Hey there, future IB mathematicians! Ever wondered how pilots navigate through the skies or how ships find their way across vast oceans? It all comes down to a crucial concept: Bearing. This isn’t just about pointing north; it’s a precise, mathematical language of direction, and it’s a staple in your IB DP Math journey.
Today, we’re not just learning math; we’re embarking on a high-stakes rescue mission. Imagine this: A search and rescue helicopter (let’s call its position H) has just spotted two stranded hikers, Sarah (S) and Ben (B), in a remote, rugged area. From the helicopter, Sarah is observed at a bearing of $070^\circ$ and 8 km away. Simultaneously, Ben is sighted at a bearing of $160^\circ$ from the helicopter. The critical piece of information for the ground team is the direct distance between Sarah and Ben, which is 10 km.
Your immediate challenge: Calculate the distance of Ben from the helicopter (HB). This is a classic IB-style problem, combining geometry and trigonometry, and we’ll be circling back to solve this thrilling scenario after we’ve built up our foundational skills.
Table of Contents
The Bearing Basics: Navigating the North Star
At its heart, bearing is a universal language for direction. It’s precise, unambiguous, and absolutely vital for navigation across all forms of transport. In the IB Diploma Program, there are three golden rules you absolutely need to nail down:
Bearing Rules
- From North: All bearings are measured from a North reference line, which points directly upwards (or $000^\circ$).
- Clockwise: Measurements are always taken in a clockwise direction from this North line.
- Three Digits: Bearings are always expressed using three digits. So, $45^\circ$ becomes $045^\circ$, and $5^\circ$ is $005^\circ$. This prevents any ambiguity!
Let’s visualize a few common examples to get a feel for this:
- A bearing of $060^\circ$ means you turn $60^\circ$ clockwise from North. This would be between North and East.
- A bearing of $225^\circ$ takes you $225^\circ$ clockwise from North. This would point halfway between South and West. (Think $180^\circ$ to South, then another $45^\circ$ to the West).
- A bearing of $330^\circ$ is $330^\circ$ clockwise from North, landing you between North and West.
Right Angles to the Rescue! (Simple Triangles)
Often, your bearing problems will involve right-angled triangles. This is fantastic news because it means you can deploy your trusty SOH CAH TOA skills! Let’s walk through an example.
Example: Drone Flight
A drone (D) is flying. From its launch pad (L), it moves 7 km East and 5 km North. Calculate the bearing of the drone from the launch pad.
Solution:
- Draw a diagram! Place L at the origin. Draw a North line upwards.
- The drone is 5 km North and 7 km East. This creates a right-angled triangle where the North line is one side (adjacent to the angle we want), the East line is another side (opposite), and the line from L to D is the hypotenuse.
- We want the angle $\theta$ from the North line (vertical) clockwise to the line LD.
- Using trigonometry: $$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{7}{5}$$
- Calculate $\theta$: $$\theta = \tan^{-1}\left(\frac{7}{5}\right) \approx 54.462^\circ$$
- Since this angle is measured from North, clockwise, it is our bearing!
- Bearing of D from L: $054.5^\circ$ (to 3 significant figures).
What if we wanted to find the bearing of the launch pad from the drone? This is where the concept of a ‘back bearing’ comes in, which we’ll cover in more detail soon. But for now, remember that North lines at different points are parallel. This is a geometry superpower!
Quick Dive into Back Bearings (from Right Triangles)
Example: Back Bearing
Using our drone example: What is the bearing of the Launch Pad (L) from the Drone (D)?
Solution:
- The bearing of D from L was $054.462^\circ$.
- Draw a North line from D. This line is parallel to the North line from L.
- The angle between the North line at D and the line DL (pointing towards L) will be the sum of $180^\circ$ and an alternate interior angle related to our original $\theta$.
- Specifically, the angle formed by the South line at D and the line DL will be $54.462^\circ$. (Imagine a ‘Z’ shape with the North lines and the transversal line LD).
- So, the bearing of L from D is $180^\circ + 54.462^\circ = 234.462^\circ$.
- Bearing of L from D: $234.5^\circ$ (to 3 s.f.).
Pro Tip: Diagram Power!
Always, always, ALWAYS start with a clear diagram. Draw your North lines, mark your angles, and label distances. This visual representation is half the battle won, helping you identify right angles, parallel lines, and the appropriate trigonometric rule to use.
Beyond 90 Degrees: Sine and Cosine Rule in Action
Not every triangle in a bearing problem will be a neat right-angled one. This is where your knowledge of the Sine Rule and Cosine Rule becomes your best friend. They’re indispensable for finding unknown sides and angles in general triangles.
Example: Cyclist’s Journey
Two cyclists, Anne (A) and Chris (C), start from the same Park Entrance (P). Anne cycles 15 km on a bearing of $055^\circ$. Chris cycles 12 km on a bearing of $135^\circ$.
Part 1: Calculate the direct distance between Anne and Chris (AC).
Solution Part 1:
- Draw P as the origin with a North line. Plot A and C based on their bearings and distances. This forms triangle PAC.
