Trigonometry

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8

THE LAW OF SINES

Statement of the law of sines

The sine of an obtuse angle

The ambiguous case

Proof of the law of sines

This is a topic in traditional trigonometry. It does not come up in calculus.

THE LAW OF SINES allows us to solve triangles that are not right-angled, and are called oblique triangles. It states the following:

The sides of a triangle are to one another
in the same ratio as the sines of their opposite angles.

Oblique triangle ABC

a : b : c = sin A : sin B : sin C.

Specifically, side a is to side b as the sine of angle A is to the sine of angle B.

a
b
  =   sin A
sin B
.

Similarly,

b
c
  =   sin B
sin C
.

And so on, for any pair of sides and their opposite angles.

The law of sines is a theorem about the geometry of any triangle. As any theorem of geometry, it can be enunciated. The algebraic statement of the law --

sin A
  a
= sin B
  b
= sin C
  c

-- cannot be verbalized. sin A moreover, which is a number, does not have a ratio to a, which is a length.

Secondly, to prove that algebraic form, it is necessary to state and prove it correctly geometrically, and then transform it algebraically.

Example 1.   a)  The three angles of a triangle are 40°, 75°, and 65°.  In what ratio are the three sides? Sketch the figure and place the ratio numbers.

Solution.  To find the ratios of the sides, we must evaluate the sines of their opposite angles.  From the Table,

sin 40° = .643

sin 75° = .966

sin 65° = .906

These are the ratios of the sides opposite those angles:

The Law of Sines

Notice that we may express the ratios as ratios of whole numbers; we may ignore the decimal points.  Why?  Because we have multiplied each side by the same number, namely 1000. (The theorem of the same multiple.)

b)  When the side opposite the 75° angle is 10 cm, how long is the side opposite the 40° angle?

The Law of Sines

Solution.  Let us call that side x.  Now, according to the Law of Sines, in every triangle with those angles, the sides are in the ratio 643 : 966 : 906. Therefore,

 x 
10
= 643
966
x = 10·643
966

With the aid of a calculator,

xapproximately6.656 cm.

Problem 1.   The three angles of a triangle are A = 30°, B = 70°, and C = 80°.

   a)    In what ratio are the three sides?  Sketch the triangle and place those ratio numbers.   (Table)

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").

The law of sines

sin 30° = .500

sin 70° = .940

sin 80° = .985

   b)    If side a = 5 cm, find sides b and c.

In every triangle with those angles, the sides are in the ratio 500 : 940 : 985.  But the side corresponding to 500 has been divided by 100.  Therefore, each side will be divided by 100.  Side b will equal 9.4 cm, and side c = 9.85 cm.

The sine of an obtuse angle

The trigonometric functions (sine, cosine, etc.) are defined in a right triangle in terms of an acute angle.  What, then, shall we mean by the sine of an obtuse angle ABC?

An obtuse angle

The sine of an obtuse angle is defined to be the sine of its supplement.

For example,

sin 120°  =  sin 60°  =  square root of 3
 2
.

(Topic 4.)

To see why we make this definition, let ABC be an obtuse angle, and

The sine of an obtuse angle

draw CD perpendicular to AB extended.

Then we define the sine of angle ABC as follows:

sin ABC =  CD
CB
.

But that is the sine of angle CBD -- opposite-over-hypotenuse.  And angle CBD is the supplement of angle ABC.  So, by the sine of an obtuse angle we mean the sine of its supplement.

For example,

sin 150 = sin 30
sin 110° = sin (180° − 110°)
  = sin 70°
  = .940,   from the Table.

Problem 1.   Evaluate the following:

a)  sin 135° = sin 45° = ½square root of 2   (Topic 4, Example 1)

b)  sin 127° = sin (180° − 127°) = sin 53° = .799
                  (From the Table)

Problem 2.

a)  The three angles of a triangle are 105°, 25°, and 50°.  In what ratio
a)  are the sides?  Sketch the triangle.

sin 105° = sin (180° - 105°)
  = sin 75°
  = .966,   from the Table.
 
sin 25° = .423
 
sin 50° = .766

Therefore, the sides opposite those angles are in the ratio

966 : 423 : 766

The law of sines

b)  If the side opposite 25° is 10 cm, how long is the side opposite 50°?

The law of sines

 x 
10
= 766
423
.

With the aid of a calculator, this implies:

x = 10·  766
423
approximately 18.1 cm

The ambiguous case

The so-called ambiguous case arises from the fact that an acute angle and an obtuse angle have the same sine.  If we had to solve

sin x = ½square root of 2,

for example, we would have

x = 45°  or  x = 135°.

