D. Trigonometric Functions · Trigonometric Ratios, Identities and LawsThe Cosine Law in Oblique Triangles
D1.6About 70 minutes
Learning goals- I can use the cosine law to find a side when I know two sides and the contained angle, including an obtuse angle.
- I can use the cosine law to find any angle, acute or obtuse, when I know three sides.
- I can explain how the sign of cos A tells whether a² is less than, equal to or greater than b² + c².
- I can solve problems with obtuse triangles, such as parallelogram diagonals and navigation.
Words to know
cosine law
In any △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 (SAS).
SAS and SSS
Two sides and the contained angle; three sides. Both need the cosine law to start.
Remember from Grade 10
You used the cosine law to solve acute triangles when there was no complete side-angle pair:
a² = b² + c² − 2bc cos A
- SAS (two sides and the angle between them): find the third side.
- SSS (three sides): rearrange to find an angle: cos A = (b² + c² − a²)/(2bc).
The law is true for every triangle, including obtuse ones. (Challenge question 1 below proves it with coordinates for any angle A.) The only difference is that cos A is negative when A is obtuse.
What the cosine term does
The cosine law is the Pythagorean theorem with a correction term, −2bc cos A.
- If A is acute, cos A > 0, so you subtract: a² is less than b² + c².
- If A = 90°, cos A = 0, and the law becomes a² = b² + c², the Pythagorean theorem.
- If A is obtuse, cos A < 0, so −2bc cos A is positive: a² is more than b² + c².
This makes sense: the wider the angle opens, the farther apart the ends of sides b and c are.
Worked exampleExample 1: SAS with an obtuse angle
In △ABC, b = 7 cm, c = 10 cm and ∠A = 118°. Find a.
- a² = 7² + 10² − 2(7)(10) cos 118°.
- cos 118° ≈ −0.4695, so −140 cos 118° ≈ +65.73.
- a² ≈ 49 + 100 + 65.73 = 214.73.
- a ≈ 14.7 cm. (Check: a is longer than both b and c, as it should be opposite the obtuse angle.)
Calculator care
Type the whole right side in one line, with brackets: 7² + 10² − 2 × 7 × 10 × cos(118). A common error is to subtract first and then multiply: (149 − 140) cos 118° is wrong. Multiplication comes before subtraction.
No ambiguity when finding angles
When you find an angle with cos⁻¹, the calculator gives an angle from 0° to 180°. That covers every possible angle in a triangle, and cosine is positive for acute angles and negative for obtuse ones. So the cosine law always tells you correctly whether an angle is acute or obtuse.
Strategy for SSS: find the largest angle (opposite the longest side) first with the cosine law. Then the other two angles must be acute, and you can safely use the sine law for them.
Worked exampleExample 2: SSS with an obtuse angle
Solve △ABC with a = 6 cm, b = 9 cm and c = 13 cm.
- The longest side is c, so find ∠C first: cos C = (6² + 9² − 13²) ÷ (2 × 6 × 9) = (36 + 81 − 169) ÷ 108 = −52/108 ≈ −0.4815.
- ∠C = cos⁻¹(−0.4815) ≈ 118.8°. (Negative cosine: obtuse.)
- Sine law for ∠A: sin A = 6 sin 118.8° ÷ 13 ≈ 0.4045, so ∠A ≈ 23.9° (acute, since ∠C is obtuse).
- ∠B ≈ 180° − 118.8° − 23.9° = 37.4°.
∠A ≈ 23.9°, ∠B ≈ 37.4°, ∠C ≈ 118.8°
Example 2 drawn to scale. Finding the largest angle first with the cosine law reveals that it is obtuse.Worked exampleExample 3: Snowmobile trails
Two straight snowmobile trails leave a junction. The angle between them is 135°. Ali rides 4.2 km along one and Bea rides 6.5 km along the other. How far apart are they, in a straight line?
- SAS: the 135° angle is between the two known sides.
- d² = 4.2² + 6.5² − 2(4.2)(6.5) cos 135°.
- d² ≈ 17.64 + 42.25 + 38.61 = 98.50.
- d ≈ 9.9 km.
Worked exampleExample 4: Diagonals of a parallelogram
A parallelogram has sides 8 cm and 12 cm and one angle of 60°. Find the lengths of both diagonals.
- Adjacent angles of a parallelogram add to 180°, so the angles are 60° and 120°.
- The short diagonal is opposite the 60° angle: d² = 64 + 144 − 192 cos 60° = 208 − 96 = 112, so d ≈ 10.6 cm.
- The long diagonal is opposite the 120° angle: D² = 64 + 144 − 192 cos 120° = 208 + 96 = 304, so D ≈ 17.4 cm.
About 10.6 cm and 17.4 cm (exactly √112 = 4√7 and √304 = 4√19).
Each diagonal of the parallelogram is the third side of a triangle with sides 8 cm and 12 cm. The angle between those sides is 60° for one diagonal and 120° for the other.Just for funThe cosine law is the Pythagorean theorem with a correction term. Think of −2bc cos A as the theorem's way of saying "it's complicated" whenever the angle isn't exactly 90°.
Hands-on activityOpen a book and predict the gap
You need: a hardcover book (or a laptop, opened carefully), a ruler or tape measure, a protractor and a calculator.
- Stand the book on a table and open the covers to a wide angle, somewhere between 100° and 150°.
- Measure the width of each cover from the spine to its outer edge (they are probably equal, say w).
- Measure the angle between the covers at the spine with the protractor (look straight down from above).
- Predict the straight-line gap between the two outer edges: d² = w² + w² − 2w² cos θ.
- Measure the actual gap at the bottom edge and compare.