- Find the angle at P (angle APC) within the triangle. This is the difference between the two bearings: $135^\circ – 055^\circ = 80^\circ$.
- We now have two sides (PA = 15 km, PC = 12 km) and the included angle (APC = $80^\circ$). This is perfect for the Cosine Rule!
- Cosine Rule: $$AC^2 = PA^2 + PC^2 – 2(PA)(PC)\cos(\angle APC)$$
- Substitute values: $$AC^2 = 15^2 + 12^2 – 2(15)(12)\cos(80^\circ)$$
- $$AC^2 = 225 + 144 – 360(0.1736)$$
- $$AC^2 = 369 – 62.50 \approx 306.50$$
- $$AC = \sqrt{306.50} \approx 17.507 \text{ km}$$
- The distance between Anne and Chris is approximately $17.5 \text{ km}$ (to 3 s.f.).
Part 2: Find the bearing of Chris (C) from Anne (A).
Solution Part 2:
- First, we need an angle inside triangle PAC. Let’s find $\angle PAC$ using the Sine Rule.
- Sine Rule: $$\frac{PC}{\sin(\angle PAC)} = \frac{AC}{\sin(\angle APC)}$$
- Substitute values: $$\frac{12}{\sin(\angle PAC)} = \frac{17.507}{\sin(80^\circ)}$$
- Rearrange: $$\sin(\angle PAC) = \frac{12 \cdot \sin(80^\circ)}{17.507} \approx \frac{12 \cdot 0.9848}{17.507} \approx 0.6759$$
- Calculate angle: $$\angle PAC = \sin^{-1}(0.6759) \approx 42.53^\circ$$
- Now, draw a North line from A. The bearing of A from P was $055^\circ$. The line AP points northeast from P.
- To find the bearing of P from A (the back bearing): $055^\circ + 180^\circ = 235^\circ$. This is the angle from the North line at A, clockwise to the line AP.
- We want the bearing of C from A, which is the angle from the North line at A, clockwise to the line AC.
- From our diagram, the line AC is positioned such that the angle $\angle PAC = 42.53^\circ$ is formed between AP and AC. Specifically, line AC is ‘less clockwise’ than line AP, relative to the North line at A.
- Bearing of C from A = (Bearing of P from A) – $\angle PAC$
- Bearing of C from A = $235^\circ – 42.53^\circ = 192.47^\circ$.
- Bearing of C from A: $192.5^\circ$ (to 3 s.f.).
Choosing Your Rule Wisely
- Use the Cosine Rule when you know two sides and the included angle (SAS) to find the third side, or when you know all three sides (SSS) to find an angle.
- Use the Sine Rule when you know two angles and any side (AAS or ASA), or two sides and a non-included angle (SSA, though beware the ambiguous case!) to find other sides or angles.
In bearing problems, you often use the Cosine Rule first to find an unknown side, and then the Sine Rule to find an internal angle necessary for calculating a final bearing.
The Path Back: Understanding Back Bearings
The concept of a ‘back bearing’ is super practical: it’s the bearing from your destination back to your starting point. It’s not just a mathematical trick; it’s essential for return journeys or establishing relative positions from multiple vantage points.
Back Bearing Rule
If you have a bearing from point A to point B (let’s call it $\theta$), the back bearing from B to A is:
- If $\theta < 180^\circ$: Back Bearing = $\theta + 180^\circ$
- If $\theta \ge 180^\circ$: Back Bearing = $\theta – 180^\circ$
This rule works because the two North lines at A and B are parallel, creating alternate interior angles that differ by exactly $180^\circ$ on a straight line.
Example: Calculating Back Bearings
- Scenario 1: The bearing from a campsite (C) to a mountain peak (M) is $085^\circ$. What is the bearing of the campsite from the mountain peak?
- Solution 1: Since $085^\circ < 180^\circ$, we add $180^\circ$. Bearing of C from M = $085^\circ + 180^\circ = 265^\circ$.
- Scenario 2: A ship (S) observes a buoy (B) at a bearing of $295^\circ$. What is the bearing of the ship from the buoy?
- Solution 2: Since $295^\circ \ge 180^\circ$, we subtract $180^\circ$. Bearing of S from B = $295^\circ – 180^\circ = 115^\circ$.
True Bearing vs. Relative Bearing
In real-world navigation, you might hear the term “relative bearing.” This refers to an angle measured relative to your own heading (e.g., “target is 30 degrees to my starboard”). However, in IB Math (and generally in most academic contexts), when we say “bearing,” we almost always mean the True Bearing – measured from North, clockwise, using three digits. If a question intends otherwise, it will explicitly state it.
Solving the High Seas Challenge! (Our IB Problem)
Alright, mathematicians, it’s time to put all our newly acquired skills to the test and solve our initial rescue mission! We need to find the distance of Ben (B) from the helicopter (H), which we’ll call HB.
IB Exam Challenge: The Rescue Mission
From a search and rescue helicopter (H), Sarah (S) is observed at a bearing of $070^\circ$ and 8 km away. Ben (B) is simultaneously sighted at a bearing of $160^\circ$ from H. The direct distance between Sarah and Ben (SB) is 10 km.
