Precalculus · G.SRT.D.11 · 7.G.A.2

The ambiguous case

Two sides and an angle that is not between them sometimes build no triangle, sometimes one, sometimes two. How do you tell which before you start, and how do you find them all?

You need

Reference anglesPrecalculus Law of sinesPrecalculus

Where you are 6 in Laws of sines and cosines

Trigonometry7

Laws of sines and cosines6

Law of sines Area from two sides and the included angle

The ambiguous case

Law of cosines Heron's formula Applications of oblique triangles

When you reach for it

You know two sides of a triangle and an angle that is not between them: side–side–angle, or SSA. In triangle ABC that means you know the angle A, the side a across from it, and a second side b. (Why AAA and SSA fail showed with compass and ruler that these parts do not pin down one triangle.) This lesson settles the question with trigonometry: how many triangles fit, zero, one or two, and what each one is. The law of sines does the solving, but on its own it hides the answer to “how many”, because it finds the angle B from its sine, and one sine belongs to two angles. The height test answers “how many” first, from one number.

Because the count can be two, SSA is called the ambiguous case.

The procedure

The lesson runs in two parts. Part 1 is the height test with an acute given angle: zero, one or two triangles. Part 2 adds an obtuse given angle, solves both triangles when there are two, and reads the same count straight off the law of sines.

Part 1: the side that swings

Figure 1The lead problem, A = 35^\circ, b = 12, a = 8. The side a swings from C; because h < a < b it reaches the ray twice.

Draw what is known. The angle A is two rays from the vertex A. Side b runs from A along one ray to the vertex C, so C is fixed. The vertex B must lie on the other ray, and it must be exactly a from C. Picture side a as a rod hinged at C, swinging; every place its free end touches the ray is a triangle. In the language of circles: the circle of radius a around C meets the ray zero, one or two times.

How far C is from the ray decides it. Drop the perpendicular from C to the line of that ray; its length is the height h. In the right triangle with hypotenuse b and angle A, the side opposite A is h, so

h = b \sin A.

Now walk along the ray from A and watch the distance to C. With A acute, the foot of the perpendicular lies on the ray. At A the distance is b. It shrinks to its smallest value, h, at the foot, and then grows without end. Each distance between h and b is passed once before the foot and once after it; each distance above b is passed only after the foot. That walk gives every case.

Acute A	Triangles	Why
a < h	0	the swinging side is too short to reach the ray
a = h	1	it just touches, at the foot: a right angle at B
h < a < b	2	it lands once on each side of the foot
a \ge b	1	the landing before the foot would be at or behind A

When a = b the would-be first landing is A itself, which collapses the triangle to a segment, so it does not count; the one triangle left is isosceles.

Part 2: an obtuse angle, and reading the sine

Figure 2An obtuse given angle, A = 115^\circ, b = 13, a = 20. The second crossing lands behind A, off the ray, so only one triangle closes.

An obtuse or right given angle. If A \ge 90^\circ, the foot of the perpendicular from C lies at A or behind it, on the extension of the ray. Walking out along the ray from A, the distance to C only grows, starting from b. So the swinging side lands on the ray once if a > b, and never if a \le b:

Obtuse or right A	Triangles
a \le b	0
a > b	1

This matches a fact from geometry: an obtuse or right angle is the largest angle of its triangle, so the side across from it must be the longest side.

The same count from the law of sines. The law gives \dfrac{\sin B}{b} = \dfrac{\sin A}{a}, so

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

The height is already inside the formula, so the sine reports the count on its own:

\sin B > 1 means a < h. No angle has that sine: no triangle.
\sin B = 1 means a = h. Then B = 90^\circ: one right triangle.
\sin B < 1 gives two candidates. The inverse sine returns the acute one, B_1. The other is its supplement, B_2 = 180^\circ - B_1, because \sin(180^\circ - x) = \sin x (Unit circle symmetry and even-odd identities; reference angles give the same fact). Supplement check: a candidate is a real angle of the triangle only if A + B < 180^\circ, so that a positive angle C is left over. Two survivors give two triangles, one gives one, none gives none.

Each surviving B finishes like an AAS triangle: C = 180^\circ - A - B, then c = \dfrac{a \sin C}{\sin A}.

The height test and the sine reading always agree, since \sin B = \dfrac{h}{a}. Use the height test to know what you are looking for, and the law to find it. Keep full calculator values and round at the end; answers are to two decimal places.

