Free textbooks › Year 9 Maths textbook
KS3 Maths · Year 9Chapter 8 · Pythagoras and trigonometry
Year 9 · Chapter 8

Pythagoras and trigonometry

By the end of this chapter you will be able to

  • Find the hypotenuse or a shorter side of a right-angled triangle with Pythagoras, and give an exact answer
  • Use sin, cos and tan to find a missing side, given the angle and one side
  • Use inverse sine, cosine and tangent to find a missing angle from two sides
  • Write exact values of sin, cos and tan at 30°, 45° and 60°, and use them in calculations
  • Solve problems in context, and combine Pythagoras and trigonometry in a multi-step question

You should already know

  • Pythagoras to find a hypotenuse or shorter side in simple triangles (Year 8, Chapter 9)
  • Angles in a triangle and the angle sum (Year 8, Chapter 9)
  • Surds and simplifying √50 = 5√2 (Year 9, Chapter 1)

Where this leads

  • Vectors, bearings and compound measures (Year 9, Chapter 10)
  • Similar triangles and scale factors (Year 9, Chapter 9)
  • The sine and cosine rules and 3-D trigonometry in GCSE

National Curriculum for England, Key Stage 3 mathematics, Year 9. Content: Use Pythagoras’ theorem in two-dimensional problems, including composite shapes and the diagonal of a cuboid, Define and use sine, cosine and tangent in right-angled triangles to find unknown sides and angles, Know the exact values of sin, cos and tan for 30°, 45° and 60°, Solve problems in context with trigonometry, such as angles of elevation and depression, ladders and slopes, Combine Pythagoras and trigonometry in multi-step problems.

Exercise P  Do you remember?

Non-calculator
1
FWork out 42 + 32.
2
FWhat is the square root of 49?
3
FThe angles in a triangle add up to how many degrees?
4
FWhich side of a right-angled triangle is opposite the right angle?
5
FWork out 180 − 90 − 35.
8.1

Pythagoras' theorem in two dimensions

Year 9
Use Pythagoras' theorem in two-dimensional problems, including composite shapes
Video lesson: Pythagoras' theorem in two dimensionsComing soon
A ladder 10 m long leans against a wall. Its base is 6 m from the wall. How high up the wall does it reach? The ladder, the wall and the ground form a right-angled triangle.

Pythagoras' theorem says that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: c2 = a2 + b2. The hypotenuse c is always the longest side, opposite the right angle.

To find a shorter side, rearrange: a2 = c2 − b2.

Key ideas

  • Pythagoras: hypotenuse2 = side2 + side2.
  • To find the hypotenuse, add the squares and take the square root.
  • To find a shorter side, subtract the squares and take the square root.
  • Give an exact answer as a surd if the question does not ask for a decimal.
Worked example 8.1.1
A right-angled triangle has shorter sides 3 cm and 4 cm. Find the hypotenuse.

Method. Use Pythagoras’ theorem to find the missing side.

  1. Write the formula. In a right-angled triangle, hypotenuse² = a² + b², where the hypotenuse is opposite the right angle.
    c2 = a2 + b2
  2. Substitute the sides. Put the two shorter sides in.
    c2 = 32 + 42
  3. Work out the squares and add.
    c2 = 9 + 16 = 25
  4. Take the square root. The hypotenuse is the positive square root.
    c = √25 = 5

Check. Check: 5 squared is 25, which equals 32 + 42.

Watch out. The hypotenuse is always the longest side. Do not add the squares when you need a shorter side.

Worked example 8.1.2
A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other shorter side.

Method. Use Pythagoras’ theorem to find the missing side.

  1. Write the formula. Rearrange to find a shorter side: a2 = c2 − b2.
    a2 = c2 − b2
  2. Substitute the values. The hypotenuse is 13 and the other side is 5.
    a2 = 132 − 52
  3. Work out.
    a2 = 169 − 25 = 144
  4. Take the square root.
    a = √144 = 12

Check. Check: 12 squared is 144, and 144 + 52 = 132.

Watch out. The hypotenuse is always the longest side. Do not add the squares when you need a shorter side.

Worked example 8.1.3
A rectangle is 8 cm by 6 cm. Find the length of its diagonal.

Method. Use Pythagoras’ theorem to find the missing side.