- Now measure the gap first, and use cos θ = (w² + w² − d²)/(2w²) to calculate the angle. Compare with the protractor.
What to look for: for a 20 cm cover opened to 120°, d² = 400 + 400 + 400 = 1200, so d ≈ 34.6 cm. Predictions within 1 cm (or angles within about 3°) are good.
Your turn
1In △ABC, b = 5 cm, c = 8 cm and ∠A = 130°. Find a.
a² = 25 + 64 − 80 cos 130° ≈ 89 + 51.42 = 140.42, so a ≈ 11.9 cm.
2In △PQR, q = 12 m, r = 15 m and ∠P = 102°. Find p.
p² = 144 + 225 − 360 cos 102° ≈ 369 + 74.85 = 443.85, so p ≈ 21.1 m.
3A triangle has sides 7, 10 and 15. Find its largest angle.
Opposite 15: cos θ = (49 + 100 − 225)/(2 × 7 × 10) = −76/140 ≈ −0.5429, so θ ≈ 122.9°.
4A triangle has sides 5, 6 and 10. Find its largest angle.
cos θ = (25 + 36 − 100)/60 = −0.65, so θ ≈ 130.5°.
5Without finding any angles, decide whether a triangle with sides 4, 5 and 7 is acute or obtuse.
Compare the longest side squared with the sum of the other two squares: 49 > 16 + 25 = 41, so the largest angle has cos < 0. Obtuse.
6What does the cosine law a² = b² + c² − 2bc cos A become when ∠A = 90°?
cos 90° = 0, so a² = b² + c², the Pythagorean theorem.
More practice: warm-up
2Write the cosine law to find side b.
3Rearrange a² = b² + c² − 2bc cos A to make cos A the subject.
cos A = (b² + c² − a²)/(2bc)
4Find a exactly: b = 3, c = 5, ∠A = 120°.
a² = 9 + 25 − 30(−1/2) = 34 + 15 = 49, so a = 7.
5If ∠A is obtuse, is cos A positive or negative?
More practice: on your own
1Solve △ABC: a = 11 cm, b = 7 cm, ∠C = 140°.
c² = 121 + 49 − 154 cos 140° ≈ 287.97, so c ≈ 17.0 cm. sin A = 11 sin 140° ÷ 16.97 ≈ 0.4167, ∠A ≈ 24.6° (acute). ∠B ≈ 15.4°.
2Solve the triangle with sides 8 cm, 11 cm and 17 cm.
Largest angle (opposite 17): cos = (64 + 121 − 289)/176 ≈ −0.5909, angle ≈ 126.2°. Opposite 8: sin = 8 sin 126.2° ÷ 17 ≈ 0.3797, angle ≈ 22.3°. Opposite 11: ≈ 180° − 126.2° − 22.3° = 31.5°.
3Two boats leave a dock at the same time on straight courses 110° apart. One travels at 15 km/h and the other at 20 km/h. How far apart are they after 2 h?
Distances: 30 km and 40 km. d² = 900 + 1600 − 2400 cos 110° ≈ 2500 + 820.85 = 3320.85, so d ≈ 57.6 km.
4A parallelogram has sides 6 cm and 9 cm and an angle of 70°. Find both diagonals.
Opposite 70°: √(36 + 81 − 108 cos 70°) ≈ 8.9 cm. Opposite 110°: √(36 + 81 − 108 cos 110°) ≈ 12.4 cm.
5Ana found a in Example 1 by computing (49 + 100 − 140) cos 118°. Explain her error.
She subtracted before multiplying. The term 2bc cos A = 140 cos 118° must be calculated first, then subtracted: 149 − 140 cos 118° ≈ 214.7, so a ≈ 14.7 cm.
Challenge
1Prove the cosine law for any angle A using coordinates: put A at (0, 0), B at (c, 0) and C at (b cos A, b sin A).
a² = BC² = (b cos A − c)² + (b sin A)² = b²cos²A − 2bc cos A + c² + b²sin²A = b²(cos²A + sin²A) + c² − 2bc cos A = b² + c² − 2bc cos A. The coordinates of C work for acute and obtuse A, so the law holds for both.
2Two sides of a triangle are 7 and 8, and the angle between them is 120°. Find the third side exactly.
x² = 49 + 64 − 112 cos 120° = 113 + 56 = 169, so x = 13.
3A clock has a 12 cm minute hand and an 8 cm hour hand. How far apart are the tips at exactly 4:00?
At 4:00 the hands are 4 × 30° = 120° apart. d² = 144 + 64 − 192 cos 120° = 208 + 96 = 304, so d ≈ 17.4 cm.
Quiz
1. If ∠A is obtuse, then in △ABC:
2. In △ABC, b = 7, c = 10 and ∠A = 118°. What is a (to one decimal place)?
3. Which formula finds ∠C from three sides?
4. cos C = −0.4815. What is ∠C?
5. Why is it good to find the largest angle of an SSS triangle first?
6. A parallelogram has sides 8 and 12 and an angle of 60°. What is its longer diagonal?
Big idea (D1.6): the cosine law from Grade 10 works in every triangle. With an obtuse angle, cos A is negative, so the "− 2bc cos A" term actually adds length. When finding angles, the cosine law never gives the wrong kind of angle, which makes it the safest first step for SSS triangles.
How to help: Watch your teen's calculator entry. The whole expression b² + c² − 2bc cos A should be typed in one go, and the square root taken at the end.
Common difficulties:
- Calculating (b² + c² − 2bc) × cos A instead of b² + c² − (2bc cos A).
- Forgetting to take the square root at the end.
- Using the sine law first for the largest angle in an SSS triangle and getting an acute angle that should be obtuse.