Unit 7 — Coastal Rescue Navigation

10:10–10:25 · bearings

A boat travels from P to Q on a bearing of 072°.

Where should Q be? Sketch the direction before reading the rules.

Start at north. Turn clockwise through 72°.

Start at north

Every bearing begins from the north line.

Turn clockwise

The size of the turn gives the direction.

Write three digits

Use leading zeros for angles below 100°.

7° = 007°
38° = 038°

Now state the bearing from Q back to P. Reverse a bearing by adding or subtracting 180°.

Check

The reverse bearing is 252°. Bearings belong to directed journeys, so the direction matters.

10:25–11:05 · main investigation

Locate vessel C.

Observation stations A and B are 6.0 km apart on an east–west coastline. B is due east of A.

From A, vessel C is on bearing 038°.
From B, vessel C is on bearing 312°.

Draw north lines at both stations. Convert the bearings into the two interior triangle angles.

Question 1Coastal navigation

A rescue craft R is 4.5 km from A on a bearing of 318°. The craft can respond directly to C when CR is no more than 5.5 km.

(a)Find angles BAC, ABC and ACB.[3]

(b)Calculate AC and BC.[4]

(c)Find angle CAR.[2]

(d)Calculate the direct rescue route CR.[3]

(e)Find the area of triangle ACR.[2]

(f)Determine whether the craft can respond directly.[2]

[Total: 16 marks]

Check answers after completing the question

∠A = 52°, ∠B = 42°, ∠C = 86°. AC ≈ 4.02 km and BC ≈ 4.74 km. ∠CAR = 80°. CR ≈ 5.49 km, so the direct response is just within range. Area ACR ≈ 8.92 km².

11:05–11:20 · SL / HL

How certain is the location?

SL transfer

Check the navigation data

Suppose the bearing from A was recorded as 041° instead of 038°, while the B reading remains 312°.

Redraw the triangle.
Recalculate AC and BC.
Explain which reported distance changes more and why a small bearing error matters.

HL extension

One sine value, two possible angles — two possible triangles?

Triangle ABC has ∠A = 52°, a = 4.74 cm and c = 6.00 cm.

Use the sine rule to find sin C.
Find both possible values of C.
Draw both triangles.
Find both possible values of b.

Check answers

SL: the new triangle angles are A = 49°, B = 42°, C = 89°; AC ≈ 4.02 km and BC ≈ 4.53 km. HL: C ≈ 85.9° or 94.1°, giving b ≈ 4.03 cm or 3.36 cm.

11:20–11:40 · 3D geometry

From a 2D map to a 3D flight path

A relay drone must reach D, 1.0 km directly above vessel C. It can launch from A or B.

AC and BC are horizontal distances on the sea-level plane.
CD is perpendicular to the plane.
AD and BD are the two flight paths.
∠CAD and ∠CBD measure each path’s angle above the plane.

line–plane angle
= angle between the flight path and its horizontal projection

Find the two right triangles in the diagram. Which sides are known from Question 1?

12:20–12:48 · 3D route decision

Launch from A or B?

The drone must reach D, 1.0 km above C. Its maximum safe climb angle is 12.5° and its battery permits a direct route of at most 5.0 km.

Launch	Horizontal projection	Actual 3D route	Line–plane angle	Safe?
A	AC ≈ 4.02 km	Calculate AD	Calculate ∠CAD	Decide
B	BC ≈ 4.74 km	Calculate BD	Calculate ∠CBD	Decide

Recommend a launch point. Check both the angle constraint and the distance constraint.

Check the two routes

From A: AD ≈ 4.15 km and the line–plane angle ≈ 14.0°, so the climb is too steep. From B: BD ≈ 4.84 km and the angle ≈ 11.9°, so both conditions are met. Launch from B.

12:48–12:58 · transfer

Can you see the hidden right triangle?

A warehouse is 24 m long, 18 m wide and 8 m high. A cable runs from one bottom corner to the opposite top corner.

Find the cable’s horizontal projection across the floor.
Find the cable length.
Find the angle between the cable and the floor.
Explain why using 24 m as the adjacent side would be incorrect.

Check

Floor diagonal = 30 m. Cable length = √(30² + 8²) ≈ 31.0 m. Angle with the floor = tan⁻¹(8/30) ≈ 14.9°.

12:58–1:05 · holiday task

One integrated flight plan

A drone travels from station S toward a target on bearing 072°. The target is 3.8 km from S, and the drone must finish 0.60 km directly above it.

Draw a plan view with a north line and state the reverse bearing.
Draw the vertical side view.
Find the direct flight distance.
Find the angle between the flight path and the horizontal plane.
Write one sentence distinguishing a bearing from a line–plane angle.
HL: write an SSA data set that produces two valid triangles, then sketch both.

Keep it short: one careful page. Bring it to the first lesson after the holiday.

Open after completing the holiday task

Reverse bearing = 252°. Direct flight distance ≈ 3.85 km. Line–plane angle ≈ 9.0°. A bearing describes horizontal direction clockwise from north; a line–plane angle describes steepness above the horizontal projection.