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Cambridge IGCSE0607

Right-angled triangles

International Mathematics 0607 Chapter Notes

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Right-angled triangles
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Right-angled triangles notes

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In this chapter

  1. 1Sine, cosine and tangent
  2. 2Pythagoras' theorem
  3. 3Two-dimensional problems
  4. 4Bearings
  5. 5Shortest distance from a point to a line
  6. 6Angles of elevation and depression

1. Sine, cosine and tangent

Name the sides from the angle you are using. The hypotenuse is opposite the right angle. The opposite side faces the angle. The adjacent side touches the angle.

Pick the ratio that joins the side you have to the side you want. Use the inverse key to find an angle. Angles are given in degrees. Write an angle answer in degrees, correct to one decimal place.

On the graphic display calculator, set degrees, enter tan⁻¹(8.5 ÷ 5) and read off the angle.

sin x = opposite/hypotenuse

cos x = adjacent/hypotenuse

tan x = opposite/adjacent

Examiner insight

“Only about half the candidates used the correct trigonometric ratio the correct way around.” “Some candidates divided 5 by 8.5 before using the inverse tangent.”

Worked example 12 marks

The diagram shows a right-angled triangle.

Find the value of x.

  1. 1

    8.5 cm is opposite x and 5 cm is adjacent to x, so tan x = 8.5/5 [1]

  2. 2

    x = tan⁻¹(8.5 ÷ 5) = 59.5° [1]

2. Pythagoras' theorem

In a right-angled triangle the square of the hypotenuse is the sum of the squares of the other two sides.

To find the hypotenuse, add the squares. To find a shorter side, subtract the square of the other short side from the square of the hypotenuse.

Keep the exact value, such as √(441) or the full calculator display, until the last line.

c² = a² + b²

a² = c² - b²

Examiner insight

“Many candidates correctly used Pythagoras’ theorem to calculate the third side of the triangle although there were those who added the squares of 100 and 80.” “The only error seen was that made by candidates who added the squares in Pythagoras instead of subtracting.”

Worked example 14 marks

Work out the perimeter of this triangle.

  1. 1

    29 cm is the hypotenuse, so 29² = 20² + b² [1]

  2. 2

    b² = 841 - 400 = 441 [1]

  3. 3

    b = 21 cm [1]

  4. 4

    perimeter = 20 + 21 + 29 = 70 cm [1]

3. Two-dimensional problems

Draw a labelled sketch first. Split the shape into right-angled triangles and mark every length and angle you know.

Work one triangle at a time. Each length you find may be a side of the next triangle. Use the unrounded value in the next step and round only the final answer.

Opposite sides of a rectangle are equal, so a length in one triangle may be a side of the outline.

Examiner insight

“The suggestion to include a labelled sketch in their solution was overlooked by many, which made it difficult for them to make a correct trigonometrical statement in order to find the length of the ladder.” “However those who chose a trigonometrical ratio involving the length they had found in part (a) often used its approximate value of 57, thus losing out on the accuracy mark.”

Worked example 16 marks

The diagram shows the trapezium ACDE.

B lies on AC and angle ABE is 90°.

BCDE is a rectangle.

Angle AEB = 39°.

AB = 7.6 m and BC = 12.4 m.

Work out the perimeter of trapezium ACDE.

  1. 1

    tan 39° = 7.6/BE, so BE = 7.6 ÷ tan 39° [1]

  2. 2

    BE = 9.385…, which is also CD [1]

  3. 3

    sin 39° = 7.6/AE, so AE = 7.6 ÷ sin 39° [1]

  4. 4

    AE = 12.07… [1]

  5. 5

    AC = 7.6 + 12.4 = 20 and DE = BC = 12.4 [1]

  6. 6

    perimeter = 20 + 9.385… + 12.4 + 12.07… = 53.9 m [1]

4. Bearings

A bearing is an angle measured clockwise from north. Write it with three figures, so 90° is 090°.

Draw a north arrow at the point you are measuring from. Then mark the angle on your sketch before you calculate.

A bearing question often hides a right-angled triangle: north and east lines meet at a right angle.

Examiner insight

“Many candidates gave 90 instead of a three-figure bearing.” “The first step was to draw a north arrow at Q and to work out which bearing was being asked for.”

Worked example 14 marks

The diagram shows four points A, B, C and D.