Worked examples

Examples 1 to 3 belong to Part 1 (counting with the height). Examples 4 to 7 belong to Part 2 (solving, the obtuse angle, and the supplement check).

Example 1

Counting the lead triangle

In triangle ABC, A = 35^\circ, a = 8 and b = 12. How many triangles have these parts?

Solution

The angle A is acute, so compare a with the height and with b. The height from C to the other ray is

h = b \sin A = 12 \sin 35^\circ \approx 6.88.

Then 6.88 < 8 < 12, that is h < a < b. The side a is long enough to reach the ray but shorter than b, so it lands once on each side of the foot of the height, as in Figure 1. Two triangles. Example 4 solves both.

Example 2

One angle, five lengths

Keep A = 30^\circ and b = 14. Count the triangles for a = 5, a = 7, a = 10, a = 14 and a = 20.

Solution

The height is exact here: h = 14 \sin 30^\circ = 14 \cdot \tfrac{1}{2} = 7. Compare each a with h = 7 and b = 14.

a = 5: 5 < 7, too short to reach the ray. No triangle.
a = 7: a = h, the side just touches at the foot. One triangle, with a right angle at B.
a = 10: 7 < 10 < 14. Two triangles.
a = 14: a = b. The first landing would be A itself, which is no triangle. One triangle, isosceles with a = b, so B = A = 30^\circ.
a = 20: 20 > 14, the first landing would fall behind A. One triangle.

Only lengths strictly between 7 and 14 give two.

Example 3

Placing a regatta mark

Bamidele is laying out a triangular regatta course. The windward mark W is 1{,}200 meters from the start buoy S. The third mark L must lie on a line from S that makes an angle of 25^\circ with SW. For which lengths of the leg WL are there two possible places for L? For which is there exactly one?

Solution

In triangle SWL the known angle is S = 25^\circ, the side from S to W is 1{,}200, and WL is the side across from S: SSA, with WL swinging from W. The height from W to the line of L is

h = 1200 \sin 25^\circ \approx 507.14 \text{ meters}.

Two places need h < WL < 1200: WL between about 507.14 and 1{,}200 meters. Exactly one place comes when WL \approx 507.14 meters (the leg meets the line at a right angle) or when WL \ge 1200 meters. A shorter leg cannot reach the line.

Example 4

Solving both triangles

Solve the triangle with A = 35^\circ, a = 8 and b = 12 from Example 1.

Solution

The law of sines gives the angle across from b:

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

This is less than 1, so there are two candidates. The inverse sine gives B_1 \approx 59.36^\circ, and its supplement is B_2 = 180^\circ - B_1 \approx 120.64^\circ. Both pass the supplement check: 35^\circ + 59.36^\circ and 35^\circ + 120.64^\circ are under 180^\circ. Two triangles, as Example 1 said.

Triangle 1. C_1 = 180^\circ - 35^\circ - 59.36^\circ = 85.64^\circ, and

c_1 = \frac{8 \sin 85.64^\circ}{\sin 35^\circ} \approx 13.91.

Triangle 2. C_2 = 180^\circ - 35^\circ - 120.64^\circ = 24.36^\circ, and

c_2 = \frac{8 \sin 24.36^\circ}{\sin 35^\circ} \approx 5.75.

So B \approx 59.36^\circ, C \approx 85.64^\circ, c \approx 13.91, or B \approx 120.64^\circ, C \approx 24.36^\circ, c \approx 5.75. In Figure 1 the acute B_1 is the far landing and the obtuse B_2 the near one. Check: the two landings are \displaystyle \sqrt{a^2 - h^2} on either side of the foot, so they are \displaystyle 2\sqrt{8^2 - h^2} \approx 8.155 apart, and indeed, to three places, c_1 - c_2 \approx 13.907 - 5.752 = 8.155.

Example 5

Reading the sine

With A = 50^\circ and b = 10, use the law of sines to count and solve the triangles for (a) a = 6, (b) a = 12.

Solution

(a) \sin B = \dfrac{10 \sin 50^\circ}{6} \approx 1.2767. No angle has a sine above 1: no triangle. The height test agrees, since h = 10 \sin 50^\circ \approx 7.66 and a = 6 is shorter.

