The Law of Sines and the Ambiguous Case, precalculus lesson

You want the distance to a tree on the far bank of a river, and you cannot cross. You can pace off a straight stretch on your own bank and measure the angle from each end of it to the tree. That gives a triangle with no right angle, so the right triangle ratios do not apply directly. The law of sines handles this triangle and many others, although in one kind of problem it can hand you two different answers.

Key terms

Oblique triangle

Definition An oblique triangle is a triangle with no right angle. Its three angles are positive and add up to $180^\circ$ , or $\pi$ radians, so either all three are acute, which makes an acute triangle, or exactly one is obtuse, which makes an obtuse triangle.

In plain words It is any triangle that is not a right triangle. Ratios such as opposite over hypotenuse need a right angle and a hypotenuse, so they cannot be used on it directly.

Example Angles of $50^\circ$ , $60^\circ$ and $70^\circ$ form an acute oblique triangle. Angles of $25^\circ$ , $35^\circ$ and $120^\circ$ form an obtuse one. Angles of $30^\circ$ , $60^\circ$ and $90^\circ$ form a right triangle, so it is not oblique. Two angles of $100^\circ$ never occur in the same triangle, since together they already pass $180^\circ$ .

Sides opposite angles

Definition In triangle $ABC$ the letters $A$ , $B$ and $C$ name the vertices and also the angles at those vertices. The side opposite angle $A$ is the side $BC$ , which does not touch vertex $A$ , and its length is written $a$ . In the same way $b$ is the length of side $CA$ , opposite $B$ , and $c$ is the length of side $AB$ , opposite $C$ . In every triangle the longer of two sides lies opposite the larger angle,

a > b \quad \text{exactly when} \quad A > B,

and equal sides lie opposite equal angles.

In plain words Each angle looks across the triangle at one side, and that side gets the same letter in lower case. A wider angle opens onto a longer side.

Example A triangle has $A = 40^\circ$ , $B = 75^\circ$ and $C = 65^\circ$ . The largest angle is $B$ , so the longest side is $b$ . The smallest angle is $A$ , so the shortest side is $a$ . In order of length, $a < c < b$ . This ordering is a quick check on any side lengths you compute later.

The law of sines

Definition In any triangle $ABC$ with sides $a$ , $b$ and $c$ opposite the angles $A$ , $B$ and $C$ ,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.

The common value equals $2R$ , where $R$ is the radius of the circle through the three vertices. To see the first equality, drop the height $h$ from $C$ to the line through $A$ and $B$ . The two right triangles it creates give $h = b \sin A$ and $h = a \sin B$ , so $b \sin A = a \sin B$ . This also holds when $A$ or $B$ is obtuse, because $\sin(180^\circ - \theta) = \sin \theta$ , and when one of them is a right angle, because its sine is $1$ and the height is then a side of the triangle. When you solve for an angle, the same law with every ratio turned upside down is handier,

\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.

In plain words Each side divided by the sine of the angle across from it gives the same number, in every triangle. So once you know one side together with its opposite angle, any other side follows as soon as its opposite angle is known. Going the other way, a known side gives only the sine of the angle across from it, and that can leave two choices for the angle.

Example Check the law on the triangle with angles $30^\circ$ , $60^\circ$ and $90^\circ$ , whose opposite sides are $1$ , $\sqrt{3}$ and $2$ .

\begin{gathered} \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2, \\[0.8em] \frac{\sqrt{3}}{\sin 60^\circ} = \frac{\sqrt{3}}{\sqrt{3}/2} = 2, \\[0.8em] \frac{2}{\sin 90^\circ} = 2. \end{gathered}

All three ratios equal $2$ . The circle through the vertices of a right triangle has the hypotenuse as a diameter, so $2R = 2$ here, just as the law says.

AAS and ASA

Definition In the AAS case you know two angles and a side that is not between them. In the ASA case you know two angles and the side between them, called the included side. In both cases, if the two known angles are positive with a sum less than $180^\circ$ and the known side is positive, there is exactly one triangle. Find the third angle from

A + B + C = 180^\circ,

then find each unknown side with the law of sines, using the known side and its opposite angle as the partner pair.

In plain words Two angles fix the shape of the triangle and one side fixes its size. Once all three angles are known, every side follows from the law of sines.

Example Back at the river, mark points $A$ and $B$ on your bank $120$ meters apart and let $C$ be the tree. You measure $A = 72^\circ$ and $B = 63^\circ$ , and the known side $c = 120$ lies between them, so this is ASA. The third angle is $C = 45^\circ$ , since the angles add to $180^\circ$ . The side $b$ runs from $A$ to the tree and $a$ runs from $B$ to the tree, so

\begin{gathered} b = \frac{120 \sin 63^\circ}{\sin 45^\circ} \approx 151.2, \\[0.8em] a = \frac{120 \sin 72^\circ}{\sin 45^\circ} \approx 161.4. \end{gathered}

The tree is about $151.2$ meters from $A$ and about $161.4$ meters from $B$ . The order $c < b < a$ matches the order $C < B < A$ of the angles.

The SSA case

Definition In the SSA case you know two sides and an angle opposite one of them, say $A$ , $a$ and $b$ , with $0^\circ < A < 180^\circ$ . The law of sines gives

\sin B = \frac{b \sin A}{a}.

