The shape of the lesson
By the end of the lesson, learners can
- Looking up, looking down (Extended): In the lab, walk the observer towards the tower and watch the angle of elevation grow.
- Bearings with trigonometry: A bearing is an angle measured clockwise from north, written with three figures (\(060^\circ\), not \(60^\circ\)).
- The shortest distance (Extended): The shortest distance from a point to a straight line is along the perpendicular.
- Sketch: draw a clear diagram with the horizontal ground or north lines marked.
- Elevation and depression: both measured from the horizontal, and equal by alternate angles.
- Bearings: clockwise from north, three figures; back bearing is \(\pm 180^\circ\).
- Journeys: split into north-south and east-west parts, or spot a right-angle turn.
- Shortest distance: the perpendicular from the point to the line.
One picture every idea returns to
Angles of elevation and depression are measured from the HORIZONTAL, and by alternate angles they are equal. Bearings are measured clockwise from NORTH and written with three figures. Find the right-angled triangle, then use SOH CAH TOA or Pythagoras.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Trigonometry in the real world.. Surveyors measure angles up to rooftops, coastguards measure angles down to boats, and navigators steer by bearings. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Looking up, looking down (Extended) | In the lab, walk the observer towards the tower and watch the angle of elevation grow. Project tab 5 of the lab (Pythagoras and trigonometry lab, elevation and depression) and let a student drive it while the class predicts. Model the worked example Height from an angle of elevation (Extended) on the board, one line at a time. Your turn: (Extended) From a point 30 m from the base of a tree, the angle of elevation of the top is \(35^\circ\). Find the height of the tree. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Bearings with trigonometry | A bearing is an angle measured clockwise from north, written with three figures (\(060^\circ\), not \(60^\circ\)). Model the worked example Splitting a journey on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: The shortest distance (Extended) | The shortest distance from a point to a straight line is along the perpendicular. Model the worked example Closest approach (Extended) on the board, one line at a time. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 18 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The shortest distance (Extended), never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Measuring the angle of depression from the vertical cliff face. | It is always measured from the horizontal. If you put \(18^\circ\) at the top of the cliff against the vertical, you get \(80\tan 18^\circ = 26.0\) m, which is far too close. |
| Measuring anticlockwise, or from the wrong point. 'The bearing of \(B\) from \(A\)' means stand at \(A\), face north, turn clockwise until you face \(B\). | Return to Bearings with trigonometry: ask a pair to state the correct version and show it on the lab before moving on. |
| Using the distance along the path, \(12\cos 20^\circ = 11.3\) km, or the original 12 km. | The shortest distance is the perpendicular one, and it is always less than the distance to \(H\). |
Elevation, depression and bearings: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
(a) Write down the three-figure bearing of due west.
(b) Write down the three-figure bearing of south-east.
(c) The bearing of \(B\) from \(A\) is \(312^\circ\). Find the bearing of \(A\) from \(B\).
(Extended) A building is 24 m tall. From a point \(P\) on level ground, the angle of elevation of the top of the building is \(37^\circ\). Calculate the distance from \(P\) to the foot of the building.
A ship sails 40 km from port on a bearing of \(150^\circ\). Calculate how far (a) south and (b) east of the port the ship is.
A plane flies from \(A\) 120 km due east to \(B\), then 90 km due north to \(C\).
(a) Calculate the distance \(AC\).
(b) Find the bearing of \(C\) from \(A\).
(c) Find the bearing of \(A\) from \(C\).
(Extended) From the top of a lighthouse 35 m above sea level, the angles of depression of two boats are \(12^\circ\) and \(20^\circ\). The boats and the foot of the lighthouse are in a straight line, with both boats on the same side.
Calculate the distance between the two boats.
(Extended) A rock \(R\) is 10 km due east of a harbour \(H\). A boat leaves \(H\) and sails in a straight line on a bearing of \(058^\circ\).
(a) Calculate the shortest distance between the rock and the boat's path.
(b) How far has the boat sailed when it is closest to the rock?
(c) The boat must stay at least 5 km from the rock. Does it?
Answer key and marking notes
- (a) \(270^\circ\)
- (b) \(135^\circ\)
- (c) \(312^\circ - 180^\circ = 132^\circ\)
- \(\tan 37^\circ = \dfrac{24}{d}\)
- \(d = \dfrac{24}{\tan 37^\circ} = 31.8\) m (3 s.f.)
- The angle between the path and the south line is \(180^\circ - 150^\circ = 30^\circ\).
- (a) South: \(40\cos 30^\circ = 34.6\) km (3 s.f.)
- (b) East: \(40\sin 30^\circ = 20\) km
- (a) \(AC = \sqrt{120^2 + 90^2} = \sqrt{22\,500} = 150\) km
- (b) Angle north of east at \(A\): \(\tan^{-1}\left(\dfrac{90}{120}\right) = 36.87^\circ\). Bearing \(= 90^\circ - 36.87^\circ = 053.1^\circ\).
