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Cambridge IGCSE0607

Non-right-angled triangles

International Mathematics 0607 Chapter Notes

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Non-right-angled triangles
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In this chapter

  1. 1Area of a triangle
  2. 2Cosine rule: finding a side
  3. 3Cosine rule: finding an angle, obtuse angles
  4. 4Sine rule and bearings
  5. 5The ambiguous case

1. Area of a triangle

The area of a triangle is half the product of two sides and the sine of the included angle. The included angle is the angle between those two sides.

The formula is given in the exam. The angle must sit between the two sides you use.

If the area is given, substitute it, then rearrange for the side or angle you want.

area = 1/2 absin C

Examiner insight

“Most candidates used their calculated angle in the formula for the area using the sine of this angle, but many had rounded this angle prematurely.” “Some candidates started with a correct equation, and were then unable to rearrange correctly.”

Worked example 13 marks

The diagram shows triangle ABC. The area of triangle ABC is 30 cm². Find the length of BC.

  1. 1

    area = 1/2 × AB × BC × sin 60^(°) [1]

  2. 2

    30 = 1/2 × 4√3 × BC × (√3)/2 [1]

  3. 3

    30 = 3 × BC, so BC = 10 cm [1]

2. Cosine rule: finding a side

Use the cosine rule when you know two sides and the included angle. It gives the third side, opposite that angle.

On a calculator paper, type the whole expression b² + c² - 2bccos A in one go with the calculator in degree mode. Read off a², then take the square root. Keep the unrounded value.

On a non-calculator paper, use an exact value such as cos 60^(°) = 1/2 and keep the working exact.

a² = b² + c² - 2bccos A

Examiner insight

“The only real challenge with the cosine rule is to ensure that (2 × 10 × 12 × cos 120) is a product that needs to be calculated before being subtracted from the two squares.” “Some candidates assumed that angle BAC was 50°.”

Worked example 13 marks

The diagram shows the positions of A, B and C. AB = 252 km and BC = 428 km. Angle ABC = 110°. The bearing of B from A is 040°. Calculate AC.

  1. 1

    AC² = 252² + 428² - 2 × 252 × 428 × cos 110^(°) [1]

  2. 2

    AC² = 320 465.8… [1]

  3. 3

    AC = 566.09… = 566 km [1]

3. Cosine rule: finding an angle, obtuse angles

With all three sides known, rearrange the cosine rule to find an angle. The angle you find is opposite the side whose square is subtracted.

An obtuse angle is between 90^(°) and 180^(°). Its cosine is negative, so a negative fraction gives the obtuse angle straight away. Keep the minus sign.

On the calculator, enter the fraction in brackets, then use inverse cosine in degree mode.

cos A = (b² + c² - a²)/2bc

Examiner insight

“The advantage of using the cosine rule here was that the calculated quotient was negative, leading to the correct obtuse angle immediately, without having to subtract from 180 degrees.” “Some candidates appeared to misunderstand the term ‘obtuse angle’ and attempted to find a reflex angle instead, by subtracting a correctly obtained acute angle from 360 degrees.”

Worked example 12 marks

The diagram shows triangle PQR. Show that cos Q = 1/5.

  1. 1

    7² = 6² + 5² - 2 × 6 × 5 × cos Q [1]

  2. 2

    cos Q = (36 + 25 - 49)/60 = 12/60 = 1/5 [1]

4. Sine rule and bearings

Use the sine rule when you know a side and the angle opposite it, and one more side or angle. Each side is paired with the angle opposite it.

To find an angle, put the sines on top. To find a side, put the sides on top. Keep unrounded values until the end.

For a bearing, first find the angle inside the triangle. Then add it to, or subtract it from, the bearing you already know.

a/(sin A) = b/(sin B) = c/(sin C)

(sin A)/a = (sin B)/b = (sin C)/c

Examiner insight

“The most common error was to omit the addition of 40°.” “The most common approach was to apply the sine rule, but weaker candidates had the pairings of sides and angles mixed up.”

Worked example 14 marks

The diagram shows the positions of A, B and C. AB = 252 km and BC = 428 km. Angle ABC = 110°. The bearing of B from A is 040°. Find the bearing of C from A.