  1. Write the formula. In a right-angled triangle, hypotenuse² = a² + b², where the hypotenuse is opposite the right angle.
    c2 = a2 + b2
  2. Substitute the sides. Put the two shorter sides in.
    c2 = 82 + 62
  3. Work out the squares and add.
    c2 = 64 + 36 = 100
  4. Take the square root. The hypotenuse is the positive square root.
    c = √100 = 10

Check. Check: 10 squared is 100, which equals 82 + 62.

Watch out. The hypotenuse is always the longest side. Do not add the squares when you need a shorter side.

Watch out

Watch out: when finding a shorter side, subtract the squares. Adding them gives the hypotenuse, which is longer.

Exam tip

Exam tip. Draw the triangle and write the hypotenuse squared first, so you do not mix up which side is which.

Exercise 8A  Pythagoras' theorem in two dimensions

Non-calculator
1
FA right-angled triangle has shorter sides 3 cm and 4 cm. Find the hypotenuse, giving an exact answer if needed.
2
FA right-angled triangle has shorter sides 5 cm and 12 cm. Find the hypotenuse, giving an exact answer if needed.
3
FA right-angled triangle has shorter sides 6 cm and 8 cm. Find the hypotenuse, giving an exact answer if needed.
4
FA right-angled triangle has shorter sides 2 cm and 3 cm. Find the hypotenuse, giving an exact answer if needed.
5
FA right-angled triangle has shorter sides 1 cm and 1 cm. Find the hypotenuse, giving an exact answer if needed.
6
FA right-angled triangle has shorter sides 4 cm and 6 cm. Find the hypotenuse, giving an exact answer if needed.
7
FA right-angled triangle has shorter sides 5 cm and 7 cm. Find the hypotenuse, giving an exact answer if needed.
8
FA right-angled triangle has shorter sides 8 cm and 15 cm. Find the hypotenuse, giving an exact answer if needed.
9
FA right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other shorter side, giving an exact answer if needed.
10
FA right-angled triangle has hypotenuse 10 cm and one shorter side 6 cm. Find the other shorter side, giving an exact answer if needed.
11
FA right-angled triangle has hypotenuse 7 cm and one shorter side 3 cm. Find the other shorter side, giving an exact answer if needed.
12
FA right-angled triangle has hypotenuse 9 cm and one shorter side 4 cm. Find the other shorter side, giving an exact answer if needed.
13
PSA ladder 10 m long leans against a wall with its base 6 m from the wall. How high up the wall does it reach?
14
PSA cuboid has sides 2 cm, 3 cm and 6 cm. Find the length of the diagonal from one corner to the opposite corner. Use the diagonal of the base first.
15
FA right-angled triangle has shorter sides 4 cm and 4 cm. Find the hypotenuse, giving an exact answer.
8.2

Sine, cosine and tangent: finding sides

Year 9
Define and use sine, cosine and tangent to find unknown sides
Video lesson: Sine, cosine and tangent: finding sidesComing soon
A ramp is at an angle to the ground. A builder needs the length of the ramp and its height, so she uses trigonometry to connect an angle with the sides of the triangle.

In a right-angled triangle, label the sides relative to the angle θ: the opposite side is across from θ, the adjacent side is next to θ (and not the hypotenuse), and the hypotenuse is opposite the right angle.

SOH CAH TOA: sin θ = opposite / hypotenuse, cos θ = adjacent / hypotenuse, tan θ = opposite / adjacent.

Key ideas

  • SOH: sin θ = opposite ÷ hypotenuse.
  • CAH: cos θ = adjacent ÷ hypotenuse.
  • TOA: tan θ = opposite ÷ adjacent.
  • Set the calculator to degree mode, and do not round until the last step.
Worked example 8.2.1
Find the side x when θ = 35° and the hypotenuse is 10 cm, where x is the side opposite θ.

Method. Choose the ratio from the sides you know, then rearrange and use the calculator.

  1. Label the sides. Mark the opposite, adjacent and hypotenuse sides for the angle θ.
    θ = 35°
  2. Choose the ratio. Use sin θ = opposite / hypotenuse.
    sin 35° = opposite / hypotenuse
  3. Write the equation. Use x = known side × ratio.
    x = 10 × sin 35°
  4. Use the calculator. sin 35° = 0.5736, correct to 4 d.p.
    x = 5.7358
  5. Round the answer. Give the answer to 1 decimal place.
    5.7 cm

Check. Check: the ratio of your answer and the known side gives sin 35° = 0.5736.

