Chapter 8 of 8

All 8 chapters
  1. 1Pythagoras' theorem, readunderstood
  2. 2Using Pythagoras, readunderstood
  3. 3Opposite, adjacent and hypotenuse, readunderstood
  4. 4Sine, cosine and tangent, readunderstood
  5. 5Finding angles, readunderstood
  6. 6Exact values, readunderstood
  7. 7The sine rule and the cosine rule, readunderstood
  8. 8Area of any triangle, and bearings, readunderstood

M4.3.8 Chapter 8

Area of any triangle, and bearings

A walker leaves camp A and walks to B on a bearing of 065°. What bearing takes her back to A?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Find the area of a triangle as 1/2 ab sin C
  • Give a direction as a three figure bearing, like 065°
  • Solve a journey in two parts with bearings and the cosine rule

You know that the area of a triangle is 1/2 × base × height. The height is not always given. Sometimes you know two sides and the angle between them. Then trigonometry finds the height for you.

A triangle ABC lying on its side a, the base, with the height h drawn dashed from A to the base and a small square where it meets the base. The side a is labelled 10 cm, the side b is labelled 8 cm and the angle at C is labelled 50°. The corners are named A, B and C.ABC10 cm8 cm50°h
Take a as the base. The height h stands on it.

In the right angled triangle at the right, sin C = h ÷ b, so h = b × sin C. Then area = 1/2 × a × h = 1/2 × a × b × sin C. This is the area rule: area = 1/2 ab sin C, where C is the angle between the sides a and b.

The angle must be the one between the two sides you use. Our height fell inside the triangle. If it falls outside, the sine of an obtuse angle gives the same height. The last chapter showed why, so the rule still holds.

Worked example

A triangle has sides a = 10 cm and b = 8 cm, and the angle between them is C = 50°. Find its area, to 1 decimal place.

  1. The angle C = 50° is between the sides a = 10 and b = 8. Use area = 1/2 × a × b × sin C.
  2. To 4 decimal places, sin 50° = 0.7660.
  3. Area = 0.5 × 10 × 8 × 0.7660 = 30.64.
  4. Round to 1 decimal place: area = 30.6 cm².

Answer: To 1 decimal place, the area is 30.6 cm².

Now bearings. A bearing gives a direction as an angle measured clockwise from north. It is always written with three figures, so 65° is written 065°. North is 000°, east is 090°, south is 180° and west is 270°.

A circle with a spoke to each of the four main compass points, starting at north at the top. The labels are N 000°, E 090°, S 180° and W 270°.N 000°E 090°S 180°W 270°
Bearings turn clockwise, the way a clock hand turns.

To draw a bearing, stand at the start. Draw a line north. Turn clockwise by the angle. Here is 065° from A to B, 5 km.

A journey from A to B of 5 km. A dashed line points north from A, marked N. An arc runs clockwise from the north line to the line AB and is labelled 065°.ABN065°5 km

Going back from B to A is the opposite direction, a half turn of 180°. The north lines at A and B are parallel, so the bearing at B is 065° + 180° = 245°. If the sum is 360° or more, take 360° away. For a bearing of 250°, 250° + 180° = 430°. Take 360° away to get 070°. That is the same as taking 180° away, which is quicker for any bearing of 180° or more.

A journey from B back to A of 5 km. A dashed line points north from B, marked N. A large arc runs clockwise from the north line to the line BA and is labelled 245°.BAN245°5 km

Worked example

The bearing of B from A is 065°. Find the bearing of A from B.

  1. A to B is a bearing of 065°. Going back from B to A turns the direction through a half turn, 180°.
  2. The north lines at A and B are parallel, so the half turn works at B too.
  3. Add 180°: 65 + 180 = 245.
  4. The result is below 360, so write it with three figures: 245°.

Answer: The bearing of A from B is 245°.

