The Law of Sines & Cosines — Print

Two rules extend trigonometry to every triangle, not just the right-angled ones.

What the practice questions cover

These questions use 6 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Two surveyors stand at either end of a lake and sight the same tree.

They can measure the distance between themselves, and the angle each one turns through. That is all.

how wide is the lake? far tree 180 m, measured 62° 41° no right angle anywhere — SOH-CAH-TOA has nothing to grip

There is no right angle anywhere in that triangle, so SOH-CAH-TOA has nothing to grip.

Yet the triangle is completely pinned down — two angles and a side leave no freedom at all.

If the shape is determined, some rule must recover the rest. The law of sines is that rule, and it works on every triangle there is.

Labelling an oblique triangle: side a faces angle A A B C c a b a sin A = b sin B each side sits opposite the angle with the same letter

Labelling comes first

Name the angles A, B, C and give each opposite side the matching lower-case letter.

Side a faces angle A. Every formula in this unit depends on that pairing being right.

Mislabelling is the single biggest source of wrong answers here, and it happens before any calculating starts.

Label the picture, not the formula

Write the letters onto the diagram before you reach for a rule. A side paired with the wrong angle gives a confident, wrong answer.

The law of sines: when a side and its angle come as a pair

asin A = bsin B holds in every triangle, so one complete pair unlocks the rest.

Use it when you know an angle and the side opposite it, plus one more piece.

That covers two angles and any side, or two sides and a non-included angle.

The law of sines: in both cases side a and the angle A facing it are known as a pair two angles + a sidegiven A, B and a — find bABCab = ?ca / sin A = b / sin Bany side will do with two anglestwo sides + an anglegiven a, b and A — find BABCabc?a / sin A = b / sin BA is not between a and b blue: a side and the angle facing it — one complete pair unlocks the rest

Look for a matching pair

If you can see a side and the angle facing it, the law of sines applies. If you cannot, you need the law of cosines instead.

The law of cosines: when no pair is available

c2 = a2 + b2 - 2abcos C is Pythagoras with a correction term for the angle not being square.

Use it with two sides and the included angle, or with all three sides and no angles.

When C is obtuse, cos C is negative and the correction adds, making c longer — exactly as the picture suggests.

The law of cosines: two sides and the angle between them, or all three sides — no side comes with the angle facing it two sides + the angle betweengiven a, b and C — find cABCabc = ?c² = a² + b² − 2ab cos Cthree sidesgiven a, b and c — find CABCabc?cos C = (a² + b² − c²) ÷ 2ab no side comes with its own angle; at C = 90° the last term is 0 and it is Pythagoras

The term that does the work

At C = 90° the correction vanishes and you are back to a2 + b2. The whole law is Pythagoras plus a lean.

Choosing between them

Two angles and a side, or a matching side-angle pair: law of sines.

Two sides with the angle between them, or three sides: law of cosines.

Start with the law of cosines when it applies, then finish with the law of sines, which is less arithmetic.

Find the largest angle first

After the law of cosines, use the sine rule on a smaller angle. That avoids the ambiguity, since only one angle in a triangle can be obtuse.

The ambiguous case

Given two sides and an angle not between them, two different triangles can sometimes fit the same measurements.

The reason is the one from the previous unit: sin 30° and sin 150° are equal, so the sine rule cannot tell them apart.

Always check whether the obtuse option still leaves the angles under 180°. If it does, both triangles are real.

Check the second answer

If the sine rule gives B = 40°, also test B = 140°. If A + 140° is still under 180°, there are two valid triangles and the problem has two answers.

Worked examples

Law of sines

In triangle ABC, A = 40°, B = 75° and a = 12. Find b to one decimal place.

  1. a and A are a matching pair, so the law of sines applies. A side with its opposite angle.
  2. bsin 75° = 12sin 40°. Set the two ratios equal.
  3. b = 12 sin 75°sin 40° ≈ 11.590.643 ≈ 18.0. Larger than a, as it should be — B is the bigger angle.

b ≈ 18.0

Law of cosines

A triangle has sides 8 and 11 with an included angle of 37°. Find the third side to one decimal place.

  1. No side-angle pair is available, so use the law of cosines. Two sides and the angle between them.
  2. c2 = 82 + 112 - 2(8)(11)cos 37°= 64 + 121 - 176(0.7986). Substitute directly.
  3. c2 ≈ 185 - 140.6 = 44.4, so c ≈ 6.7. Shorter than both sides, which a narrow included angle should give.

c ≈ 6.7

Finding an angle from three sides

A triangle has sides 5, 7 and 9. Find the largest angle to the nearest degree.

  1. The largest angle faces the longest side, so it is opposite 9. Biggest side, biggest angle.
  2. 81 = 25 + 49 - 2(5)(7)cos C, so 81 = 74 - 70cos C. Rearranged law of cosines.
  3. cos C = -770 = -0.1, giving C ≈ 96°. Negative cosine means obtuse — consistent with 81 > 25 + 49.

