Year 11 Extended · Unit 2 Trigonometric functions and any triangle · Lesson 6 of 6

Two ships leave a harbour on different bearings. How far apart are they when there is no right angle to use?

Pythagoras needs a right angle. Split any triangle by a height and a correction term appears: the cosine rule. With the area formula and bearings it solves navigation problems.

Geometry and TrigonometryScientific calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Lesson 5: in triangle ABC, A = 30°, B = 90° and a = 6. Find b.
    Answer12. 6 ÷ sin 30° = 6 ÷ 0.5.
  2. From Year 9: the legs of a right-angled triangle are 9 and 12. Find the hypotenuse.
    Answer15. 81 + 144 = 225.
  3. From Lesson 1: what is cos 120°?
    Answer−0.5. The tip is on the left of the vertical axis.
Words used in this lesson
cosine rule
a rule linking the three sides and one angle of any triangle. For example: c² = a² + b² − 2ab cos C.
bearing
a direction measured clockwise from north, with three digits. For example: 090° is due east.
included angle
the angle between two given sides. For example: the angle at C between a and b.
area formula
half the product of two sides and the sine of the angle between them. For example: ½ ab sin C.
obtuse
an angle between 90° and 180°. For example: an angle of 130°.

Two ships from Mombasa

Two ships leave the harbour at Mombasa at the same time. Ship P sails 12 km on a bearing of 040°. Ship Q sails 9 km on a different bearing. A bearing is an angle measured clockwise from north, written with three digits. A radio operator wants to know how far apart the ships are, but there is no right angle in the picture.

Predict first: will the distance between the ships be more or less than the 12 km that P has sailed? Can it ever be as large as 12 + 9? What bearing for Q would make the distance straightforward to find with Pythagoras? Use the first figure to see how the cosine rule arises, and the second to test your predictions.

60
6.9height a sin C
9.8third side c

C = 60°: c² = 8² + 11² − 2 × 8 × 11 × cos 60° = 185 − 88 = 97, so c = 9.8.

Show me the steps
  1. C = 60°: the height is 8 sin 60° = 6.93 and the foot is 8 cos 60° = 4 from C. The rest of side b is 11 − 4 = 7, and c² = 7² + 6.93² = 97, so c = 9.8. This is 185 − 88.
  2. C = 90°: cos 90° = 0, so the foot of the height is at C itself, and c² = 11² + 8² = 185, which is Pythagoras, c = 13.6.
  3. C = 130°: cos 130° = −0.643, so the foot is a cos C = −5.1 from C, on the other side of C and outside b. The horizontal part is 11 + 5.1 = 16.1 and c² = 185 + 113.1 = 298.1, so c = 17.3.

The cosine rule. In any triangle, c² = a² + b² − 2ab cos C, where C is the angle between the sides a and b.

  • Why it works: the height from B splits side b into a cos C and b − a cos C. Pythagoras on the right-angled triangle gives c² = (b − a cos C)² + (a sin C)². Expand, and use sin² C + cos² C = 1: c² = a² + b² − 2ab cos C.
  • For an angle: cos C = (a² + b² − c²) ÷ 2ab. The inverse cosine gives the correct angle directly, obtuse or acute.
  • Area = ½ ab sin C, with C between the sides a and b.
  • Bearings: the angle between two courses from one point is the difference of their bearings. At a turn, the angle inside the triangle is 180° minus the angle you turned through.
  • Which rule: two sides and the angle between them, or all three sides, needs the cosine rule. An angle and its opposite side needs the sine rule.
100
60angle between the courses (degrees)
10.8distance between the ships (km)

Bearing 100°: the courses are 100° − 40° = 60° apart, so PQ² = 12² + 9² − 2 × 12 × 9 × cos 60° = 117 and PQ = 10.8 km.