  1. 1

    (sin BAC)/428 = (sin 110^(°))/(566.09…) (the unrounded AC from part (a)) [1]

  2. 2

    sin BAC = (428 × sin 110^(°))/(566.09…) = 0.7104… [1]

  3. 3

    angle BAC = 45.27…^(°) [1]

  4. 4

    bearing of C from A = 40^(°) + 45.27…^(°) = 085.3^(°) [1]

5. The ambiguous case

Between 0^(°) and 180^(°), an angle and its supplementary angle have the same sine. So the sine rule can give an acute angle B and an obtuse angle, 180^(°) - B.

This happens when you know two sides and an angle that is not between them. Sketch both triangles. Keep the obtuse angle only if it and the given angle add up to less than 180^(°).

A question may say which one it wants, for example by stating that the angle is obtuse.

Examiner insight

“No candidates gave a diagram to illustrate the two possible answers, which would have helped them to visualise the situation.” “Many thought the other possibility was one of the other angles of the triangle.”

Worked example 14 marks

In triangle ABC, angle A = 40^(°), BC = 7 cm and AC = 9 cm. Find the two possible sizes of angle ABC.

  1. 1

    (sin B)/9 = (sin 40^(°))/7 [1]

  2. 2

    sin B = (9 sin 40^(°))/7 = 0.8264… [1]

  3. 3

    acute angle B = 55.7^(°) [1]

  4. 4

    obtuse angle B = 180^(°) - 55.7^(°) = 124.3^(°) (valid, as 40^(°) + 124.3^(°) < 180^(°)) [1]

Recap

  • Area of a triangle = 1/2 absin C, with C the angle between sides a and b.
  • Cosine rule for a side: a² = b² + c² - 2bccos A. Work out the product before subtracting.
  • Cosine rule for an angle: cos A = (b² + c² - a²)/2bc. A negative value means an obtuse angle.
  • Sine rule: each side is paired with the angle opposite it.
  • Ambiguous case: sin B gives an acute angle and the obtuse angle 180^(°) - B.
  • Bearings: find the angle in the triangle, then add it to or subtract it from the known bearing.
  • Keep unrounded values until the final answer.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Triangle ABC has AB = 7 cm, BC = 8 cm and AC = 10 cm. No angle is known. State which rule you use first to find angle B.1 mark
  2. Do not use a calculator. Two sides of a triangle are 8 cm and 5 cm and the angle between them is 60^(°). Find the exact area of the triangle.2 marks
  3. Do not use a calculator. A triangle has sides of 6 cm and 8 cm with an acute angle between them. Its area is 12√3 cm². Find the angle.3 marks
  4. Do not use a calculator. Two sides of a triangle are 3 cm and 5 cm and the angle between them is 120^(°). Find the third side.3 marks
  5. Triangle PQR has PQ = 7 cm, QR = 9 cm and PR = 12 cm. Calculate angle PQR.3 marks
  6. In triangle ABC, angle A = 48^(°), angle B = 71^(°) and BC = 13.5 cm. Calculate AC.3 marks
  7. In triangle ABC, angle A = 35^(°), BC = 6 cm and AB = 9 cm. Find the two possible sizes of angle ACB.3 marks
  8. A field ABC has AB = 85 m, AC = 120 m and BC = 100 m. Calculate the area of the field. Give your answer correct to 3 significant figures.4 marks
  9. Do not use a calculator. A triangular flower bed has sides of 4 m, 5 m and 7 m. Angle θ is the angle between the 4 m side and the 5 m side. (a) Show that cosθ = -1/5. (b) Find the exact area of the flower bed. Give your answer in the form k√6. (c) A path runs from the corner at θ straight to the 7 m side, by the shortest route. Find the exact length of the path.7 marks
  10. A lighthouse L is 8.5 km from a harbour H on a bearing of 050^(°). A ship S is due east of H and 6 km from L. Angle HSL is obtuse. Calculate (a) angle HSL, (b) the bearing of L from S, (c) the distance HS.9 marks

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