96°

Words to know

Oblique triangle
Any triangle with no right angle.
Law of sines
asin A = bsin B = csin C.
Law of cosines
c2 = a2 + b2 - 2abcos C.
Included angle
The angle between two given sides.
Ambiguous case
Two sides and a non-included angle, which can describe two different triangles.

Mistakes to watch for

Pairing a side with the wrong angle. Side a must face angle A. Using asin B produces an answer that looks reasonable and is not, so label the diagram before calculating.

Reaching for the law of sines with two sides and the included angle. There is no complete side-angle pair there, so the sine rule cannot start. That case needs the law of cosines.

Missing the second triangle in the ambiguous case. If sin B = 0.64 gives B = 40°, then B = 140° may also fit. Check whether the angles still total under 180° before discarding it.

Dropping the minus sign in the law of cosines. It is a2 + b2 - 2abcos C, and the subtraction is what makes a narrow angle give a short third side.

Trusting Pythagoras on a triangle with no right angle. 5, 7 and 9 do not satisfy a2+b2=c2, and using it anyway gives a length that no triangle has.

This is how every map was made before satellites. Surveyors walked a baseline, measured angles to distant landmarks, and triangulated outward across whole countries.

The Great Trigonometrical Survey of India ran for seventy years on these two rules, and measured the height of Everest to within a few metres.

The law of cosines contains Pythagoras. Put C = 90° and cos C = 0, and c2 = a2 + b2 - 2abcos C collapses to the familiar form.

Problem 1

In triangle ABC, A = 40° and B = 75°. Find C.

Worked answer

65

180-40-75 = 65

Problem 2

Two angles and one side are known. Which rule applies?

  1. Law of sines
  2. Law of cosines
  3. Pythagoras
  4. Neither
Worked answer

Law of sines

The third angle follows, giving a complete pair

Problem 3

All three sides are known and no angles. Which rule applies?

  1. Law of cosines
  2. Law of sines
  3. Pythagoras
  4. Neither
Worked answer

Law of cosines

No angle is known, so the sine rule cannot start

Problem 4

In triangle ABC, A = 30°, a = 10 and B = 45°. Find b to one decimal place.

Worked answer

14.1

b = 10 sin 45°sin 30°

= 7.0710.5 ≈ 14.1

Problem 5

Can a triangle have sides 3, 4 and 8?

  1. No
  2. Yes
  3. Only if it is obtuse
  4. Only if it is right-angled
Worked answer

No

3+4 = 7, which is less than 8

The short sides cannot reach across

Problem 6

In triangle ABC, A = 50°, B = 60° and a = 8. Find b to one decimal place.

Worked answer

9.0

b = 8sin 60°sin 50°

= 6.9280.766 ≈ 9.0

Problem 7

A triangle has sides 5 and 6 with an included angle of 60°. Find the third side to one decimal place.

Worked answer

5.6

25+36-2(5)(6)(0.5)

= 61-30 = 31

c = √31 ≈ 5.6

Problem 8

A triangle has sides 2 and 3 with an included angle of 120°. Find the third side to one decimal place.

Worked answer

4.4

4+9-2(2)(3)(-0.5)

= 13+6 = 19

c = √19 ≈ 4.4

Problem 9

A triangle has sides 4, 5 and 6. Find the angle opposite the side of length 6, to the nearest degree.

Worked answer

83

36 = 16+25-2(4)(5)cos C

cos C = 540 = 0.125

C ≈ 83°

Problem 10

A triangle has sides 3, 4 and 5. Find the largest angle in degrees.

Worked answer

90

9+16 = 25

So it is right-angled

Problem 11

Find the area of a triangle with sides 6 and 8 and an included angle of 30°.

Worked answer

12

12(6)(8)(0.5)

= 12

Problem 12

In triangle ABC, a = 8, A = 40° and b = 10. Find the acute value of B to the nearest degree.

Worked answer

53

sin B = 10sin 40°8 = 0.8035

B ≈ 53°

Problem 13

With a = 5, b = 8 and A = 30°, how many triangles are possible?

  1. Two
  2. One
  3. None
  4. Three
Worked answer

Two

sin B = 8(0.5)5 = 0.8, so B = 53° or 127°

30+53 and 30+127 are both under 180, so both work

Problem 14

A triangle has sides 7, 8 and 12. Find the largest angle to the nearest degree.

Worked answer

106

144 = 49+64-2(7)(8)cos C

112cos C = 113-144 = -31

cos C ≈ -0.277, so C ≈ 106°

Problem 15

Two observers 100 m apart both sight a tree, at 40° and 65° from the line joining them. Find the distance from the first observer to the tree, to one decimal place.

Worked answer

93.8

Third angle = 180-40-65 = 75°

d = 100sin 65°sin 75° = 90.6396.59 × 100 ≈ 93.8

Problem 16

A triangle has sides 7 and 9 with an included angle of 90°. Find the third side to one decimal place.

Worked answer

11.4

49+81 = 130

c = √130 ≈ 11.4

Exactly Pythagoras

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