C. Trigonometric Functions · Trigonometry of Acute TrianglesDiscovering and Verifying the Cosine Law
C1.3 C1.4About 75 minutes
Learning goals- I can verify the cosine law with measurements and with technology.
- I can explain how the cosine law comes from two right triangles and the Pythagorean theorem.
- I can explain why the cosine law becomes the Pythagorean theorem when the angle is 90°.
- I can use the cosine law to find the third side when I know two sides and the angle between them (SAS).
Words to know
cosine law
In any acute △ABC: a² = b² + c² − 2bc cos A (and similarly b² = a² + c² − 2ac cos B, c² = a² + b² − 2ab cos C).
contained angle
The angle between two given sides (the angle where they meet).
SAS
Two sides and the contained angle.
correction term
The −2bc cos A part of the cosine law. It adjusts the Pythagorean theorem for a non-right angle.
When the sine law gets stuck
A surveyor stands on one side of a pond. She can measure her distance to each end of the pond and the angle between those two lines, but she cannot walk across the water. That is two sides and the angle between them (SAS). The sine law can't start, because no side is known together with its opposite angle.
The cosine law handles this case. It looks like the Pythagorean theorem with an extra piece. (If you took MPM2D, this is a review with a new focus: verifying the law with measurements and technology.)
The cosine law
In any acute △ABC:
a² = b² + c² − 2bc cos A
The angle on the right (A) is across from the side on the left (a). The other two sides (b and c) are the ones that form that angle. By changing letters:
- b² = a² + c² − 2ac cos B
- c² = a² + b² − 2ab cos C
Investigate: does it work?
A student drew an acute triangle and measured b = 7.0 cm, c = 8.0 cm, the contained angle ∠A = 47° and the opposite side a = 6.0 cm. Then she tested the formula with a calculator:
b² + c² − 2bc cos A = 49 + 64 − 2(7.0)(8.0) cos 47° = 113 − 112 × 0.6820... ≈ 36.62
So a ≈ √36.62 ≈ 6.05 cm. The measured value was 6.0 cm: they agree to within measuring error.
With technology: in a spreadsheet, put b, c and ∠A (in degrees) in cells B2, C2 and D2. The formula =SQRT(B2^2+C2^2-2B2C2*COS(RADIANS(D2))) predicts a. In GeoGebra, build a triangle, measure everything, and compare a² with b² + c² − 2bc cos(α) while you drag a vertex: they stay equal.
| b (cm) | c (cm) | ∠A | a measured (cm) | a from cosine law (cm) |
|---|
| 7.0 | 8.0 | 47° | 6.0 | 6.05 |
| 10.0 | 12.0 | 70° | 12.7 | 12.72 |
| 9.0 | 9.0 | 60° | 9.0 | 9.00 |
Why it works
Draw the altitude h from C to AB, meeting it at D. Let AD = x, so DB = c − x.
- In right △ACD: h² = b² − x², and cos A = x/b, so x = b cos A.
- In right △BCD: a² = h² + (c − x)².
- Substitute h²: a² = b² − x² + c² − 2cx + x². The x² terms cancel: a² = b² + c² − 2cx.
- Replace x with b cos A: a² = b² + c² − 2bc cos A.
Connection to Pythagoras: if ∠A were 90°, then cos 90° = 0 and the law becomes a² = b² + c². The "−2bc cos A" term is a correction for angles that are not 90°. For an acute angle, cos A is positive, so a² is a bit less than b² + c².
Worked exampleExample 1: SAS
In △ABC, b = 11 cm, c = 14 cm and ∠A = 52°. Find a.
- ∠A is between b and c, so this is SAS. Use a² = b² + c² − 2bc cos A.
- a² = 11² + 14² − 2(11)(14) cos 52°
- a² = 121 + 196 − 308 × 0.6157...
- a² ≈ 317 − 189.62 ≈ 127.38
- a ≈ √127.38 ≈ 11.3 cm
Order of operations
Multiply 2bc cos A first, then subtract it from b² + c². A very common mistake is to calculate (b² + c² − 2bc) and then multiply by cos A. On a calculator, type it all in one line, or find 2bc cos A first and write it down.
Worked exampleExample 2: Across the pond
A surveyor at C measures CA = 350 m and CB = 420 m to the two ends of a pond. The angle ∠ACB = 63°. How long is the pond (AB)?
- The known angle C is between the two known sides: SAS. Use the version with C: c² = a² + b² − 2ab cos C.
- Here a = CB = 420 and b = CA = 350.
- c² = 420² + 350² − 2(420)(350) cos 63° = 176 400 + 122 500 − 294 000 × 0.4540...
- c² ≈ 165427
- c ≈ 406.7 m
The pond is about 407 m long.
Example 2. The surveyor at C knows two distances and the angle between them, the classic SAS set-up for the cosine law. (Drawn to scale.)Worked exampleExample 3: Same law, different letters
In △ABC, a = 7.5 m, b = 9.2 m and ∠C = 48°. Find c.
- ∠C is between a and b. Use c² = a² + b² − 2ab cos C.
- c² = 56.25 + 84.64 − 2(7.5)(9.2) cos 48° ≈ 140.89 − 92.34 ≈ 48.55
- c ≈ 6.97 m
Just for funThe cosine law is Pythagoras with a correction term. Think of it as the Pythagorean theorem after it went to Grade 11 and learned to handle more complicated relationships.
Hands-on activityTest the cosine law with a door
You need: a door that swings, a tape measure, a protractor (or a phone angle app) and a calculator.