Watch out. Make sure your calculator is in degree mode, and do not round until the last step.

Worked example 8.2.2
Find the side x when θ = 65° and the opposite side is 15 cm, where x is the adjacent side.

Method. Choose the ratio from the sides you know, then rearrange and use the calculator.

  1. Label the sides. Mark the opposite, adjacent and hypotenuse sides for the angle θ.
    θ = 65°
  2. Choose the ratio. Use tan θ = opposite / adjacent.
    tan 65° = opposite / adjacent
  3. Write the equation. Use x = known side ÷ ratio.
    x = 15 ÷ tan 65°
  4. Use the calculator. tan 65° = 2.1445, correct to 4 d.p.
    x = 6.9946
  5. Round the answer. Give the answer to 1 decimal place.
    7.0 cm

Check. Check: the ratio of your answer and the known side gives tan 65° = 2.1445.

Watch out. Make sure your calculator is in degree mode, and do not round until the last step.

Worked example 8.2.3
Find the hypotenuse x when θ = 30° and the adjacent side is 9 cm.

Method. Choose the ratio from the sides you know, then rearrange and use the calculator.

  1. Label the sides. Mark the opposite, adjacent and hypotenuse sides for the angle θ.
    θ = 30°
  2. Choose the ratio. Use cos θ = adjacent / hypotenuse.
    cos 30° = adjacent / hypotenuse
  3. Write the equation. Use x = known side ÷ ratio.
    x = 9 ÷ cos 30°
  4. Use the calculator. cos 30° = 0.8660, correct to 4 d.p.
    x = 10.3923
  5. Round the answer. Give the answer to 1 decimal place.
    10.4 cm

Check. Check: the ratio of your answer and the known side gives cos 30° = 0.8660.

Watch out. Make sure your calculator is in degree mode, and do not round until the last step.

Watch out

Watch out: check which side is opposite and which is adjacent before you choose a ratio. The adjacent side is never the hypotenuse.

Exam tip

Exam tip. Write the ratio you are using, such as tan 65° = 15 / x, before you rearrange.

Exercise 8B  Finding sides with sine, cosine and tangent (calculator)

Calculator allowed
1
FFind x, where x is opposite the angle 35° and the hypotenuse is 10 cm.
2
FFind x, where x is adjacent to the angle 40° and the hypotenuse is 12 cm.
3
FFind x, where x is opposite the angle 25° and the adjacent side is 8 cm.
4
FFind x, where x is the hypotenuse, the angle is 50° and the opposite side is 7 cm.
5
FFind x, where x is the hypotenuse, the angle is 30° and the adjacent side is 9 cm.
6
FFind x, where x is adjacent to the angle 65° and the opposite side is 15 cm.
7
PSFind x, where x is the hypotenuse, the angle is 20° and the opposite side is 6 cm.
8
PSFind x, where x is opposite the angle 33° and the hypotenuse is 20 cm.
9
FFind x, where x is adjacent to the angle 45° and the hypotenuse is 4 cm.
10
FFind x, where x is opposite the angle 70° and the adjacent side is 3 cm.
11
PSFind x, where x is adjacent to the angle 58° and the hypotenuse is 11 cm.
12
RA right-angled triangle has an angle of 40°, with the hypotenuse 13 cm. Explain why you would use cos to find the adjacent side, and find it.
8.3

Finding angles with inverse trigonometry

Year 9
Use inverse sine, cosine and tangent to find an unknown angle
Video lesson: Finding angles with inverse trigonometryComing soon
A skateboard ramp rises 5 m over a horizontal distance of 12 m. The angle the ramp makes with the ground is found by using the inverse tangent of the ratio of the two sides.

If you know two sides of a right-angled triangle, use the ratio of those sides to find the angle. Use the inverse function to undo the ratio: sin−1, cos−1 or tan−1.

Key ideas

  • Choose the ratio from the two sides you know: sin, cos or tan.
  • Use the inverse function (shift or 2nd) to find the angle.
  • An angle found this way is less than 90°.
  • Round the angle to the number of decimal places asked for.
Worked example 8.3.1
Find θ when the opposite side is 5 and the hypotenuse is 8.

