the Law of Cosines

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.

The Law of Cosines is valid when the included angle is obtuse. But in that case, the cosine is negative. See Topic 15.

Proof of the Law of Cosines

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:  bx.

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 + (bx

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.


Next Topic:  The Law of Sines


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