Now a journey in two parts. A boat leaves A and sails 8 km on a bearing of 070° to B. It turns and sails 6 km on a bearing of 150° to C. How far is C from A, and on what bearing?

A journey from A to B to C. From A the boat sails 8 km to B, with the bearing 070°. From B it sails 6 km to C, with the bearing 150°. Dashed north lines at A and B are marked N.ABCNN070°150°8 km6 km

First find the angle ABC inside the triangle. At B the way back to A is the bearing 070° + 180° = 250°. The way on to C is 150°. The angle between the two lines is 250° − 150° = 100°.

The journey drawn as a triangle ABC, with north at the top. Dashed north lines at A and B are marked N. At A the bearing 070° is marked. At B the bearing 150° towards C and the bearing 250° back to A are marked, and the angle ABC between them is marked 100°. The side AB is labelled 8 km and the side BC is labelled 6 km. The dashed side AC is labelled x. The corners are named A, B and C.NN070°150°250°100°ABC8 km6 kmx
The angle inside the triangle at B is the difference of the two bearings: 250° − 150° = 100°.

Worked example

A boat sails 8 km on a bearing of 070° from A to B. Then it sails 6 km on a bearing of 150° to C. How far is C from A, to 1 decimal place?

  1. Draw the journey. A is the start, B is the turn and C is the end.
  2. At B the bearing of A is 70 + 180 = 250°. The bearing of C is 150°.
  3. The angle ABC is the difference: 250 − 150 = 100°.
  4. We know two sides, 8 and 6, and the angle between them. Use the cosine rule: AC² = 8² + 6² − 2 × 8 × 6 × cos 100°.
  5. To 4 decimal places, cos 100° = −0.1736. The cosine of an obtuse angle is negative, as the last chapter showed.
  6. Use all the digits of cos 100°. To 4 decimal places, AC² = 116.6702.
  7. To 4 decimal places, AC = 10.8014.
  8. Round to 1 decimal place: AC = 10.8 km.

Answer: To 1 decimal place, C is 10.8 km from A.

Now the bearing of C from A. Find the angle BAC with the sine rule. Then add it to 070°, because from A the point C lies clockwise of B.

The same journey drawn as a triangle ABC, with north at the top. A dashed north line at A is marked N. An arc from north to the dashed line AC is labelled 103°, and the small angle between AB and AC at A is marked 33.2°. The sides are labelled 8 km, 6 km and 10.8 km. The corners are named A, B and C.N103°33.2°ABC8 km6 km10.8 km
The bearing of C from A is the bearing of B, 070°, plus the angle BAC, 33.2°.

Worked example

Use the same journey. Find the bearing of C from A, to the nearest degree.

  1. We want the angle BAC. BC = 6 is opposite it, and AC = 10.8014 is opposite the angle 100°. Use the sine rule.
  2. sin A ÷ 6 = sin 100° ÷ 10.8014, so sin A = 6 × sin 100° ÷ 10.8014.
  3. Use all the digits on your calculator. To 4 decimal places, sin A = 0.5470.
  4. The inverse sin key gives, to 2 decimal places, A = 33.16.
  5. BC is the shortest side, so A is acute and 33.16° is the only answer.
  6. From A, the point B is on 070° and C is clockwise of B. The bearing of C is 70 + 33.16 = 103.16.
  7. Round to the nearest degree: the bearing of C from A is 103°.

Answer: The bearing of C from A is 103°.

See it three ways

With things

A path leaves a hut and runs north east. North east is half way between north and east.

As a picture

A path from H to P of 5 km. A dashed line points north from H, marked N. An arc runs clockwise from the north line to the path and is labelled 045°.HPN045°5 km

In symbols

North east = half of 090° = 045°.

Try it

Draw a bearing

You need: paper, a ruler, a protractor.

  1. Mark a point A. Draw a straight line up the page from A, 3 cm long. This is north.
  2. Place the protractor at A. Measure 65° clockwise from the north line, and mark it.
  3. Draw a line 5 cm long from A through the mark. Label the end B.

