1. The working height is 9m. As Ben explains, the fire-fighters place the foot of the ladder roughly a third of the working height away from the wall, giving an angle of about 71 degrees. Students could suggest similar triangles that would be created by applying this rule of thumb to different working heights. The length of the ladder was 9.5m.

2. The theta symbol is used throughout the programme. The definition of the opposite and the adjacent sides by their relation to the working angle could be reinforced here. Other ways of naming and labelling angles could be discussed, and comparisons made with classroom texts and examination papers.

3. The length is 63.4cm. The working height was 60cm. Both Ben (using the values given in Question 1 above) and Katie found the value of the sine ratio and obtained the same result to two decimal places. Students could work out the sine ratio for other triangles similar to the original one and establish that it remains constant. Differences in the answers due to rounding the lengths could also be discussed.

4. ‘sin = opposite / hypotenuse’. The idea that this formula deals with three values, and that knowing any two of them allows the third to be found, could be developed here. Some students may know, or like to invent, mnemonics for remembering the trigonometric ratios.

5. Katie keys ‘15 x sin 65’ into her calculator. The problem is to find the height that a ladder of length 15m could reach up a wall when placed at an angle of 65 degrees. The answer is 13.6m. Students could find out how to use their own calculators to achieve the same result. Some groups may benefit from looking in more detail at the rearrangement that was shown prior to this calculation.

6. Rachel starts off wearing a turquoise and black wet suit. For the jumps she changes to black and white.

7. The ramps are always inclined at 14 degrees and the take-off height is 1.6m. Students could suggest why they need to be kept constant. What would happen if the angle was steeper, for example? The term ‘angle of elevation’ could be introduced and defined. Ben calculated the length of the ramp as 6.6m. Again, the process of rearrangement and use of a calculator could be covered in more detail.

Ben’s mistake was to identify the sides incorrectly. The triangle could be drawn in other orientations and its sides named. The discussion could be extended to applications of trigonometry with other figures — for example, triangles where the altitude is known or required, or right-angled triangles within quadrilaterals.

9. The angles of elevation of the three ramps were 15, 10 and 5 degrees. Given that all the ramps were 150cm long — the same length as the final one that Katie measured — students could work out the kerb heights that each of the ramps would reach. They could also investigate how long a ramp would need to be for different kerb heights if the angle was required to be 5 degrees.

10. The ramp is 150cm long. Katie asks the viewers to find the angle of this ramp given that the height is 20cm. Students could do the calculation and check that they understand and can use the inverse-sine function. Real ramps around the school or in public places could be measured, to see if they are likely to be manageable.

Worksheet 2: Tick or Trash

Question 1

Student A is correct. Student B has done the division the wrong way round and then failed to use the inverse-sine function. The final answer is rounded to one significant figure and not to the nearest degree. Clearly the answer is not sensible.

Question 2

Student B is correct. Student A has not added on the height of the bottom of the string. Student A marked h on the initial diagram incorrectly. Student A has also omitted the units in the final answer.

Worksheet 3: Exam Practise Questions (Edexcel)

This is a selection of past exam questions from Edexcel, targeted at the middle ability range. You can use the sequence as it stands or select individual questions to suit your needs.

Given below is the mark scheme for each question, which will help you to see how marks are allocated and what the examiner is especially looking out for. The mark schemes use the following abbreviations:
oe
or equivalent
cao
correct answer only
ft
follow-through marks
dep
dependent
indep
independent
M
method marks
A
accuracy marks
B
benefit-of-doubt mark
SC
special case

Question 1

cos 23º = ⁶/_d

d = ⁶ / _{cos 23º} (=6.518) = 6.518

= 6.52 cm

Notes:

M1. M1 (dep) for making the d the subject. A1 accept 6.5 with working.

Question 2

(a) root(16² — 4²) = 15.5

(b) cos angle = ⁴/₁₆; angle = 75.5º

Question 3

(a) DC ÷ 8.2 = tan 37º

DC = 8.2 x tan 37º = 6.179...

(b) sin DAB = 8.2 ÷ 17.9 = 0.4581...

so DAB = 27.26...º =27.3º

Question 4

(a) x ÷ 1.4 = tan 62º

x = 1.4 x tan 62º (= 1.4 x 1.8807...) = 2.63(30...)

(b) cos x = 1.4 ÷ 3.5

x = 66.4218...

x = 66.4º

Question 5

(a) root(8² — 5²) = root(39)

Answer = 6.24

(b) i) cos C = ⁵/₈ gives 51.3º

OR sin^—1 ^‘6.24’/₈

OR tan^—1 ^‘6.24’/₅

Answer = 51.3º

ii) 270º + 51.3º = 321º

iii) 90º + 51.3º = 141º

Notes:

(a) M1 8² — 5². A1 for root(39) or root(64 — 25). A1 cao 6.24 or better (6.244998).

(b) i) M1 =cos C = ⁵/₈ (oe): any unambiguous indication of the use of cos with ⁵/₈. M1 cos^—1 (oe) (dep). A1 cao 51.3º rounds to... B1 270º + ‘51.3º’ ft accept answers given as 3 figures. B1 90º + ‘51.3º’ ft accept answers given as 3 figures. Or (ii) — 180º ft.

SC: give B1 for 302 ± 2º in (ii) AND 122 ± 2º in (iii)