- Measure the door's width w (from the hinge side to the outer edge).
- Open the door to some angle θ and measure θ at the hinge, between the door and the closed position (use the edge of the door frame).
- Predict the straight-line distance d from the outer edge of the door to where that edge sits when closed: d² = w² + w² − 2w·w cos θ.
- Measure d with the tape and compare. Repeat for a few angles less than 90°.
What to look for: predictions within a centimetre or two. The triangle here is isosceles (two sides are the door width).
Your turn
1Write the cosine law to find side q in △PQR.
2In △ABC, b = 8 cm, c = 10 cm and ∠A = 60°. Find a.
a² = 64 + 100 − 160 cos 60° = 164 − 80 = 84, a ≈ 9.2 cm
3In △ABC, a = 15 m, c = 12 m and ∠B = 70°. Find b.
b² = 225 + 144 − 360 cos 70° ≈ 245.87, b ≈ 15.7 m
4In △XYZ, x = 4.2 km, y = 5.6 km and ∠Z = 55°. Find z.
z² = 17.64 + 31.36 − 47.04 cos 55° ≈ 22.02, z ≈ 4.69 km
5Measured: b = 6.5 cm, c = 9.0 cm, ∠A = 64°, a = 8.6 cm. Does the cosine law agree?
Predicted a = √(42.25 + 81 − 117 cos 64°) ≈ 8.48 cm. Measured 8.6 cm. They agree within about 1 mm, so yes.
6What does the cosine law become when the angle is 90°? Why?
cos 90° = 0, so a² = b² + c²: the Pythagorean theorem.
7Two hikers leave a trail junction on straight trails that make a 72° angle. One walks 3.5 km and the other walks 4.0 km. How far apart are they?
d² = 12.25 + 16 − 28 cos 72° ≈ 19.60, d ≈ 4.43 km
More practice: warm-up
1Evaluate 2 × 5 × 6 × cos 60°.
2Evaluate 9 + 16 − 24 cos 90°.
3In a² = b² + c² − 2bc cos A, which side is across from ∠A?
4Is SAS a case for the sine law or the cosine law?
More practice: on your own
1Two sides of a triangular garden are 8.0 m and 6.5 m with a 75° angle between them. How much edging is needed for the third side?
√(64 + 42.25 − 104 cos 75°) ≈ 8.91 m
2Ana calculated a² = (11² + 14² − 2 × 11 × 14) cos 52° for Example 1. What did she get and what went wrong?
(317 − 308) × cos 52° = 9 × 0.616 ≈ 5.54, so a ≈ 2.4 cm, which is impossibly small. She multiplied cos 52° by everything instead of only by 2bc.
3A spreadsheet row has b = 12 in B5, c = 15 in C5 and ∠A = 40° in D5. Write a formula for a, and give its value.
=SQRT(B5^2+C5^2-2B5C5*COS(RADIANS(D5))); a ≈ 9.66
4A ship leaves port and sails 24 km. A second ship leaves at the same time and sails 30 km. Their courses make an angle of 58°. How far apart are the ships?
d² = 576 + 900 − 1440 cos 58°, d ≈ 26.7 km
5Show that the cosine law gives a² < b² + c² whenever ∠A is acute.
For an acute angle, cos A > 0, so 2bc cos A > 0. Subtracting a positive amount makes a² = b² + c² − 2bc cos A less than b² + c².
Challenge
1An equilateral triangle has sides of length s. Use the cosine law to confirm that each angle is 60°.
s² = s² + s² − 2s·s cos A, so s² = 2s² − 2s² cos A. Then 2s² cos A = s², cos A = 1/2, ∠A = 60°.
2Write a derivation of b² = a² + c² − 2ac cos B using the altitude from C. (Let DB = y.)
In right △BCD, h² = a² − y² and y = a cos B. In right △ACD, b² = h² + (c − y)² = a² − y² + c² − 2cy + y² = a² + c² − 2cy = a² + c² − 2ac cos B.
3Two forces pull on a box with strengths 50 N and 70 N. The angle between them is 40°. The size of the combined pull is the third side of a triangle whose sides are 50 and 70 with an angle of 180° − 40° = 140° between them. Explain why we can't use the cosine law from this unit directly, and what changes.
140° is obtuse, and this unit only uses acute triangles. The cosine law still works for obtuse angles, but cos 140° is negative, which you will study with angles beyond 90° in the Functions course (MCR3U) or in Grade 12. (The combined pull is about 113 N.)
Quiz
1. Which is the cosine law for side a?
2. The cosine law is the right tool when you know:
3. In △ABC, b = 11, c = 14, ∠A = 52°. What is a?
4. In the derivation of the cosine law, AD = x is replaced by:
5. If ∠A = 90°, the cosine law becomes:
6. A common calculator error is to compute:
Big idea: Expectation C1.3 (verify the cosine law through investigation, for example with dynamic geometry software or by measuring drawn triangles) and the start of C1.4 (use it to calculate sides in acute triangles). Students do not need to memorize the derivation, but seeing it as "Pythagoras plus a correction" helps them remember the formula.
How to help: Ask your teen to point to the angle and the two sides that form it. Watch the calculator entry: 2bc cos A must be calculated as one piece.
Common difficulties:
- Order of operations: (b² + c² − 2bc) cos A instead of b² + c² − 2bc cos A.
- Forgetting the square root at the end.
- Using an angle that is not between the two known sides.
- Spreadsheet COS without RADIANS.