What you will see: The line AB points in the direction 065°. Now stand at B, draw a north line, and measure clockwise to A. You should find 245°.

If you cannot try it, this is what happens.
A line from A to B of 5 cm. A dashed line points north from A, marked N. An arc runs clockwise from the north line to the line AB and is labelled 065°.ABN065°5 cm

A wrong idea many people have

Bearings turn anticlockwise, like angles in a maths class.

Why it feels right: Angles in a graph are measured anticlockwise from the right, so it seems natural to do the same with directions.

The test that breaks it: West is 270° on a bearing. If bearings turned anticlockwise, a quarter turn from north would be west, 090°. The compass picture shows east is 090°.

What is true: A bearing starts at north and turns clockwise, the way a clock hand does, and it has three figures.

A wrong idea many people have

The area rule works with any angle in the triangle.

Why it feels right: There are three angles and sin C looks like a general part of the rule.

The test that breaks it: For sides 10 cm and 8 cm with 50° between them the area is 30.6 cm². Using the 51.6° angle at B, which is not between those sides, gives 31.4 cm². That is a different answer, and it is wrong.

What is true: The angle must be between the two sides in the formula. If you have sides a and b, use angle C. If you have b and c, use A.

Key words

area rule
area = 1/2 ab sin C, with C the angle between sides a and b.
bearing
a direction given as an angle clockwise from north, written with three figures.

Back to your guess

The answer is B: 245°, because turning round adds a half turn: 65° + 180° = 245°.

The bearing back is 245°. A bearing is the angle clockwise from north. Going back is a half turn, 180°, from the way you came, so you add 180°. This chapter explains bearings, and finds areas of triangles with no right angle.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A ship sails due west. What is its bearing?

    Tap your answer.

    1. Not this one. 090° is due east. A bearing turns clockwise from north, so east comes after a quarter turn, and west comes after three quarters. Go to M4.3.8. Try another.

    2. Yes, that is right. West is three quarters of a full turn clockwise from north: 3 ÷ 4 × 360 = 270°.

    3. Not this one. 180° is due south, a half turn from north. West is three quarters of the way round. Go to M4.3.8. Try another.

    4. Not this one. 000° is due north. Bearings start at north and turn clockwise, so west is not at the start. Go to M4.3.8. Try another.

    See the answer

    270°

    West is three quarters of a full turn clockwise from north: 3 ÷ 4 × 360 = 270°.

    • If you chose A: 090° is due east. A bearing turns clockwise from north, so east comes after a quarter turn, and west comes after three quarters. Go to M4.3.8.
    • If you chose C: 180° is due south, a half turn from north. West is three quarters of the way round. Go to M4.3.8.
    • If you chose D: 000° is due north. Bearings start at north and turn clockwise, so west is not at the start. Go to M4.3.8.

  2. The bearing of Q from P is 130°. What is the bearing of P from Q?

    A journey from P to Q of 5 km. A dashed line points north from P, marked N. An arc runs clockwise from the north line to the line PQ and is labelled 130°. The line PQ runs down to the right.PQN130°5 km
    Tap your answer.

    1. Yes, that is right. Going back is a half turn: 130 + 180 = 310. The sum is below 360, so the bearing is 310°.

    2. Not this one. You took 130° from 180°. Going back is a half turn, so add 180° to the bearing. Go to M4.3.8. Try another.

    3. Not this one. You took 130° from 360°. A reversed direction is a half turn away, not the mirror image. Add 180°. Go to M4.3.8. Try another.

    4. Not this one. That is the bearing of Q from P. The way back points the opposite way, so the bearing changes by 180°. Go to M4.3.8. Try another.

    See the answer

    310°

    Going back is a half turn: 130 + 180 = 310. The sum is below 360, so the bearing is 310°.