B is due North of C.

C is due East of A.

(a) Find the bearing of D from A.

(b) Find BC.

  1. 1

    (a) AC points due East (090°) and D is 34° clockwise from it: 90 + 34 = 124° [1]

  2. 2

    (b) North and East are at right angles, so angle BCA = 90° and BC² = 6.9² - 4.6² [1]

  3. 3

    BC² = 47.61 - 21.16 = 26.45 [1]

  4. 4

    BC = 5.14 km [1]

5. Shortest distance from a point to a line

The perpendicular distance from a point to a line is the shortest distance to the line.

So "nearest point" or "shortest distance" means drop a perpendicular to the line. This makes a right angle, and a right-angled triangle to work in.

Examiner insight

“This shortest distance between a line and a point is never found to be easy, largely because of its recognition rather than the calculation.” “Those that realised that the required length was the perpendicular distance from D to AB were usually successful with the trigonometry.”

Worked example 13 marks

P is the point on the line CD that is the nearest point to A.

Calculate CP.

  1. 1

    nearest point: AP is perpendicular to CD, so angle CPA = 90° [1]

  2. 2

    in triangle CPA, CA = 9 is the hypotenuse and CP is adjacent to 35°: cos 35° = CP/9 [1]

  3. 3

    CP = 9 cos 35° = 7.37 cm [1]

6. Angles of elevation and depression

An angle of elevation is measured up from the horizontal. An angle of depression is measured down from the horizontal.

Draw the horizontal through the point you are looking from, then the line of sight. Find the right-angled triangle they make with a vertical line.

Worked example 13 marks

From a point on level ground 30 m from the foot of a vertical tower, the angle of elevation of the top of the tower is 38°. Calculate the height of the tower.

  1. 1

    height is opposite 38° and 30 m is adjacent: tan 38° = h/30 [1]

  2. 2

    h = 30 tan 38° [1]

  3. 3

    h = 23.4 m [1]

Recap

  • Hypotenuse is opposite the right angle. sin = opp ÷ hyp, cos = adj ÷ hyp, tan = opp ÷ adj.
  • Use the inverse key to find an angle. Give angles to one decimal place.
  • Pythagoras: c² = a² + b². Add for the hypotenuse, subtract for a shorter side.
  • Sketch, label, split into right-angled triangles, and round only at the end.
  • A bearing is measured clockwise from north, from the point you start at, in three figures.
  • Extended: the shortest distance from a point to a line is the perpendicular distance.
  • Extended: an angle of elevation or depression is measured from the horizontal.

End-of-chapter exercise

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

  1. Do not use a calculator. In a right-angled triangle, the side opposite angle x is 3 cm, the side adjacent to x is 4 cm and the hypotenuse is 5 cm. Write down tan x as a fraction.1 mark
  2. Do not use a calculator. A rectangle is 24 cm long and 7 cm wide. Find the length of a diagonal.2 marks
  3. Do not use a calculator. A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other shorter side.2 marks
  4. A ladder 6.0 m long leans against a vertical wall. Its foot is 1.5 m from the wall on level ground. Calculate the angle the ladder makes with the ground.3 marks
  5. In a right-angled triangle, an angle of 28° is opposite a side of length 9.0 cm. Calculate the length of the hypotenuse.3 marks
  6. A ship sails 8.0 km due East and then 6.0 km due North. (a) Calculate its distance from its start. (b) Calculate the bearing of the ship from its start.4 marks
  7. From a lighthouse L, a buoy B is 12 km away on a bearing of 040°. Calculate how far East of L the buoy is.3 marks
  8. Do not use a calculator. An isosceles triangle has two equal sides of 13 cm and a base of 10 cm. (a) Find the height of the triangle. (b) Find the area of the triangle.4 marks
  9. A drone hovers at D, 12 m above level ground. A vertical mast stands on the ground. From D, the angle of depression of the foot F of the mast is 21° and the angle of elevation of the top T of the mast is 34°. Calculate the height of the mast.5 marks
  10. Do not use a calculator. A straight coastal path runs from A to B, where AB = 25 km. A lighthouse L is 15 km from A and 20 km from B, and angle ALB = 90°. (a) Find the shortest distance from L to the path. (b) The lamp can be seen from any point within 13 km of L. Find the length of the path from which the lamp can be seen.7 marks

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