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Cambridge IGCSE0580

Right-angled triangles

Mathematics 0580 Chapter Notes

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Right-angled triangles
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1. The three ratios

In a right-angled triangle, the hypotenuse is the longest side, opposite the right angle. Name the other two sides from the angle you are using: the opposite side faces the angle and the adjacent side touches it.

Choose the ratio that holds the side you know and the side you want. Write the equation first, then rearrange.

To find a side, put the unknown on top if you can. If the unknown is on the bottom, divide.

sin x = opposite/hypotenuse

cos x = adjacent/hypotenuse

tan x = opposite/adjacent

Examiner insight

“A small number of candidates used the incorrect ratio, usually sine.” “Candidates are advised to learn efficient calculator use so that they don’t have to re-type the rounded ratio.”

Worked example 13 marks

A right-angled triangle has an angle of 52°. The side adjacent to this angle is 9 cm. Calculate the hypotenuse.

  1. 1

    Adjacent and hypotenuse: use cosine. cos 52° = 9/h [1]

  2. 2

    h = 9/(cos 52°) [1]

  3. 3

    h = 14.61… = 14.6 cm [1]

2. Finding an angle

When the angle is the unknown, use the inverse function on the calculator: sin⁻¹, cos⁻¹ or tan⁻¹.

Work out the ratio and press the inverse key on the full calculator value. Do not type a rounded ratio back in.

Angles are in degrees. Give the answer correct to 1 decimal place.

x = sin⁻¹(opposite/hypotenuse)

x = tan⁻¹(opposite/adjacent)

Examiner insight

“There was some premature rounding of the ratio to e.g. 0.25 and 0.26 which led to inaccurate answers.” “The rubric states that angles should be given to 1 decimal place.”

Worked example 13 marks

A right-angled triangle has a hypotenuse of 11 cm. The side opposite the angle x is 6 cm. Calculate x.

  1. 1

    Opposite and hypotenuse: use sine. sin x = 6/11 [1]

  2. 2

    x = sin⁻¹(6/11) = 33.06… [1]

  3. 3

    x = 33.1° [1]

3. Pythagoras, trigonometry and bearings

Use Pythagoras' theorem when you know two sides and want the third: add the squares for the hypotenuse, subtract them for a shorter side. Use trigonometry when an angle is involved.

A problem may need both. Find a side with one method, then use it for the other. Keep the full calculator value between steps.

For a bearing, draw the north line first and mark the angle from it. Do not assume a line is West to East unless the question says so.

c² = a² + b²

a² = c² - b²

Examiner insight

“Some gave insufficient figures in their answer, and some gave the answer in surd form.” “The diagram was not to scale and candidates should not assume that CB was a West to East line.”

Worked example 14 marks

AC is a diameter of a circle, centre O. B is a point on the circle. AB = 14 cm and BC = 8 cm. Angle ABC = 90° (angle in a semicircle).

(b) Calculate AC.

(c) Calculate angle BAC.

  1. 1

    (b) AC² = 14² + 8² = 260 [1]

  2. 2

    AC = √(260) = 16.1 cm [1]

  3. 3

    (c) BC is opposite angle BAC and AB is adjacent: tan BAC = 8/14 [1]

  4. 4

    BAC = tan⁻¹(8/14) = 29.7° [1]

4. Shortest distance to a line

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

In a triangle, the shortest distance from a vertex to the opposite side makes a right angle with that side. It splits the triangle into right-angled triangles, so use the ratios on one of them. Do not assume it bisects the angle or the side.

Examiner insight

“Most candidates did not recognise that the shortest distance was found by using a perpendicular from side PR to the point Q.” “Quite a few found the length of AB and gave this as the answer for AC.”

Worked example 17 marks

Do not use a calculator. Triangle ABC has a right angle at B. Angle ACB = 30° and BC = 12 cm.

(a) Find the length of AC. Give your answer in the form a√b.

(b) Find the shortest distance from B to AC.