Method. Use the inverse trigonometric function on the ratio of the two sides.

  1. Choose the ratio. The two sides you know tell you which ratio to use: sin θ = opposite / hypotenuse.
    sin θ = opposite / hypotenuse
  2. Substitute. Put the two known sides in.
    sin θ = 5 ÷ 8 = 0.6250
  3. Use the inverse function. Use sin−1 (shift + sin on the calculator) to find θ.
    θ = sin−1(0.6250) = 38.6822°
  4. Round the answer. Give the angle to 1 decimal place.
    θ = 38.7°

Check. Check: sin of 38.7° is about 0.625.

Watch out. The inverse function is the button labelled sin−1, cos−1 or tan−1.

Worked example 8.3.2
Find θ when the opposite side is 7 and the adjacent side is 4.

Method. Use the inverse trigonometric function on the ratio of the two sides.

  1. Choose the ratio. The two sides you know tell you which ratio to use: tan θ = opposite / adjacent.
    tan θ = opposite / adjacent
  2. Substitute. Put the two known sides in.
    tan θ = 7 ÷ 4 = 1.7500
  3. Use the inverse function. Use tan−1 (shift + tan on the calculator) to find θ.
    θ = tan−1(1.7500) = 60.2551°
  4. Round the answer. Give the angle to 1 decimal place.
    θ = 60.3°

Check. Check: tan of 60.3° is about 1.750.

Watch out. The inverse function is the button labelled sin−1, cos−1 or tan−1.

Worked example 8.3.3
Find θ when the adjacent side is 6 and the hypotenuse is 10.

Method. Use the inverse trigonometric function on the ratio of the two sides.

  1. Choose the ratio. The two sides you know tell you which ratio to use: cos θ = adjacent / hypotenuse.
    cos θ = adjacent / hypotenuse
  2. Substitute. Put the two known sides in.
    cos θ = 6 ÷ 10 = 0.6000
  3. Use the inverse function. Use cos−1 (shift + cos on the calculator) to find θ.
    θ = cos−1(0.6000) = 53.1301°
  4. Round the answer. Give the angle to 1 decimal place.
    θ = 53.1°

Check. Check: cos of 53.1° is about 0.600.

Watch out. The inverse function is the button labelled sin−1, cos−1 or tan−1.

Watch out

Watch out: sin−1 is not 1 divided by sin. It is the angle whose sine is the value.

Exam tip

Exam tip. Check that the calculator is in degree mode before you start, and that the answer is less than 90°.

Exercise 8C  Finding angles (calculator)

Calculator allowed
1
FA right-angled triangle has opposite and hypotenuse sides 5 and 8. Find the angle θ, to 1 d.p.
2
FA right-angled triangle has adjacent and hypotenuse sides 6 and 10. Find the angle θ, to 1 d.p.
3
FA right-angled triangle has opposite and adjacent sides 7 and 4. Find the angle θ, to 1 d.p.
4
FA right-angled triangle has opposite and hypotenuse sides 3 and 10. Find the angle θ, to 1 d.p.
5
FA right-angled triangle has adjacent and hypotenuse sides 4 and 9. Find the angle θ, to 1 d.p.
6
FA right-angled triangle has opposite and adjacent sides 5 and 12. Find the angle θ, to 1 d.p.
7
FA right-angled triangle has opposite and adjacent sides 9 and 4. Find the angle θ, to 1 d.p.
8
FA right-angled triangle has opposite and hypotenuse sides 9 and 12. Find the angle θ, to 1 d.p.
9
FA right-angled triangle has adjacent and hypotenuse sides 7 and 8. Find the angle θ, to 1 d.p.
10
FA right-angled triangle has opposite and adjacent sides 3 and 3. Find the angle θ, to 1 d.p.
11
PSA ramp rises 1.5 m along a slope of length 10 m. Find the angle of the slope to the ground, to 1 d.p.
8.4

Exact values of sine, cosine and tangent

Year 9
Know the exact values of sin, cos and tan for 30°, 45° and 60°
Video lesson: Exact values of sine, cosine and tangentComing soon
A carpenter cuts a triangular shelf bracket from a square board. The triangle has angles 30°, 60° and 90°, and the lengths involve exact values, not decimals, so the carpenter keeps the surd form.