Show me the steps
  1. Bearing 100°: the angle between the courses is 100° − 40° = 60°. PQ² = 144 + 81 − 216 cos 60° = 225 − 108 = 117, so PQ = 10.8 km. That is less than 12 + 9 = 21.
  2. Bearing 130°: the angle is 90°, cos 90° = 0 and PQ² = 144 + 81 = 225, so PQ = 15 km. The cosine rule has become Pythagoras.
  3. Bearing 220°: the angle is 180°, cos 180° = −1 and PQ² = 225 + 216 = 441, so PQ = 21 km, which is 12 + 9. The ships are in a straight line with the harbour between them.

Work out 2ab cos C as one number and take it away from the sum. It is not (a² + b² − 2ab) × cos C, and it is not (a + b)². And an obtuse angle makes cos C negative, so subtracting it adds to c²: the third side is longer than Pythagoras gives.

Worked examples

Two sides of a triangle are 7 and 10 and the angle between them is 52°. Find the third side c.

  1. Two sides and the included angle: the cosine rule. c² = 49 + 100 − 2 × 7 × 10 × cos 52° = 149 − 140 × 0.6157 = 149 − 86.19 = 62.81.
  2. c = √62.81 = 7.9 to one decimal place.

A boat leaves a port and sails 14 km on bearing 070°. It then turns and sails 20 km on bearing 150°. How far is it from the port?

  1. The boat turns through 150° − 70° = 80°, so the angle inside the triangle at the turn is 180° − 80° = 100°.
  2. d² = 14² + 20² − 2 × 14 × 20 × cos 100° = 596 + 560 × 0.1736 = 596 + 97.24 = 693.24, so d = 26.3 km.

Your go. The first step is given. The sides of a triangle are 5, 8 and 10. Find its largest angle.

  1. The largest angle is opposite the longest side, 10. Call it C, so c = 10.
  2. cos C = (5² + 8² − 10²) ÷ (2 × 5 × 8). Work it out and use the inverse cosine.
Check the answer

The top is 25 + 64 − 100 = −11 and the bottom is 80, so cos C = −0.1375 and C = 97.9°. The cosine is negative, so the angle is obtuse. A common slip is to subtract in the wrong order and get +11.

With a calculator

Work out the whole of 2ab cos C first, write it down, and only then subtract it from a² + b². Press the square-root key last. Keep the unrounded numbers on the calculator and round only the final answer.

For an angle from three sides, work out the fraction (a² + b² − c²) ÷ 2ab with brackets, then use the inverse cosine key. Unlike sine, the inverse cosine gives the right angle between 0° and 180° directly: for example cos⁻¹(−0.3) shows 107.5, an obtuse angle, with no second case to test.

Check before you go on

1. In triangle PQR, PQ = 7, QR = 12 and PR = 8. Which expression gives cos Q?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Two sides of a triangle are 9 and 14 and the angle between them is 65°. Find the third side, to 1 decimal place.

3. A triangular plot has two sides of 10 m and 13 m with an angle of 48° between them. Find its area in square metres, to 1 decimal place.

Level 2 · Building

4. The sides of a triangle are 6, 7 and 11. Find its largest angle, to 1 decimal place.

5. A boat sails 15 km on a bearing of 050°, then 20 km on a bearing of 160°. How far is it from where it started, to 1 decimal place?

Level 3 · Stretch

6. A triangular sail has sides of 9 m, 12 m and 16 m. Find its area, to 1 decimal place.

7. A triangular field has two sides PQ = 120 m and PR = 150 m and an area of 7000 m². Find the acute angle at P, to 1 decimal place.

Something to think about

The cosine rule can be rearranged to find an angle: cos C = (a² + b² − c²) ÷ 2ab. What does the sign of the top of the fraction tell you about the angle C without any calculator?

One way to think about it

If a² + b² is bigger than c², the top is positive, cos C is positive and C is acute. If they are equal the top is 0 and C is a right angle, which is the converse of Pythagoras. If a² + b² is smaller than c², the top is negative and C is obtuse.

Exit ticket

8. Which of these triangle problems needs the cosine rule rather than the sine rule?