(Topic 4, Example 1.)

In the following example, we will see how this ambiguity could arise.

The law of sines

In triangle ABC, angle A = 30°, side a = 1.5 cm, and side b = 2 cm.  Let us use the law of sines to find angle B.

 sin B 
sin 30°
=  2 
1.5
 
Since sin 30° = ½ (Topic 4, Example 2),  
sin B = ½·  20
15
  = 10
15
  = 2
3
  approximately .666

On inspecting the Table for the angle whose sine is closest to .666, we find

Bapproximately42°.

But the sine of an angle is equal to the sine of its supplement.  That is, .666 is also the sine of 180° − 42° = 138°.

This problem has two solutions.  Not only is angle CBA a solution,

The law of sines

but so is angle CB'A, which is the supplement of angle CBA.  (We can see that it is the supplement by looking at the isosceles triangle CB'B; angle CB'A is the supplement of angle CB'B, which is equal to angle CBA.)

Given two sides of a triangle a, b, then, and the acute angle opposite one of them, say angle A, under what conditions will the triangle have two solutions, only one solution, or no solution?

The law of sines

Let us first consider the case a < b.  Upon applying the law of sines, we arrive at this equation:

1) sin B = sin A·  b
a
.
Now, since  h
b
 =  sin A , where h is the height of the triangle (Fig. 1),

then

b sin A = h.

On replacing this in the right-hand side of equation 1), it becomes

sin B h
a
.

There are now three possibilities:

The law of sines

h
a
< 1,  which implies  h < a (Fig. 1),
h
a
= 1,  which implies  h = a (Fig. 2),
h
a
> 1,  which implies  h > a (Fig. 3).

In the first of these -- h or b sin A < a -- there will be two triangles.

In the second -- h or b sin A = a -- there will be one right-angled triangle.

And in the third -- h or b sin A > a -- there will be no solution.

Example 2.   Let a = 2 cm, b = 6 cm, and angle A = 60°.  How many solutions are there for angle B?

Answer.   We must calculate b sin A.  If it is less than a, there will be two solutions.  If it is equal to a, there will be one solution.  And if it is greater than a, there will be no solution.

Now, sin 60° =  square root of 3
 2
.  (Topic 4.)  Therefore,
b sin A   =  6·  square root of 3
 2
  =   3square root of 3 .

Since a = 2, then  b sin A > a.  (square root of 3approximately1.732).

There is no solution.

Finally, we will consider the case in which angle A is acute, and a > b.

the law of sines

In this case, there is only one solution, namely, the angle B in triangle CBA.  For, in triangle CAB', the angle CAB' is obtuse.

Problem 3.   In each of the following, find the number of solutions.

a)  Angle A = 45°,  a = square root of 2,  b = 2.

Since square root of 2 < 2, this is the case a < b.  sin 45° = square root of 2/2. Therefore, b sin A = 2square root of 2/2 = square root of 2, which is equal to a. There is therefore one solution: angle B is a right angle.

b)  Angle A = 45°,  a = 1.8,  b = 2.

Again, a < b.  b sin A = 2square root of 2/2 = square root of 2, which is less than a. Therefore there are two solutions.

c)  Angle A = 45°,  a = 2,  b = 1.5.

Here, a > b.  Therefore there is one solution.

d)  Angle A = 45°,  a = 1.4,  b = 2.

a < b.  b sin A = 2square root of 2/2 = square root of 2, which is greater than a. Therefore there are no solutions.

Proof of the law of sines

The sides of a triangle are to one another in the same ratio as the sines
of their opposite angles.

Since the trigonometric functions are defined in terms of a right-angled triangle, then it is only with the aid of right-angled triangles that we can prove anythingexclamation

The Law of Sines

In triangle ABC, then, draw CD perpendicular to AB. Then CD is the height h of the triangle. The height now separates triangle ABC into two right triangles, CDA and CDB.

We will now show that

sin A
sin B
  =   a
b
.

Now, in triangle CDA,

sin A =  h
b
.

While in triangle CDB,

sin B =  h
a
.

Therefore,

sin A
sin B
  =   h/b
h/a
  =   h
b
·   a
h
  =   a
b
.

This is what we wanted to prove.

In the same way, we could prove that

sin B
sin C
  =   b
c

and so on, for any pair of angles and their opposite sides.


Next Topic:  The Circle


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