(b) \sin B = \dfrac{10 \sin 50^\circ}{12} \approx 0.6384, so B_1 \approx 39.67^\circ and B_2 \approx 140.33^\circ. The supplement check fails for B_2: 50^\circ + 140.33^\circ = 190.33^\circ, more than 180^\circ. One triangle, which the height test predicted because a = 12 \ge b = 10. It has B \approx 39.67^\circ, C = 180^\circ - 50^\circ - 39.67^\circ = 90.33^\circ and

c = \frac{12 \sin 90.33^\circ}{\sin 50^\circ} \approx 15.66.

Example 6

An obtuse given angle

Let A = 115^\circ. Solve the triangle for (a) b = 13 and a = 20, (b) b = 12.5 and a = 12.

Solution

(a) The angle is obtuse and a = 20 > b = 13, so the height test says one triangle (Figure 2). The law confirms it: \sin B = \dfrac{13 \sin 115^\circ}{20} \approx 0.5891, so B_1 \approx 36.09^\circ and B_2 \approx 143.91^\circ, and 115^\circ + 143.91^\circ is far over 180^\circ. With B \approx 36.09^\circ, C = 180^\circ - 115^\circ - 36.09^\circ = 28.91^\circ, and

c = \frac{20 \sin 28.91^\circ}{\sin 115^\circ} \approx 10.67.

(b) Now a = 12 \le b = 12.5: no triangle. Here the law’s warning is quieter. \sin B = \dfrac{12.5 \sin 115^\circ}{12} \approx 0.9441 is below 1, and gives B_1 \approx 70.75^\circ. But 115^\circ + 70.75^\circ = 185.75^\circ, so even the acute candidate fails the supplement check, and its supplement fails by more. With an obtuse given angle, a sine below 1 is not enough; the supplement check must be run on both candidates.

Example 7

Raising a drawbridge

Figure 3Example 7. Both leaf positions fit the measurements; only one is a position the bridge can take.

Folarin stands on the road at P, 30 meters back from the hinge H of a drawbridge whose leaf is 24 meters long. He sights the tip T of the partly raised leaf at an angle of 32^\circ above the road. At what angle \theta above the water is the leaf raised, and how high is its tip?

Solution

Triangle PHT has the angle P = 32^\circ, the side PH = 30 next to it, and the leaf HT = 24 across from it: SSA. The height test: h = 30 \sin 32^\circ \approx 15.90, and 15.90 < 24 < 30, so two triangles fit, both in Figure 3.

The angle at the tip comes from the law of sines:

\sin T = \frac{30 \sin 32^\circ}{24} \approx 0.6624,

so T_1 \approx 41.48^\circ or T_2 \approx 138.52^\circ. The angle of the triangle at the hinge is then H_1 = 180^\circ - 32^\circ - 41.48^\circ = 106.52^\circ or H_2 = 180^\circ - 32^\circ - 138.52^\circ = 9.48^\circ.

That inside angle at H is measured from the road. The leaf’s angle above the water is its supplement, \theta = 180^\circ - H: \theta_1 \approx 73.48^\circ or \theta_2 \approx 170.52^\circ. The second would put the leaf leaning back over the road, almost flat, which a leaf hinged at the water’s edge cannot do. The mathematics allows both triangles; the bridge picks one. So \theta \approx 73.48^\circ, which is also P + T_1, the exterior angle of the triangle at H.

The tip height is 24 \sin 73.48^\circ \approx 23.01 meters. Check: the sightline gives PT = \dfrac{24 \sin 106.52^\circ}{\sin 32^\circ} \approx 43.42, and 43.42 \sin 32^\circ \approx 23.01 as well.

Boundaries

The test is for SSA only. It needs a known angle and the side across from it. With SSS or SAS there is no such pair and the law of cosines takes over, and its inverse cosine is never ambiguous, since cosine separates acute angles (positive) from obtuse ones (negative). With AAS or ASA no angle is ever found from a sine.

The supplement candidate exists only as a candidate. \sin(180^\circ - x) = \sin x offers a second angle every time \sin B < 1; it becomes a triangle only when A + B_2 < 180^\circ. When \sin B = 1 the two candidates coincide at 90^\circ.

The test reads lengths, not the drawing. A sketch of SSA can look like one triangle when there are two, or close up when no triangle exists. Compute h.

A context may reject a triangle the mathematics allows, as the drawbridge did. Both triangles are correct answers to the triangle problem; only the situation chooses between them.

Rounding near the boundary. If a is within rounding of h, a calculator can show \sin B as 0.9999 or 1.0001. Work exactly when the numbers allow, as with \sin 30^\circ = \tfrac{1}{2}.

Where this goes next

The law of cosines solves the SSS and SAS triangles, and it can also solve SSA by turning the problem into a quadratic in c: the quadratic’s zero, one or two positive roots are the same count this lesson reads from the height. The area formula from two sides and the included angle uses the same height h = b \sin A. Applications of oblique triangles meet SSA whenever a distance and a sighting are known instead of two sightings, and there the second triangle is either a real second answer or, as with the drawbridge, ruled out by the setting.

Exercises

Exercises 1 to 8 go with Part 1 (Examples 1 to 3), Exercises 9 to 18 with Part 2 (Examples 4 to 7), and the reasoning group draws on both.

Part 1: counting with the height

For Exercises 1 to 5, find the height h = b \sin A and say how many triangles have the given parts.

A = 40^\circ, b = 15, a = 8.
A = 40^\circ, b = 15, a = 11.
A = 40^\circ, b = 15, a = 17.
A = 45^\circ, \displaystyle b = 10\sqrt{2}, a = 10. Work exactly.
A = 60^\circ, b = 20, a = 20. What kind of triangle is it?
Let A = 25^\circ and b = 30. For which values of a are there no triangles, exactly one, and exactly two?
Damilola checks a second regatta course. The windward mark W is 800 meters from the start S, the third mark must lie on a line from S at 38^\circ to SW, and the leg WL is to be 450 meters. How many places are there for L?
Let A = 30^\circ and a = 9. For which values of b are there two triangles?

Part 2: solving

Solve each triangle, or show that none exists. If there are two, give both.

A = 28^\circ, a = 10, b = 16.
A = 52^\circ, a = 14, b = 11.
A = 64^\circ, a = 9, b = 12.
A = 120^\circ, a = 18, b = 11.
A = 130^\circ, a = 9, b = 10. Note what the sine reports and what the supplement check reports.
A = 90^\circ, a = 13, b = 5.
A = 60^\circ, b = 6, \displaystyle a = 3\sqrt{3}. Work exactly.

Part 2: drawbridges and regattas

Damilola stands on the road 26 meters from the hinge of a drawbridge with a 20-meter leaf, and sights the tip at 35^\circ above the road. How many triangles fit, and at what angle above the water is the leaf raised?
For the bridge in Exercise 16, at what sighting angle from Damilola’s spot would the sightline meet the leaf at a right angle, giving exactly one triangle?
Bamidele’s next course has SW = 1{,}500 meters, the line for L at 30^\circ to SW, and WL = 1{,}000 meters. Find both possible distances SL, to the nearest meter.

Reasoning

Show that \sin B > 1 in the law of sines happens exactly when a < h, and \sin B = 1 exactly when a = h.
Explain, by walking along the ray, why an acute angle A with a \ge b never gives two triangles.
Explain why an obtuse given angle needs a > b for a triangle to exist, and why it can never give two.
When A is acute and a = b, the law of sines gives B_1 = A and B_2 = 180^\circ - A. Explain what goes wrong with the second.
The inverse sine returns only the acute candidate B_1. Give a set of parts from this lesson’s exercises that has a second triangle besides the one with B_1, and one for which B_1 gives the only triangle. How does the height test tell the two apart before any inverse sine is taken?

Answers

h \approx 9.64; none. h = 15 \sin 40^\circ and 8 < 9.64.
h \approx 9.64; two. 9.64 < 11 < 15.
h \approx 9.64; one. 17 \ge 15, so the landing before the foot falls behind A.
h = 10; one, a right triangle with B = 90^\circ. \displaystyle h = 10\sqrt{2} \cdot \dfrac{\sqrt{2}}{2} = 10 = a.
h \approx 17.32; one, equilateral. a = b gives B = A = 60^\circ, so C = 60^\circ too.
h = 30 \sin 25^\circ \approx 12.68. None for a < 12.68; one for a \approx 12.68 (exactly a = 30 \sin 25^\circ) or a \ge 30; two for 12.68 < a < 30.
None. h = 800 \sin 38^\circ \approx 492.53 meters, and 450 < 492.53: the leg cannot reach the line.
9 < b < 18. Two triangles need h < a < b with h = b \sin 30^\circ = \dfrac{b}{2}: \dfrac{b}{2} < 9 gives b < 18, and 9 < b.
Two triangles. \sin B = \dfrac{16 \sin 28^\circ}{10} \approx 0.7512. B \approx 48.69^\circ, C \approx 103.31^\circ, c \approx 20.73; or B \approx 131.31^\circ, C \approx 20.69^\circ, c \approx 7.53. Here c = \dfrac{10 \sin C}{\sin 28^\circ}.
One triangle: B \approx 38.25^\circ, C \approx 89.75^\circ, c \approx 17.77. \sin B = \dfrac{11 \sin 52^\circ}{14} \approx 0.6192; the supplement 141.75^\circ fails, since 52^\circ + 141.75^\circ > 180^\circ.
None. \sin B = \dfrac{12 \sin 64^\circ}{9} \approx 1.1984 > 1; equally h \approx 10.79 > 9.
One triangle: B \approx 31.95^\circ, C \approx 28.05^\circ, c \approx 9.77. A is obtuse and 18 > 11; \sin B = \dfrac{11 \sin 120^\circ}{18} \approx 0.5292, c = \dfrac{18 \sin 28.05^\circ}{\sin 120^\circ}.
None. The sine is fine, \sin B = \dfrac{10 \sin 130^\circ}{9} \approx 0.8512, giving B_1 \approx 58.34^\circ, but 130^\circ + 58.34^\circ = 188.34^\circ > 180^\circ, and B_2 fails by more. The height test says the same: obtuse A with a = 9 \le b = 10.
One triangle: B \approx 22.62^\circ, C \approx 67.38^\circ, c = 12. \sin B = \dfrac{5}{13}; c = 13 \sin C with \sin C = \cos B = \dfrac{12}{13}, as the Pythagorean theorem confirms: 5^2 + 12^2 = 13^2.
One right triangle: B = 90^\circ, C = 30^\circ, c = 3. \displaystyle \sin B = \dfrac{6 \cdot \frac{\sqrt{3}}{2}}{3\sqrt{3}} = 1, and \displaystyle c = \dfrac{3\sqrt{3} \sin 30^\circ}{\sin 60^\circ} = 3.
Two triangles fit (h = 26 \sin 35^\circ \approx 14.91 < 20 < 26); the leaf is raised about 83.21^\circ. \sin T = \dfrac{26 \sin 35^\circ}{20} \approx 0.7456, T \approx 48.21^\circ or 131.79^\circ; the hinge angle is 96.79^\circ or 13.21^\circ, so the angle above the water is 83.21^\circ or 166.79^\circ, and only the first is possible.
About 50.28^\circ. One right triangle needs the leaf to equal the height: 20 = 26 \sin P, so P = \sin^{-1}\dfrac{20}{26}.
About 1{,}960 meters or 638 meters. \sin L = \dfrac{1500 \sin 30^\circ}{1000} = 0.75, so L \approx 48.59^\circ or 131.41^\circ, the angle at W is 101.41^\circ or 18.59^\circ, and SL = \dfrac{1000 \sin W}{\sin 30^\circ}.
\sin B = \dfrac{b \sin A}{a} = \dfrac{h}{a}, with h and a positive. So \sin B > 1 exactly when h > a, and \sin B = 1 exactly when h = a.
Along the ray the distance to C is b at A, falls to h at the foot, then rises. Before the foot every distance is at most b, and equals b only at A, which gives no triangle. So a side a \ge b can land only after the foot, once.
With A obtuse, the foot of the perpendicular from C is behind A, so along the ray the distance to C starts at b and only increases. Each length above b is reached exactly once, and no length up to b is reached at all (except b at A itself, a collapsed triangle).
A + B_2 = A + (180^\circ - A) = 180^\circ, leaving C = 0^\circ. The “triangle” has B landing on A: it collapses to the segment AC. Only the isosceles triangle with B_1 = A is real.
Exercise 9 (A = 28^\circ, a = 10, b = 16) has two triangles: besides B_1 \approx 48.69^\circ there is the one with B_2 \approx 131.31^\circ. Exercise 10 (A = 52^\circ, a = 14, b = 11) has one, and B_1 is it. The height test tells them apart before any inverse sine: h < a < b in the first, a \ge b in the second.