    • If you chose B: You took 130° from 180°. Going back is a half turn, so add 180° to the bearing. Go to M4.3.8.
    • If you chose C: You took 130° from 360°. A reversed direction is a half turn away, not the mirror image. Add 180°. Go to M4.3.8.
    • If you chose D: That is the bearing of Q from P. The way back points the opposite way, so the bearing changes by 180°. Go to M4.3.8.

  3. A triangle has sides 7 cm and 10 cm with an angle of 40° between them. What is its area, to 1 decimal place?

    A triangle ABC with the angle 40° at C between the side a of 7 cm and the side b of 10 cm. The corners are named A, B and C.ABC7 cm10 cm40°
    Tap your answer.

    1. Not this one. You left out the half. The area is 1/2 × a × b × sin C, and you worked out a × b × sin C. A triangle is half of a parallelogram. Go to M4.3.8. Try another.

    2. Not this one. You used cos 40° instead of sin 40°. The height of the triangle is b × sin C, so the area rule uses the sine. Go to M4.3.8. Try another.

    3. Yes, that is right. Area = 1/2 × 7 × 10 × sin 40° = 22.498, which is 22.5 cm² to 1 decimal place.

    4. Not this one. You used 1/2 × 7 × 10. That is half the product of the sides. The height is less than the side, so you must multiply by sin 40°. Go to M4.3.8. Try another.

    See the answer

    22.5 cm²

    Area = 1/2 × 7 × 10 × sin 40° = 22.498, which is 22.5 cm² to 1 decimal place.

    • If you chose A: You left out the half. The area is 1/2 × a × b × sin C, and you worked out a × b × sin C. A triangle is half of a parallelogram. Go to M4.3.8.
    • If you chose B: You used cos 40° instead of sin 40°. The height of the triangle is b × sin C, so the area rule uses the sine. Go to M4.3.8.
    • If you chose D: You used 1/2 × 7 × 10. That is half the product of the sides. The height is less than the side, so you must multiply by sin 40°. Go to M4.3.8.

  4. A triangle has sides 8 cm and 7 cm, and its area is 20 cm². The angle C between the two sides is not shown. What can C be, to 1 decimal place?

    Tap your answer.

    1. Yes, that is right. 20 = 1/2 × 8 × 7 × sin C, so sin C = 40 ÷ 56 = 0.7143. The inverse sin gives 45.58°, and 180° minus that is the second angle.

    2. Not this one. Two angles have the same sine, an acute one and 180° minus it. Here sin C = 0.7143, and both 45.6° and 134.4° fit the area. Go to M4.3.7. Try another.

    3. Not this one. You took 45.6° from 90°, which gives 44.4°. The two angles with the same sine add to 180°, not 90°: 45.6° and 134.4°. Go to M4.3.7. Try another.

    4. Not this one. You stopped at sin C = 0.7143, which is 40 ÷ 56. Press the inverse sin key to get the angle, and give both angles. Go to M4.3.5. Try another.

    See the answer

    45.6° or 134.4°

    20 = 1/2 × 8 × 7 × sin C, so sin C = 40 ÷ 56 = 0.7143. The inverse sin gives 45.58°, and 180° minus that is the second angle.

    • If you chose B: Two angles have the same sine, an acute one and 180° minus it. Here sin C = 0.7143, and both 45.6° and 134.4° fit the area. Go to M4.3.7.
    • If you chose C: You took 45.6° from 90°, which gives 44.4°. The two angles with the same sine add to 180°, not 90°: 45.6° and 134.4°. Go to M4.3.7.
    • If you chose D: You stopped at sin C = 0.7143, which is 40 ÷ 56. Press the inverse sin key to get the angle, and give both angles. Go to M4.3.5.

  5. A ship sails 5 km on a bearing of 040°. It then sails 7 km on a bearing of 130°. How far is it from where it started, to 1 decimal place?