Some ratios have exact values that you should learn:

anglesincostan
30°1/2√3/21/√3
45°1/√21/√21
60°√3/21/2√3

Key ideas

  • sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3.
  • sin 45° = cos 45° = 1/√2, and tan 45° = 1.
  • sin 60° = √3/2, cos 60° = 1/2, and tan 60° = √3.
  • Give exact answers in surd form unless a decimal is asked for.
Worked example 8.4.1
Work out 10 sin 30° exactly.

Method. Use the exact value of sin 30°, then multiply out.

  1. Recall the exact value. The exact value of sin 30° is 1/2.
    sin 30° = 1/2
  2. Substitute. Put the exact value in place of the trigonometric ratio.
    10 × 1/2
  3. Simplify. Simplify the answer exactly, leaving any surd in its simplest form.
    5

Check. Check with a calculator: the value is about 5.

Watch out. Exact values come from the 30°, 45° and 60° triangles. They are not calculator decimals.

Worked example 8.4.2
Work out 4 sin 60° exactly.

Method. Use the exact value of sin 60°, then multiply out.

  1. Recall the exact value. The exact value of sin 60° is √3/2.
    sin 60° = √3/2
  2. Substitute. Put the exact value in place of the trigonometric ratio.
    4 × √3/2 = 2√3
  3. Simplify. Simplify the answer exactly, leaving any surd in its simplest form.
    2√3

Check. Check with a calculator: the value is about 2√3.

Watch out. Exact values come from the 30°, 45° and 60° triangles. They are not calculator decimals.

Worked example 8.4.3
Work out 3 tan 45° exactly.

Method. Use the exact value of tan 45°, then multiply out.

  1. Recall the exact value. The exact value of tan 45° is 1.
    tan 45° = 1
  2. Substitute. Put the exact value in place of the trigonometric ratio.
    3 × 1
  3. Simplify. Simplify the answer exactly, leaving any surd in its simplest form.
    3

Check. Check with a calculator: the value is about 3.

Watch out. Exact values come from the 30°, 45° and 60° triangles. They are not calculator decimals.

Watch out

Watch out: cos 60° = 1/2, not √3/2. The values for 30° and 60° are swapped between sine and cosine.

Exam tip

Exam tip. If a question asks for an exact answer, do not use a calculator decimal.

Exercise 8D  Exact values (non-calculator)

Non-calculator
1
FWork out 10 sin 30° exactly.
2
FWork out 8 cos 60° exactly.
3
FWork out 3 tan 45° exactly.
4
FWork out 4 sin 60° exactly, giving your answer as a surd.
5
FWork out 6 cos 30° exactly, giving your answer as a surd.
6
FWork out 2 tan 60° exactly.
7
FWork out 6 sin 45° exactly, giving your answer as a surd.
8
FWork out 4 cos 45° exactly, giving your answer as a surd.
9
PSWork out 3 tan 30° exactly.
10
PSWork out sin2 30° + cos2 60° exactly.
8.5

Angles of elevation and depression, ladders and slopes

Year 9
Solve problems in context with trigonometry, such as angles of elevation and depression
Video lesson: Angles of elevation and depression, ladders and slopesComing soon
From the top of a lighthouse, a boat is seen below the horizon. The angle of depression to the boat is measured. From the boat, the angle of elevation to the top is the same as the angle of depression from the top.

The angle of elevation is measured up from the horizontal. The angle of depression is measured down from the horizontal. They are equal when the two lines are drawn between the same two points.

A ladder leaning against a wall makes an angle with the ground. The slope of a ramp is the tangent of its angle.

Key ideas

  • Elevation is up from the horizontal. Depression is down from the horizontal.
  • Alternate angles: the angle of depression equals the angle of elevation from the other point.
  • Use the side you know and the side you need to choose the ratio.
  • Draw a diagram first, and mark the right angle and the known values.
Worked example 8.5.1
A tower is 12 m high. Find the angle of elevation from a point 20 m from its base.

Method. Use tan to find the angle of elevation from the height and the horizontal distance.

  1. Draw a diagram. The angle of elevation is measured up from the horizontal.
    height 12 m, distance 20 m
  2. Choose the ratio. The height is opposite the angle, and the distance along the ground is adjacent: tan θ = opposite / adjacent.
    tan θ = 12 ÷ 20
  3. Use the inverse function.
    θ = tan−1(0.6000) = 30.9638°
  4. Round the answer. Give the angle to 1 decimal place.
    31.0°

Check. A steeper angle means a larger height for the same distance.

Watch out. Angle of elevation is the angle above the horizontal. Angle of depression is the angle below it, and is equal to the matching elevation.

Worked example 8.5.2
A 6 m ladder leans against a wall at 70° to the ground. Find the height it reaches up the wall.

Method. Choose the ratio from the sides you know, then rearrange and use the calculator.

  1. Label the sides. Mark the opposite, adjacent and hypotenuse sides for the angle θ.
    θ = 70°
  2. Choose the ratio. Use sin θ = opposite / hypotenuse.
    sin 70° = opposite / hypotenuse
  3. Write the equation. Use x = known side × ratio.
    x = 6 × sin 70°
  4. Use the calculator. sin 70° = 0.9397, correct to 4 d.p.
    x = 5.6382
  5. Round the answer. Give the answer to 1 decimal place.
    5.6 m

Check. Check: the ratio of your answer and the known side gives sin 70° = 0.9397.

Watch out. Make sure your calculator is in degree mode, and do not round until the last step.

Worked example 8.5.3
From the top of a 20 m cliff, the angle of depression of a boat is 25°. Find the horizontal distance to the boat.

Method. Choose the ratio from the sides you know, then rearrange and use the calculator.

  1. Label the sides. Mark the opposite, adjacent and hypotenuse sides for the angle θ.
    θ = 25°
  2. Choose the ratio. Use tan θ = opposite / adjacent.
    tan 25° = opposite / adjacent
  3. Write the equation. Use x = known side ÷ ratio.
    x = 20 ÷ tan 25°
  4. Use the calculator. tan 25° = 0.4663, correct to 4 d.p.
    x = 42.8901
  5. Round the answer. Give the answer to 1 decimal place.
    42.9 m

Check. Check: the ratio of your answer and the known side gives tan 25° = 0.4663.

Watch out. Make sure your calculator is in degree mode, and do not round until the last step.

Watch out

Watch out: the angle at the top of a ladder is not the angle to the ground. Check which angle the question gives.

Exam tip

Exam tip. Mark the horizontal line on your diagram. The angles of elevation and depression are both measured from it.

Exercise 8E  Elevation, depression, ladders and slopes (calculator)

Calculator allowed
1
PSA point is 12 m above the ground and 20 m from the foot of a tower. Find the angle of elevation of the top, to 1 d.p.
2
PSA point is 8 m above the ground and 15 m from the foot of a tower. Find the angle of elevation of the top, to 1 d.p.
3
PSA point is 30 m above the ground and 40 m from the foot of a tower. Find the angle of elevation of the top, to 1 d.p.
4
PSA point is 5 m above the ground and 12 m from the foot of a tower. Find the angle of elevation of the top, to 1 d.p.
5
PSA point is 50 m above the ground and 30 m from the foot of a tower. Find the angle of elevation of the top, to 1 d.p.
6
PSA point is 7 m above the ground and 25 m from the foot of a tower. Find the angle of elevation of the top, to 1 d.p.
7
PSA 6 m ladder leans against a wall at 70° to the ground. Find how high up the wall it reaches, to 1 d.p.
8
PSThe same ladder is 6 m long at 70° to the ground. Find the distance of the foot of the ladder from the wall, to 1 d.p.
9
PSFrom the top of a 20 m cliff, the angle of depression of a boat is 25°. Find the horizontal distance from the boat to the foot of the cliff, to 1 d.p.
10
PSA ramp is 10 m long and rises 1.5 m. Find the angle of the ramp to the ground, to 1 d.p.
11
PSA ramp rises 1 m for every 8 m along the ground. Find the angle of the ramp to the ground, to 1 d.p.
12
PSFrom the top of a building, the angle of depression of a car is 30°. The building is 25 m high. Find the distance of the car from the foot of the building, to 1 d.p.
13
RExplain why the angle of depression from the top of a cliff to a boat equals the angle of elevation from the boat to the top.
14
PSA pole is 8 m high. The angle of elevation of its top from a point on the ground is 55°. Find the distance from the point to the foot of the pole, to 1 d.p.
8.6

Combining Pythagoras and trigonometry

Year 9
Combine Pythagoras and trigonometry in multi-step problems
Video lesson: Combining Pythagoras and trigonometryComing soon
A triangular garden has a right angle and a known hypotenuse. To find the fence needed, the gardener finds each side and adds them. Some questions need Pythagoras first, then trigonometry for an angle.