If this number is greater than $1$ , no angle $B$ exists. If it equals $1$ , the only candidate is $B = 90^\circ$ . If it lies strictly between $0$ and $1$ , two angles between $0^\circ$ and $180^\circ$ have this sine,

\begin{aligned} B_1 &= \arcsin\left(\frac{b \sin A}{a}\right) \quad \text{and} \\[0.8em] B_2 &= 180^\circ - B_1. \end{aligned}

A candidate $B$ gives a triangle exactly when $A + B < 180^\circ$ . The inverse sine on a calculator returns only $B_1$ , which lies between $0^\circ$ and $90^\circ$ , so you have to test $B_2$ yourself.

In plain words Draw angle $A$ and mark $C$ on one of its arms at distance $b$ . Side $a$ then swings around $C$ like the arm of a compass, and its tip has to land on the other arm of angle $A$ . It may miss that arm, touch it, or cross it at one point or at two, which is why this case is called ambiguous.

Example Take $A = 30^\circ$ , $b = 10$ and $a = 6$ . Then

\sin B = \frac{10 \sin 30^\circ}{6} = \frac{5}{6},

so $B_1 \approx 56.44^\circ$ and $B_2 \approx 123.56^\circ$ . Both pass the test, since even $30^\circ + 123.56^\circ$ is less than $180^\circ$ . Finishing each triangle with the law of sines gives the table below.

Quantity	First triangle	Second triangle
$B$	$56.44^\circ$	$123.56^\circ$
$C = 150^\circ - B$	$93.56^\circ$	$26.44^\circ$
$c = 6 \sin C / \sin 30^\circ$	$11.98$	$5.34$

Both triangles have $A = 30^\circ$ , $b = 10$ and $a = 6$ , yet they have different shapes.

Zero, one or two triangles

Definition Suppose $A$ , $a$ and $b$ are given with $0^\circ < A < 180^\circ$ , $a > 0$ and $b > 0$ , and let $h = b \sin A$ , the distance from $C$ to the line along the other arm of angle $A$ . The number of triangles is decided by these rules.

If $A$ is acute and $a < h$ , there is no triangle.
If $A$ is acute and $a = h$ , there is exactly one triangle, a right triangle with $B = 90^\circ$ .
If $A$ is acute and $h < a < b$ , there are exactly two triangles.
If $A$ is acute and $a \ge b$ , there is exactly one triangle.
If $A$ is right or obtuse, there is exactly one triangle when $a > b$ and none when $a \le b$ .

In plain words For an acute angle, compare $a$ with the height $h$ and with $b$ . Shorter than $h$ gives no triangle, exactly $h$ gives a right triangle, between $h$ and $b$ gives two, and $b$ or longer gives one. For a right or obtuse angle, the side across from it must be the longest side.

Example Fix $A = 40^\circ$ and $b = 8$ , so $h = 8 \sin 40^\circ \approx 5.14$ , and try three values of $a$ .

$a$	Comparison	Triangles
$4$	$a < h$	none, and indeed $\sin B \approx 1.29 > 1$
$6$	$h < a < b$	two, with $B \approx 58.99^\circ$ or $B \approx 121.01^\circ$
$9$	$a \ge b$	one, with $B \approx 34.85^\circ$

For $a = 9$ the second candidate $180^\circ - 34.85^\circ = 145.15^\circ$ fails, because $40^\circ + 145.15^\circ$ is more than $180^\circ$ .

Check yourself

In triangle $ABC$ the angles are $A = 40^\circ$ and $B = 65^\circ$ , and the side $a = 10$ . Find $C$ , $b$ and $c$ .

Show the answer

This is AAS. The third angle is $C = 75^\circ$ , and the law of sines gives

\begin{gathered} b = \frac{10 \sin 65^\circ}{\sin 40^\circ} \approx 14.10, \\[0.8em] c = \frac{10 \sin 75^\circ}{\sin 40^\circ} \approx 15.03. \end{gathered}

The order $a < b < c$ matches $A < B < C$ .

How many triangles have $A = 120^\circ$ , $a = 5$ and $b = 5.5$ ?

Show the answer

None. The angle $A$ is obtuse, so $a$ would have to be the longest side, but $a < b$ . The law of sines agrees. It gives $\sin B \approx 0.953$ , so $B \approx 72.3^\circ$ or $B \approx 107.7^\circ$ , and either value makes $A + B$ more than $180^\circ$ .

Take $A = 35^\circ$ , $b = 12$ and $a = 8$ . How many triangles are there? Find every possible value of $B$ to two decimal places.

Show the answer

The height is $h = 12 \sin 35^\circ \approx 6.88$ , and $h < a < b$ , so there are two triangles. From

\sin B = \frac{12 \sin 35^\circ}{8} \approx 0.8604

the candidates are $B \approx 59.36^\circ$ and $B \approx 120.64^\circ$ . Both work, since $35^\circ + 120.64^\circ$ is less than $180^\circ$ .

The Law of Sines and the Ambiguous Case

Key terms

Oblique triangle

Sides opposite angles

The law of sines

AAS and ASA

The SSA case

Zero, one or two triangles

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus