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Amy Ferguson Moncure

IB AA · Unit 2 · 18B Trigonometric Functions

The Ambiguous Case

Two sides and an angle that is not between them can make two triangles, one triangle, or none.

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Key Ideas Easy Medium Hard

asin A = bsin B

h = b sin A

Area = 12 ab sin C

What to Remember

Compare the side with the height

h = b sin A

If a is shorter than h, there is no triangle.

If a equals h, angle B is 90°.

If h < a < b, there are two triangles.

If a is at least b, and A is acute, there is one triangle.

The supplement

sin(180° − B) = sin B, so a calculator’s acute angle has a partner.

Keep the partner only when the three angles still add to 180°.

An obtuse given angle leaves room for only an acute partner.

Area gives two angles

sin C = 2 × areaa b

Both the acute angle and its supplement can be the included angle.

Easy

One triangle

The opposite side is longer than the other given side, so only the acute angle fits.

Example
In triangle ABC, angle A = 35°, side a = 11 cm, and side b = 8 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.

h = b sin A

h = 8 × sin 35°

h = 4.59 cm

sin Bb = sin Aa

sin B = 8 × sin 35°11

sin B = 0.417

B = 24.7°

To the nearest degree, B = 25°.

The supplement is 155.3°.

To the nearest degree, that supplement is 155°.

180° − 35° − 24.7° = 120.3°

35° + 155° = 190°, which is more than 180°.

Only the acute angle fits.

Answer: 25°

1
In triangle ABC, angle A = 42°, side a = 13 cm, and side b = 10 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 10 × sin 42°

h = 6.69 cm

sin Bb = sin Aa

sin B = 10 × sin 42°13

sin B = 0.515

B = 31.0°

To the nearest degree, B = 31°.

The supplement is 149.0°.

To the nearest degree, that supplement is 149°.

180° − 42° − 31.0° = 107.0°

42° + 149° = 191°, which is more than 180°.

Only the acute angle fits.

Answer: 31°

2
In triangle ABC, angle A = 48°, side a = 11 cm, and side b = 8 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 8 × sin 48°

h = 5.95 cm

sin Bb = sin Aa

sin B = 8 × sin 48°11

sin B = 0.540

B = 32.7°

To the nearest degree, B = 33°.

The supplement is 147.3°.

To the nearest degree, that supplement is 147°.

180° − 48° − 32.7° = 99.3°

48° + 147° = 195°, which is more than 180°.

Only the acute angle fits.

Answer: 33°

3
In triangle ABC, angle A = 55°, side a = 13 cm, and side b = 10 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 10 × sin 55°

h = 8.19 cm

sin Bb = sin Aa

sin B = 10 × sin 55°13

sin B = 0.630

B = 39.1°

To the nearest degree, B = 39°.

The supplement is 140.9°.

To the nearest degree, that supplement is 141°.

180° − 55° − 39.1° = 85.9°

55° + 141° = 196°, which is more than 180°.

Only the acute angle fits.

Answer: 39°

4
In triangle ABC, angle A = 30°, side a = 12 cm, and side b = 10 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 10 × sin 30°

h = 5 cm

sin Bb = sin Aa

sin B = 10 × sin 30°12

sin B = 0.417

B = 24.6°

To the nearest degree, B = 25°.

The supplement is 155.4°.

To the nearest degree, that supplement is 155°.

180° − 30° − 24.6° = 125.4°

30° + 155° = 185°, which is more than 180°.

Only the acute angle fits.

Answer: 25°

5
In triangle ABC, angle A = 40°, side a = 14 cm, and side b = 9 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 9 × sin 40°

h = 5.79 cm

sin Bb = sin Aa

sin B = 9 × sin 40°14

sin B = 0.413

B = 24.4°

To the nearest degree, B = 24°.

The supplement is 155.6°.

To the nearest degree, that supplement is 156°.

180° − 40° − 24.4° = 115.6°

40° + 156° = 196°, which is more than 180°.

Only the acute angle fits.

Answer: 24°

Medium

None, one right angle, or two

Start with the height. Then test the supplement.

Example
In triangle ABC, angle A = 28°, side a = 5 cm, and side b = 9 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.

h = b sin A

h = 9 × sin 28°

h = 4.23 cm

sin Bb = sin Aa

sin B = 9 × sin 28°5

sin B = 0.845

B = 57.7°

To the nearest degree, B = 58°.

The supplement is 122.3°.

To the nearest degree, that supplement is 122°.

180° − 28° − 57.7° = 94.3°

180° − 28° − 122.3° = 29.7°

Both third angles are positive, so both triangles work.

Answer: 58° or 122°

1
In triangle ABC, angle A = 40°, side a = 5 cm, and side b = 12 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 12 × sin 40°

h = 7.71 cm

Side a is 5 cm, which is shorter than the height.

The side does not meet the base. There is no triangle.

Answer: no triangle

2
In triangle ABC, angle A = 50°, side a = 6 cm, and side b = 11 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 11 × sin 50°

h = 8.43 cm

Side a is 6 cm, which is shorter than the height.

The side does not meet the base. There is no triangle.

Answer: no triangle

3
In triangle ABC, angle A = 30°, side a = 5 cm, and side b = 10 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 10 × sin 30°

sin 30° = 12

h = 10 × 12

h = 5 cm

Side a equals the height, so angle B is a right angle.

B = 90°

C = 180° − 30° − 90°

C = 60°

Answer: 90°

4
In triangle ABC, angle A = 28°, side a = 7 cm, and side b = 13 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 13 × sin 28°

h = 6.10 cm

sin Bb = sin Aa

sin B = 13 × sin 28°7

sin B = 0.872

B = 60.7°

To the nearest degree, B = 61°.

The supplement is 119.3°.

To the nearest degree, that supplement is 119°.

180° − 28° − 60.7° = 91.3°

180° − 28° − 119.3° = 32.7°

Both third angles are positive, so both triangles work.

Answer: 61° or 119°

5
In triangle ABC, angle A = 28°, side a = 8 cm, and side b = 14 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 14 × sin 28°

h = 6.57 cm

sin Bb = sin Aa

sin B = 14 × sin 28°8

sin B = 0.822

B = 55.2°

To the nearest degree, B = 55°.

The supplement is 124.8°.

To the nearest degree, that supplement is 125°.

180° − 28° − 55.2° = 96.8°

180° − 28° − 124.8° = 27.2°

Both third angles are positive, so both triangles work.

Answer: 55° or 125°

Hard

Check someone else’s work

A sine greater than 1 can mean the sides were swapped. An angle sum over 180° rejects a supplement.

Example
In triangle ABC, angle B = 42°, AB = 8 cm, and AC = 15 cm. AB is opposite angle C, and AC is opposite angle B. Three students found angle C. Which solution is correct, and what is wrong with the other two?

Solution 1

sin C8 = sin 42°15

sin C = 8 × sin 42°15

C = 20.9°

So C = 21°.

Solution 2

sin C15 = sin 42°8

sin C = 1.25

1.25 > 1, so there is no triangle.

Solution 3

sin C = 0.357

C = 20.9° or C = 159.1°

So C = 21° or 159°.

Pair each side with the angle opposite it.

ABsin C = ACsin B

8sin C = 15sin 42°

sin C = 8 × sin 42°15

sin C = 0.357

C = 20.9°

To the nearest degree, C = 21°.

The supplement is 159.1°.

42° + 159.1° = 201.1°, which is more than 180°.

Solution 1 is the one that fits.

Solution 2 swaps the two sides, so the sine comes out greater than 1. That is a pairing error. The triangle does exist.

Solution 3 keeps an angle that cannot sit next to 42°.

Answer: Solution 1, C = 21°

1
In triangle ABC, side a = 7 cm, side b = 10 cm, and the area is 28 cm². Find the possible sizes of the included angle C, to the nearest degree.
Show Answer

Area = 12 ab sin C

28 = 12 × 7 × 10 × sin C

sin C = 567 × 10

sin C = 45

sin C = 0.8

C = 53.1°

To the nearest degree, C = 53°.

The supplement is 126.9°.

To the nearest degree, that supplement is 127°.

Either angle can sit between the two given sides.

Answer: 53° or 127°

2
In triangle ABC, side a = 6 cm, side b = 11 cm, and the area is 24 cm². Find the possible sizes of the included angle C, to the nearest degree.
Show Answer

Area = 12 ab sin C

24 = 12 × 6 × 11 × sin C

sin C = 486 × 11

sin C = 811

sin C = 0.727

C = 46.7°

To the nearest degree, C = 47°.

The supplement is 133.3°.

To the nearest degree, that supplement is 133°.

Either angle can sit between the two given sides.

Answer: 47° or 133°

3
In triangle ABC, angle A = 40°, side a = 8 cm, and side c = 10 cm. Side a is opposite angle A. Find the two possible lengths of side b.
Show Answer

Angle C is opposite side c = 10 cm.

sin Cc = sin Aa

sin C = 10 × sin 40°8

sin C = 0.803

C = 53.46°

B = 180° − 40° − 53.46°

B = 86.54°

b = 8 × sin 86.54°sin 40°

b = 12.4 cm

C = 126.54°

B = 180° − 40° − 126.54°

B = 13.46°

b = 8 × sin 13.46°sin 40°

b = 2.90 cm

Answer: 12.4 cm or 2.90 cm

4
In triangle ABC, angle A = 120°, side a = 14 cm, and side b = 10 cm. Show that there is only one possible triangle, and find the length of side c.
Show Answer

h = b sin A

h = 10 × sin 120°

h = 8.66 cm

sin Bb = sin Aa

sin B = 10 × sin 120°14

sin B = 0.619

B = 38.2°

To the nearest degree, B = 38°.

The supplement is 141.8°.

To the nearest degree, that supplement is 142°.

Angle A is obtuse, so angle B must be acute.

180° − 120° − 38.2° = 21.8°

120° + 142° is more than 180°, so the obtuse choice for B does not fit.

C = 180° − 120° − 38.21°

C = 21.79°

c = a × sin Csin A

c = 14 × sin 21.79°sin 120°

c = 6.00 cm

Answer: one triangle, c = 6.00 cm

5
In triangle ABC, angle A = 100°, side a = 15 cm, and side b = 8 cm. Side a is opposite angle A. Find the possible sizes of angle B, to the nearest degree.
Show Answer

h = b sin A

h = 8 × sin 100°

h = 7.88 cm

sin Bb = sin Aa

sin B = 8 × sin 100°15

sin B = 0.525

B = 31.7°

To the nearest degree, B = 32°.

The supplement is 148.3°.

To the nearest degree, that supplement is 148°.

Angle A is obtuse, so angle B must be acute.

180° − 100° − 31.7° = 48.3°

100° + 148° is more than 180°, so the obtuse choice for B does not fit.

Answer: 32°

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