Year 11 Standard · Unit 5 Ready for the Diploma · Lesson 5 of 6

What can you do with a triangle that has no right angle?

Right-angled triangle methods need a right angle. Two rules, one with sines and one with cosines, work in any triangle, and each needs a particular set of measurements.

FunctionsScientific calculator

Warm-up

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

  1. From Year 9: in a right-angled triangle the hypotenuse is 10 and an angle is 30°. Find the side opposite it.
    Answer5. 10 × sin 30°.
  2. From Year 8: the angles of a triangle are 50° and 70°. What is the third?
    Answer60°. 180 − 120.
  3. From Year 9: in a right-angled triangle the legs are 6 and 8. Find the hypotenuse.
    Answer10. 36 + 64 = 100.
Words used in this lesson
sine rule
a side divided by the sine of its opposite angle is the same for all three sides. For example: a ÷ sin A = b ÷ sin B.
cosine rule
the square of a side equals the sum of the squares of the others minus 2ab cos C. For example: used with two sides and the angle between them.
opposite
across from: side a is opposite angle A. For example: the side that does not touch the angle.
included angle
the angle between two given sides. For example: the angle at C between a and b.
obtuse
an angle between 90° and 180°. For example: an angle of 120°.
degree mode
the calculator setting that reads angles in degrees. For example: needed for sin 30 = 0.5.

Across the lake

A surveyor wants the distance between two points A and B on opposite shores of a lake near Kisumu. She cannot cross the water. She stands at A, walks along the shore to a point C, and measures some of the distances and angles with an instrument. There is no right angle anywhere in the picture.

Predict first: from the Year 9 right-angled triangle methods you know, can you find AB? How many of the three sides and three angles do you think she needs to have measured, and which ones? Use the first figure to see a pattern in triangles with no right angle, and the second to see how a side depends on the angle between two others.

40
60
15.6a ÷ sin A, the same as b ÷ sin B and c ÷ sin C
80angle C (degrees)

A = 40°, B = 60°, C = 80°. With a = 10: b = 13.5 and c = 15.3. Each side ÷ sine of its opposite angle = 15.6.

Show me the steps
  1. A = 40°, B = 60°: the third angle is C = 80°. The sides are b = 13.5 and c = 15.3, and each side divided by the sine of its opposite angle is 15.6.
  2. A = 30°, B = 80°: C = 70°. Now a ÷ sin A = 10 ÷ 0.5 = 20, and b = 19.7, c = 18.8, with b ÷ sin B and c ÷ sin C also 20.
  3. A = 90°, B = 40°: C = 50°. The triangle has a right angle, sin 90° = 1, and the value is 10, the hypotenuse. The sine rule includes the right-angled case.

Two rules for any triangle. Side a is opposite angle A, side b is opposite B and side c is opposite C.

  • The sine rule: a ÷ sin A = b ÷ sin B = c ÷ sin C. Use it when you know an angle and the side opposite it, plus one more angle, and want a side.
  • Why it works: the height from C to side AB is b sin A, and it is also a sin B. So b sin A = a sin B, which rearranges to a ÷ sin A = b ÷ sin B.
  • The cosine rule: c2 = a2 + b2 − 2ab cos C. Use it when you know two sides and the angle between them, and want the third side.
  • Area = ½ ab sin C, from two sides and the angle between them.

The angles of a triangle add to 180°, so two angles give the third.

60
8.2the third side c
63 lessc squared compared with 7 squared + 9 squared

Sides 7 and 9 with C = 60°: c² = 130 − 63 = 67, so c = 8.2. The square of c is 63 less than 7² + 9².

Show me the steps
  1. C = 60°: c2 = 72 + 92 − 2 × 7 × 9 × cos 60° = 130 − 63 = 67, so c = 8.2. The square of c is smaller than 130 by 63.
  2. C = 90°: cos 90° = 0, so the term disappears and c2 = 130, the Pythagoras value. c = 11.4.
  3. C = 120°: cos 120° = −0.5, so subtracting a negative adds: c2 = 130 + 63 = 193 and c = 13.9. The square of c is now larger than 130.

Check the pairs, and check the mode. In the sine rule the angle and side must be opposite each other. In the cosine rule the angle is between the two known sides. With an angle of 90° the term 2ab cos C is zero and the cosine rule becomes Pythagoras. And if the calculator is in radian mode, sin 30 gives about −0.99, not 0.5, so check it first.

Worked examples

In triangle ABC, A = 50°, B = 70° and a = 8. Find b.

  1. Side a is opposite A and side b is opposite B, so the sine rule applies: b ÷ sin 70° = 8 ÷ sin 50°.
  2. b = 8 × sin 70° ÷ sin 50° = 8 × 0.9397 ÷ 0.7660 = 9.8 to one decimal place.

Two sides are a = 6 and b = 10, and the angle between them is C = 70°. Find the third side c.

  1. Two sides and the angle between them: the cosine rule. c2 = 36 + 100 − 2 × 6 × 10 × cos 70°.
  2. cos 70° = 0.3420, so c2 = 136 − 41.04 = 94.96.
  3. c = √94.96 = 9.7 to one decimal place.

Your go. The first step is given. Two sides of a triangle are 7 and 12 and the angle between them is 35°. Find its area.

  1. The area is ½ × 7 × 12 × sin 35°.
  2. Work out sin 35°, then multiply and halve.
Check the answer

½ × 7 × 12 = 42, and sin 35° = 0.5736, so the area is 42 × 0.5736 = 24.09, which is 24.1 square units. A common slip is to use cos 35° or to leave out the half.

With a calculator

Put the calculator in degree mode before anything else. Check with sin 30, which should give 0.5. In radian mode it gives a different number (about −0.99) and every later answer is wrong.

For the sine rule, type the top of the fraction (for example 15 × sin 25) and then divide by sin 65, using brackets or the equals key, so that the division applies to the whole top. For the cosine rule, work out the whole of a2 + b2 − 2ab cos C first, write it down, and only then press the square-root key.

Check before you go on

1. In triangle PQR, PQ = 8, QR = 11 and angle Q = 40°. You want the length PR. Which rule do you use?

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. In triangle ABC, A = 30°, B = 45° and a = 6. Use the sine rule to find b, to 1 decimal place.

3. Side a = 7 is opposite angle A = 30°. What is the value of a ÷ sin A?

Level 2 · Building

4. Two sides of a triangle are 5 and 8 and the angle between them is 50°. Find the third side, to 1 decimal place.

5. In triangle ABC, B = 35°, C = 80° and c = 12. Find b, to 1 decimal place.

Level 3 · Stretch

6. Two sides of a triangle are 9 and 14 and the angle between them is 40°. Find its area, to 1 decimal place.

7. Two sides of a triangle are 5 and 7 and the angle between them is 120°. Find the third side, to 1 decimal place.

Something to think about

In the cosine rule, what happens to the third side as the angle C gets closer and closer to 180°? What shape is the triangle becoming, and what is c in terms of a and b?

One way to think about it

The triangle flattens into a straight line, with A and B on opposite sides of C. Then c is a + b, because cos 180° = −1 and c2 = a2 + b2 + 2ab = (a + b)2.

Exit ticket

8. In triangle ABC, A = 62°, B = 48° and a = 10. Which expression gives side c?