The law of sines, including the ambiguous case.

THE LAW OF SINES

WE USE THE LAW OF SINES AND THE LAW OF COSINES to solve triangles that are not right-angled. Such triangles are called oblique or scalene triangles.

The law of sines states the following:

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

This means that in the oblique triangle ABC, side a, for example, 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.

Strictly, A, B, C are points. They are the vertices of the triangle. "Angle A" is a brief way of saying "The angle at the point A." By "sin A," then, we mean "The sine of the angle at the point A."

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:

Notice that we may express the ratios as ratios of whole numbers; we may ignore the decimal points. (Equivalently, 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?

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,

x6.656 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)

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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 mean by the sine of an obtuse angle ABC?

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

For example,

sin 120°

sin 60°

(Topic 7.)

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

draw CD perpendicular to AB extended.

Then the sine of angle ABC is defined as follows:

sin ABC =

CD
BC

But this 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 110°	=	sin (180° − 110°)
	=	sin 70°
	=	.940, from the Table.

Problem 1. Evaluate the following:

a) sin 135° = sin 45° = ½ (Topic 6, 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 are
a) the sides? Sketch the triangle.

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

With the aid of a calculator, this implies:

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 = ½,

for example, we would have

x = 45° or x = 135°.

(Topic 6, Example 1.)

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

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 7, Example 2),
sin B	=	½·	20 15
	=	10 15
	=	2 3
		.666

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

B42°.

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,

but so is angle CB'A, which is the supplement of angle CBA. (We can see that it is the supplement in the isosceles triangle CBB'; 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?

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

b
a

h
b

, 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:

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° =

. (Topic 7.) Therefore,

b sin A

= 6·

Since a = 2, then b sin A > a. (1.732).

There is no solution.

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

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 = , b = 2.

Since < 2, this is the case a < b. sin 45° = /2. Therefore, b sin A = 2/2 = , 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 = 2/2 = , 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 = 2/2 = , 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.

In triangle ABC, 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.

(This was necessary, because the trigonometric functions are defined in terms of a right triangle.)

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 Law of Cosines

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