- (c) \(53.1^\circ + 180^\circ = 233.1^\circ\)
- Far boat (\(12^\circ\)): distance \(= \dfrac{35}{\tan 12^\circ} = 164.66\ldots\) m
- Near boat (\(20^\circ\)): distance \(= \dfrac{35}{\tan 20^\circ} = 96.16\ldots\) m
- Distance between boats \(= 164.66\ldots - 96.16\ldots = 68.5\) m (3 s.f.)
- Angle between \(HR\) (bearing \(090^\circ\)) and the path: \(90^\circ - 58^\circ = 32^\circ\).
- (a) Perpendicular distance \(= 10\sin 32^\circ = 5.30\) km (3 s.f.)
- (b) Distance along the path \(= 10\cos 32^\circ = 8.48\) km (3 s.f.)
- (c) \(5.30 > 5\), so yes, the boat stays more than 5 km from the rock.
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The bearing of B from A is \(057^\circ\). Find the bearing of A from B.
- A\(237^\circ\)
- B\(123^\circ\)
- C\(303^\circ\)
- D\(147^\circ\)
A ship sails 32 km from a port on a bearing of \(115^\circ\). How far south of the port is the ship?
- A13.5 km
- B29.0 km
- C14.9 km
- D75.7 km
Town B is 7 km east and 4 km north of town A. Find the bearing of B from A.
- A\(060.3^\circ\)
- B\(029.7^\circ\)
- C\(119.7^\circ\)
- D\(240.3^\circ\)
(Extended) From a point on level ground 65 m from the foot of a building, the angle of elevation of the top of the building is \(27^\circ\). Find the height of the building.
- A33.1 m
- B29.5 m
- C128 m
- D57.9 m
(Extended) From the top T of a cliff 60 m high, the angles of depression of two boats P and Q are \(35^\circ\) and \(16^\circ\). The boats and the foot F of the cliff are in a straight line. Find the distance PQ between the boats.
- A124 m
- B24.8 m
- C295 m
- D174 m
(Extended) A rock R is 14 km due east of a harbour H. A boat leaves H and sails in a straight line on a bearing of \(048^\circ\). Find the shortest distance between the rock and the boat's path.
- A9.37 km
- B10.4 km
- C12.6 km
- D14 km
Quiz key
The back bearing differs by \(180^\circ\): \(057^\circ + 180^\circ = 237^\circ\). The tempting \(123^\circ\) is \(180^\circ - 57^\circ\), which reflects the direction instead of reversing it. \(303^\circ\) is \(360^\circ - 57^\circ\), measured anticlockwise, and \(147^\circ\) adds \(90^\circ\).
The path is \(180^\circ - 115^\circ = 65^\circ\) from due south. In the right-angled triangle, the distance south is adjacent to the \(65^\circ\) angle: \(32\cos 65^\circ = 13.52\ldots = 13.5\) km. The tempting 29.0 km is \(32\sin 65^\circ\), the distance EAST. 14.9 km uses \(32\tan 25^\circ\), and 75.7 km divides by \(\cos 65^\circ\), longer than the journey itself.
The angle between AB and east is \(\tan^{-1}\dfrac{4}{7} = 29.74\ldots^\circ\). The bearing is measured from north, so it is \(90^\circ - 29.7^\circ = 060.3^\circ\). The tempting \(029.7^\circ\) gives the angle from EAST as if it were the bearing. \(119.7^\circ\) adds to \(90^\circ\) as if B were south of east, and \(240.3^\circ\) is the bearing of A from B.
\(\tan 27^\circ = \dfrac{h}{65}\), so \(h = 65\tan 27^\circ = 33.11\ldots = 33.1\) m. The tempting 29.5 m uses \(65\sin 27^\circ\), treating 65 m as the line of sight (the hypotenuse). 128 m divides by \(\tan 27^\circ\), and 57.9 m uses cosine.
\(FP = \dfrac{60}{\tan 35^\circ} = 85.69\) m and \(FQ = \dfrac{60}{\tan 16^\circ} = 209.24\) m, so \(PQ = 209.24 - 85.69 = 123.6 = 124\) m (3 s.f.). The tempting 24.8 m multiplies by the tangents instead of dividing. 295 m adds the distances, and 174 m uses the difference of the angles, \(\dfrac{60}{\tan 19^\circ}\), which is not an angle in any right-angled triangle here.
HR points east (bearing \(090^\circ\)), so the angle between the path and HR is \(90^\circ - 48^\circ = 42^\circ\). The perpendicular from R to the path makes a right-angled triangle with hypotenuse HR = 14 km, so the shortest distance is \(14\sin 42^\circ = 9.367\ldots = 9.37\) km. The tempting 10.4 km is \(14\cos 42^\circ\), the distance ALONG the path to the closest point. 12.6 km uses tangent, and 14 km is the distance from the harbour, not from the path.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.