  1. 1

    (a) cos 30° = 12/AC [1]

  2. 2

    AC = 12 ÷ (√3)/2 = 24/(√3) [1]

  3. 3

    Multiply top and bottom by √3: (24√3)/3 [1]

  4. 4

    AC = 8√3 cm [1]

  5. 5

    (b) The shortest distance is the perpendicular from B to AC, with BC as the hypotenuse [1]

  6. 6

    12 × sin 30° = 12 × 1/2 [1]

  7. 7

    = 6 cm [1]

5. Angles of elevation and depression

Both angles are measured from the horizontal. An angle of elevation looks up from the horizontal. An angle of depression looks down from the horizontal.

The angle of depression of B from A is the angle at A between the horizontal and the line down to B. It equals the angle at B between the line and the ground.

Place the given angle on the sketch before choosing a ratio.

Examiner insight

“Most of these did correct work using a misplaced 52 ° (at the top of the triangle), effectively using an angle of depression of 38 °.” “A sizeable number of candidates found the hypotenuse, giving a fairly common incorrect answer of 380.7.”

Worked example 15 marks

The point A is on the top of a vertical building. The horizontal distance from the base of the building to B is 22 m. AB = 25 m and the height of the building is h m.

(a) Find the value of h.

(b) Work out the angle of depression of B from A.

  1. 1

    (a) h² = 25² - 22² [1]

  2. 2

    = 625 - 484 = 141 [1]

  3. 3

    h = √(141) = 11.9 m [1]

  4. 4

    (b) The angle of depression equals the angle at B. Use the two given sides: cos x = 22/25 [1]

  5. 5

    x = cos⁻¹(22/25) = 28.357… = 28.4° [1]

Recap

  • Hypotenuse is opposite the right angle. Name opposite and adjacent from the angle you use.
  • sin = opp/hyp, cos = adj/hyp, tan = opp/adj.
  • To find an angle, use the inverse function on the full calculator value. Give it to 1 decimal place.
  • Pythagoras: add the squares for the hypotenuse, subtract them for a shorter side.
  • For bearings, draw the north line first. Do not assume a line is West to East.
  • Extended: the perpendicular from a point to a line is the shortest distance.
  • Extended: angles of elevation and depression are 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 the angle x is 3 cm and the side adjacent to it is 4 cm. Write down the value of tan x as a fraction.1 mark
  2. Do not use a calculator. The shorter sides of a right-angled triangle are 5 cm and 12 cm. Find the length of the hypotenuse.2 marks
  3. A right-angled triangle has an angle of 38° and a hypotenuse of 15 cm. Calculate the length of the side opposite the 38° angle.2 marks
  4. Do not use a calculator. A ladder of length 6.5 m leans against a vertical wall. Its foot is on horizontal ground 2.5 m from the wall. How high up the wall does the ladder reach?3 marks
  5. The shorter sides of a right-angled triangle are 7 cm and 9 cm. Calculate the angle between the hypotenuse and the 9 cm side.3 marks
  6. A ship leaves port P and sails 12 km due east. It then sails 5 km due north to Q. Calculate the bearing of Q from P. Give your answer correct to 1 decimal place.3 marks
  7. Do not use a calculator. The point P is 13 cm from a point A on a straight line. The shortest distance from P to the line is 5 cm. Find the distance from A to the point on the line nearest to P.2 marks
  8. From a point on level ground 40 m from the foot of a vertical tower, the angle of elevation of the top of the tower is 28°. Calculate the height of the tower.3 marks
  9. Do not use a calculator. Buoy B is 15 km due north of buoy A. Buoy C is 20 km due east of B. A ship sails in a straight line from A to C. Find the shortest distance between the ship and B during the journey.5 marks
  10. From the top of a vertical cliff 60 m high, the angle of depression of a boat is 25°. The boat sails directly towards the base of the cliff until the angle of depression is 40°. Calculate the distance the boat sails.5 marks

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