Pythagoras with a correction term, for triangles that have no right angle.
The problem
Two sides measure 8 and 11, with an angle of 40° between them. Find the third side.
c² = 8² + 11² − 2(8)(11)cos 40° = 185 − 176(0.766) ≈ 50.2, so c ≈ 7.1. D is c² left un-rooted. With only 40° between them the sides nearly fold together, so the third side is shorter than either.
The Law of Sines needs a side paired with its opposite angle. Here there is none.
two sides and the angle between (SAS)
all three sides (SSS)
a complete side-angle pair
c² = a² + b² − 2ab · cos C
c is the side you want; C is the angle facing it
Drawn to scale: the 40° between the 8 and the 11 is what fixes the far side c.
c² = 8² + 11² − 2(8)(11) cos 40°
a = 8, b = 11, C = 40°
= 64 + 121 − 176(0.766)
cos 40° = 0.766
= 185 − 134.8
the correction
c² = 50.2
not the answer yet
c ≈ 7.08
take the square root
The whole term 2ab cos C is subtracted as one lump — do the multiplication before subtracting. And the formula gives c2; forgetting the final square root leaves you with 50.2.
if C = 90°
a right angle
cos 90° = 0
the term vanishes
c² = a² + b²
the old theorem
cos is positive
c comes out shorter
cos is negative
subtracting a negative — c grows
Squeeze the angle and the far side shrinks. Open it out and the far side stretches.
Rearrange the same formula.
cos C = a² + b² − c²2ab
sides 5, 7, 9 — find the largest angle
it faces the 9
cos C = (25 + 49 − 81) ÷ (2 · 5 · 7)
c = 9
= −7 ÷ 70 = −0.1
negative, so obtuse
C = cos⁻¹(−0.1) ≈ 95.7°
just past a right angle
A negative cosine always means an obtuse angle — a useful check before you press the button.
| You are given | Use |
|---|---|
| two angles and any side | Sines |
| two sides and the angle between | Cosines |
| all three sides | Cosines |
| two sides and an angle not between | Sines — watch the ambiguity |
Cosines opens the door when nothing pairs up. After one use you usually have a complete pair, and Sines finishes the job faster.
Bearings and navigation questions almost always start with two distances and the angle between them — a Law of Cosines setup.
Two sides are 3√2 and 5, with a 45° angle between them. Find the third side.
c² = (3√2)² + 5² − 2(3√2)(5) cos 45°
SAS
= 18 + 25 − 30√2 × (√22)
(3√2)² = 9 × 2
= 43 − 30 = 13
√2 × √2 = 2, and 30 × 2 ÷ 2 = 30
c = √13
≈ 3.6
Different: a side with a root pairs with cos 45° so the roots cancel, and the answer is itself a root.
A triangle has sides 9, 12 and 15. Find its largest angle.
cos C = (81 + 144 − 225) ÷ (2 · 9 · 12)
C faces the 15
= 0 ÷ 216 = 0
the top vanishes
C = 90°
a 3-4-5 triangle, × 3
Different: a cosine of exactly 0 means a right angle — positive is acute, negative is obtuse, zero is the line between.
Can a triangle have sides 3, 4 and 8? Find the angle facing the 8.
cos C = (9 + 16 − 64) ÷ (2 · 3 · 4)
SSS
= −39 ÷ 24 = −1.625
below −1
no such triangle
a cosine lies between −1 and 1
Different: the formula flags the impossible triangle — 3 + 4 is less than 8, so the sides never meet. A calculator shows an error for cos⁻¹(−1.625); that error is the answer.
Parallelogram ABCD has AB = 6, AD = 4 and ∠A = 60°. Find both diagonals.
BD² = 6² + 4² − 2(6)(4) cos 60° = 52 − 24 = 28
BD faces the 60° at A
BD = √28 = 2√7
the shorter diagonal
AC² = 6² + 4² − 2(6)(4) cos 120° = 52 + 24 = 76
∠B = 180 − 60 = 120°, and BC = AD = 4
AC = √76 = 2√19
the longer diagonal
Different: each diagonal closes a triangle made of two sides of the parallelogram. Consecutive angles add to 180°, so one diagonal faces 60° and the other 120°.
A triangle has a 120° angle. The two sides along it are 5 and x, and the side facing it is 19. Find x. The formula still works — the unknown just lands in two terms.
19² = x² + 5² − 2(x)(5) cos 120°
the facing side goes on the left
361 = x² + 25 + 5x
cos 120° = −½, so − × − = +
x² + 5x − 336 = 0
set to zero
x = (−5 + √25 + 1344) ÷ 2 = (−5 + 37) ÷ 2
quadratic formula; √1369 = 37
x = 16
the negative root is not a length
Check: 16² + 5² + 5 × 16 = 256 + 25 + 80 = 361 = 19² ✓
The side on the left of c² = a² + b² − 2ab cos C is always the one opposite C. Putting x there instead (x² = 361 + 25 − …) builds a different triangle altogether.
Expect 60° or 120° (cos = ½ or −½) so the arithmetic is exact: two ships or two roads with the angle between, the diagonals of a parallelogram, the largest or smallest angle from three sides, or a side solved from a quadratic. Watch the sign — a 120° angle adds ab instead of subtracting it. Ten-second check: the side facing an obtuse angle is the longest, longer than √a² + b².
The problem
Two sides of 6 and 10 meet at 120°. Find the third side.
cos 120° = −0.5, so the term −2(6)(10)cos 120° comes out as +60. Then c² = 36 + 100 + 60 = 196 and c = 14. C uses +0.5 — an obtuse angle has a negative cosine, which pushes the opposite side longer.