In a multi-step problem, plan first: which side or angle do you need, and what do you know? Use Pythagoras for two known sides, and a trigonometric ratio when one side and one angle are known.

Key ideas

  • Plan the steps before you calculate.
  • Pythagoras: two sides known, third side needed.
  • Trigonometry: one side and one angle known.
  • Keep full calculator values, and round only the final answer.
Worked example 8.6.1
A right-angled triangle has a hypotenuse of 13 cm and an angle of 40°. Find its perimeter, to 1 d.p.

Method. Use sine and cosine for the two shorter sides, then add all three sides.

  1. Find the opposite side. Opposite = hypotenuse × sin θ.
    13 × sin 40° = 8.3562
  2. Find the adjacent side. Adjacent = hypotenuse × cos θ.
    13 × cos 40° = 9.9586
  3. Add the three sides. The perimeter is the sum of all three sides.
    13 + 8.3562 + 9.9586 = 31.3148
  4. Round the answer. Give the perimeter to 1 decimal place.
    31.3 cm

Check. Check: the two shorter sides satisfy Pythagoras: 8.362 + 9.962 is about 169.0.

Watch out. Use the full calculator values in the working, and round only the final perimeter.

Worked example 8.6.2
A right-angled triangle has shorter sides 5 cm and 12 cm. Find the angle opposite the 5 cm side, to 1 d.p.

Method. Use Pythagoras to find the third side, then a trigonometric ratio to find the angle.

  1. Find the hypotenuse. Use Pythagoras to find the longest side first.
    c2 = 52 + 122 = 169
  2. Write the hypotenuse. Take the square root.
    c = √169 = 13.0000
  3. Choose the ratio for the angle. The side a is opposite the angle, and b is adjacent, so use tan.
    tan θ = 5 ÷ 12
  4. Use the inverse function.
    θ = tan−1(0.4167) = 22.6199°
  5. Round the answer. Give the angle to 1 decimal place.
    22.6°

Check. Check: the hypotenuse is 13.000, which is longer than both sides.

Watch out. Use Pythagoras only when two sides are known. Use the trigonometric ratio for the angle.

Worked example 8.6.3
A right-angled triangle has shorter sides 6 cm and 8 cm. Find its hypotenuse, then the angle opposite the 6 cm side.

Method. Use Pythagoras to find the third side, then a trigonometric ratio to find the angle.

  1. Find the hypotenuse. Use Pythagoras to find the longest side first.
    c2 = 62 + 82 = 100
  2. Write the hypotenuse. Take the square root.
    c = √100 = 10.0000
  3. Choose the ratio for the angle. The side a is opposite the angle, and b is adjacent, so use tan.
    tan θ = 6 ÷ 8
  4. Use the inverse function.
    θ = tan−1(0.7500) = 36.8699°
  5. Round the answer. Give the angle to 1 decimal place.
    36.9°

Check. Check: the hypotenuse is 10.000, which is longer than both sides.

Watch out. Use Pythagoras only when two sides are known. Use the trigonometric ratio for the angle.

Watch out

Watch out: check that your calculator stays in degree mode for every step, not only the first.

Exam tip

Exam tip. Write each step on a new line with its name, such as "Pythagoras" or "tan ratio", so the marks are easy to award.

Exercise 8F  Multi-step problems (calculator)

Calculator allowed
1
PSA right-angled triangle has a hypotenuse of 13 cm and an angle of 40°. Find its perimeter, to 1 d.p.
2
PSA right-angled triangle has a hypotenuse of 10 cm and an angle of 25°. Find its perimeter, to 1 d.p.
3
PSA right-angled triangle has shorter sides 5 cm and 12 cm. Find the angle opposite the 5 cm side, to 1 d.p.
4
PSA right-angled triangle has shorter sides 4 cm and 9 cm. Find the angle opposite the 4 cm side, to 1 d.p.
5
PSA right-angled triangle has shorter sides 3 cm and 7 cm. Find the angle opposite the 3 cm side, to 1 d.p.
6
PSTwo shorter sides of a right-angled triangle are 6 cm and 8 cm. Find the angle opposite the 6 cm side, to 1 d.p.
7
PSA right-angled triangle has shorter sides 2 cm and 5 cm. Find the angle opposite the 2 cm side, to 1 d.p.
8
PSA right-angled triangle has a hypotenuse of 17 cm and an angle of 35°. Find its perimeter, to 1 d.p.

Chapter summary

SkillMethod
Pythagorasc2 = a2 + b2 for the hypotenuse c. Rearrange for a shorter side.
SOH CAH TOAsin = opp/hyp, cos = adj/hyp, tan = opp/adj. Choose the ratio with the two sides you know or need.
Finding an angleUse the inverse function (sin−1, cos−1, tan−1) on the ratio.
Exact valuessin, cos and tan at 30°, 45° and 60° are surds or fractions. Learn the table.
Elevation and depressionMeasured from the horizontal. Equal to each other by alternate angles.

Can you…?

  • I can use Pythagoras' theorem to find a hypotenuse or a shorter side
  • I can use sin, cos and tan to find a missing side
  • I can use inverse trigonometry to find a missing angle
  • I know the exact values of sin, cos and tan at 30°, 45° and 60°
  • I can solve problems with elevation, depression and slopes
  • I can combine Pythagoras and trigonometry in a multi-step problem

Key terms

hypotenuse
the longest side of a right-angled triangle, opposite the right angle
opposite
the side across from the angle being used
adjacent
the side next to the angle being used, which is not the hypotenuse
angle of elevation
the angle measured up from the horizontal
angle of depression
the angle measured down from the horizontal

Exercise 8Z  Check-up

Calculator allowed
1
A right-angled triangle has shorter sides 9 cm and 12 cm. Find the hypotenuse.
2
A right-angled triangle has hypotenuse 17 cm and one shorter side 8 cm. Find the other shorter side.
3
Find x, where x is opposite the angle 42° and the hypotenuse is 9 cm. Give your answer to 1 d.p.
4
Find x, where x is adjacent to the angle 28° and the opposite side is 10 cm. Give your answer to 1 d.p.
5
A right-angled triangle has opposite 7 cm and hypotenuse 10 cm. Find the angle, to 1 d.p.
6
Work out 6 cos 30° exactly.
7
Work out 4 tan 60° exactly.
8
A tower is 30 m high. Find the angle of elevation of its top from a point 45 m from its foot, to 1 d.p.
9
A cuboid is 3 cm by 4 cm by 12 cm. Find the length of its space diagonal, using Pythagoras twice.
10
A right-angled triangle has shorter sides 2 cm and 6 cm. Find the angle opposite the 2 cm side, to 1 d.p.

Total 24 marks. Allow about 24 minutes.

Exercise 8X  Practice test questions

Calculator allowed
1
FFind the hypotenuse of a right-angled triangle with shorter sides 5 cm and 6 cm. Give your answer as a surd.[2 marks]
2
FA right-angled triangle has hypotenuse 10 cm and one shorter side 7 cm. Find the other shorter side, to 1 d.p.[2 marks]
3
FFind x, where x is opposite the angle 52° and the hypotenuse is 12 cm. Give your answer to 1 d.p.[2 marks]
4
FFind x, where x is the hypotenuse, the angle is 38° and the adjacent side is 9 cm. Give your answer to 1 d.p.[3 marks]
5
FA right-angled triangle has opposite 4 cm and adjacent 9 cm. Find the angle, to 1 d.p.[2 marks]
6
FWork out 2 sin 45° exactly.[2 marks]
7
PSA ladder 5 m long leans against a wall with its foot 3 m from the wall. Find the height it reaches up the wall.[3 marks]
8
PSA road rises 1 m for every 10 m along it. Find the angle of the road to the horizontal, to 1 d.p.[3 marks]
9
PSA right-angled triangle has a hypotenuse of 15 cm and an angle of 30°. Find its perimeter, to 1 d.p.[3 marks]
10
RExplain why sin 30° is less than sin 60°, and state the exact values of each.[2 marks]
← Chapter 7: Sequences and iterationChapter 9: Similarity and circle theorems →

Want a tutor to go through this chapter with your child?

Same tutor, same time, every week, live online. The first session is a free introduction.

💬 Book a Free Intro on WhatsApp
Chat on WhatsApp