    A journey from A to B to C. From A the ship sails 5 km to B, with the bearing 040°. From B it sails 7 km to C, with the bearing 130°. Dashed north lines at A and B are marked N.ABCNN040°130°5 km7 km
    Tap your answer.

    1. Not this one. You added the two legs, 5 + 7. The ship does not sail straight back, so the distance from the start is shorter than the sum. Go to M4.3.8. Try another.

    2. Not this one. You used 130° as the angle in the cosine rule. The angle inside the triangle at B is the difference between the bearings of A and C from B. Go to M4.3.8. Try another.

    3. Not this one. You used 40° as the angle in the cosine rule. The angle inside the triangle at B is the difference between the bearings of A and C from B. Go to M4.3.8. Try another.

    4. Yes, that is right. At B the bearing of A is 40 + 180 = 220°, and the bearing of C is 130°. The angle ABC is 220 − 130 = 90°, a right angle. So AC² = 5² + 7² = 74, and AC = √74, which is 8.6 km to 1 decimal place.

    See the answer

    8.6 km

    At B the bearing of A is 40 + 180 = 220°, and the bearing of C is 130°. The angle ABC is 220 − 130 = 90°, a right angle. So AC² = 5² + 7² = 74, and AC = √74, which is 8.6 km to 1 decimal place.

    • If you chose A: You added the two legs, 5 + 7. The ship does not sail straight back, so the distance from the start is shorter than the sum. Go to M4.3.8.
    • If you chose B: You used 130° as the angle in the cosine rule. The angle inside the triangle at B is the difference between the bearings of A and C from B. Go to M4.3.8.
    • If you chose C: You used 40° as the angle in the cosine rule. The angle inside the triangle at B is the difference between the bearings of A and C from B. Go to M4.3.8.

  6. From an earlier chapter

    In triangle ABC, a = 7 cm, b = 5 cm and c = 8 cm. Find the angle C, to 1 decimal place.

    Tap your answer.

    1. Not this one. You have the sign the wrong way: cos C = (c² − a² − b²) ÷ 2ab. The rule is cos C = (a² + b² − c²) ÷ 2ab, which is positive here, so C is acute. Go to M4.3.7. Try another.

    2. Not this one. You stopped at cos C = 0.1429. Press the inverse cos key to turn the ratio into an angle. Go to M4.3.5. Try another.

    3. Yes, that is right. cos C = (7² + 5² − 8²) ÷ (2 × 7 × 5) = 10 ÷ 70 = 0.1429. The inverse cos gives 81.787°, which is 81.8° to 1 decimal place.

    See the answer

    81.8°

    cos C = (7² + 5² − 8²) ÷ (2 × 7 × 5) = 10 ÷ 70 = 0.1429. The inverse cos gives 81.787°, which is 81.8° to 1 decimal place.

    • If you chose A: You have the sign the wrong way: cos C = (c² − a² − b²) ÷ 2ab. The rule is cos C = (a² + b² − c²) ÷ 2ab, which is positive here, so C is acute. Go to M4.3.7.
    • If you chose B: You stopped at cos C = 0.1429. Press the inverse cos key to turn the ratio into an angle. Go to M4.3.5.

  7. A triangle has sides 9 cm and 11 cm with an angle of 55° between them. How many square centimetres is its area, to 1 decimal place?

    See the answer

    40.5

    Area = 1/2 × 9 × 11 × sin 55° = 40.548, which is 40.5 to 1 decimal place.

  8. The bearing of B from A is 250°. What is the bearing of A from B? Write the number.

    See the answer

    070

    Going back is a half turn. 250 + 180 = 430, which is 360 or more, so take 360 away: 70. As a bearing, write 070°.

Keep it

Copy this into your exercise book.

Area = half × a × b × sin C, with C between a and b. A bearing is clockwise from north, in three figures. Back bearing: add or take away 180°.

Draw: A triangle with the angle C between sides a and b. Beside it, a north line with a bearing arc.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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