Challenge: Calculate the distance of Ben from the helicopter (HB).
Step-by-Step Solution:
- Visualize and Diagram: Draw the helicopter (H) at the origin. Draw a North line. Plot Sarah (S) and Ben (B) based on their bearings. This forms a triangle HSB.
- Identify Knowns:
- HS = 8 km (distance to Sarah)
- SB = 10 km (distance between Sarah and Ben)
- Bearing of S from H = $070^\circ$
- Bearing of B from H = $160^\circ$
- Goal: Find HB (let’s call it $x$)
- Find the Included Angle: The angle inside our triangle at the helicopter, $\angle SHB$, is the difference between the two bearings: $160^\circ – 070^\circ = 90^\circ$. (Aha! This means it’s a right-angled triangle, which simplifies things, but the Cosine Rule works even for non-right angles.)
- Apply the Cosine Rule (General Approach): Even though it’s a right triangle, let’s show how the Cosine Rule applies, as it’s the general method for non-right triangles (which most IB problems often are).
$$SB^2 = HS^2 + HB^2 – 2(HS)(HB)\cos(\angle SHB)$$ - Substitute Values:
$$10^2 = 8^2 + x^2 – 2(8)(x)\cos(90^\circ)$$ - Simplify: Remember that $\cos(90^\circ) = 0$. This simplifies the equation significantly!
$$100 = 64 + x^2 – 0$$
$$100 = 64 + x^2$$ - Solve for $x$:
$$x^2 = 100 – 64$$
$$x^2 = 36$$
$$x = \sqrt{36}$$
$$x = 6 \text{ km}$$ - Final Answer: The distance of Ben from the helicopter (HB) is exactly 6 km. Mystery solved! The rescue team now has crucial information.
Notice how sometimes the bearings might give you a $90^\circ$ angle, which allows for a quick Pythagoras check! But the Cosine Rule is the robust tool for any such scenario.
Practice, Recap & Your Next Steps
What a journey! We’ve navigated the foundational principles of bearing, learned to apply SOH CAH TOA to simple right triangles, and mastered the powerful Sine and Cosine Rules for those trickier non-right-angled scenarios. We also demystified back bearings, an often-overlooked but crucial concept.
Key takeaways to stick in your mental compass:
- Bearing Rules: Always from North, always clockwise, always three digits.
- Diagrams are Gold: Draw them, label them, love them.
- Toolbox: For right triangles, SOH CAH TOA. For non-right triangles, Sine Rule and Cosine Rule.
- Back Bearings: A simple $\pm 180^\circ$ trick.
Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.
Additional Practice
Ready to solidify your understanding? Tackle these practice problems. Remember to draw diagrams!
Question 1: A hiker starts from point P and walks to point Q, which is 5 km North and 12 km East of P. What is the bearing of Q from P?
Show Answer
Solution:
- Diagram: Draw P as the origin. Point Q forms a right-angled triangle with sides 5 km (North, adjacent) and 12 km (East, opposite).
- We need the angle $\theta$ from the North line clockwise to PQ.
- Using tangent: $$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{12}{5}$$
- $$\theta = \tan^{-1}\left(\frac{12}{5}\right) \approx 67.380^\circ$$
- Bearing of Q from P: $067.4^\circ$ (3 s.f.)
Question 2: What is the back bearing of point P from point Q in Question 1?
Show Answer
Solution:
- Original bearing from P to Q is $067.380^\circ$, which is less than $180^\circ$.
- Back Bearing of P from Q = $067.380^\circ + 180^\circ = 247.380^\circ$.
- Bearing of P from Q: $247^\circ$ (3 s.f.)
Question 3: An airplane flies from Airport A on a bearing of $110^\circ$ for 200 km to Airport B. What is the bearing of Airport A from Airport B?
Show Answer
Solution:
- Original bearing from A to B is $110^\circ$, which is less than $180^\circ$.
- Back Bearing of A from B = $110^\circ + 180^\circ = 290^\circ$.
- Bearing of A from B: $290^\circ$.
Question 4: A submarine travels from point X on a bearing of $240^\circ$ to point Y. What is the bearing of X from Y?
Show Answer
Solution:
- Original bearing from X to Y is $240^\circ$, which is greater than $180^\circ$.
- Back Bearing of X from Y = $240^\circ – 180^\circ = 060^\circ$.
- Bearing of X from Y: $060^\circ$.
Question 5: A boat leaves harbour H and sails 12 km to point C on a bearing of $040^\circ$. From C, it then sails 15 km to point D on a bearing of $130^\circ$. Calculate the distance HD.
Show Answer
Solution:
- Draw H as origin. North line from H. Plot C (12 km, $040^\circ$). Draw North line from C.
- Plot D (15 km from C, $130^\circ$ from C). This forms $\triangle HCD$.
- Find $\angle HCD$: The bearing from H to C is $040^\circ$. The back bearing of H from C (angle from North at C clockwise to CH) is $040^\circ + 180^\circ = 220^\circ$.
- The bearing from C to D (angle from North at C clockwise to CD) is $130^\circ$.
- The angle $\angle HCD$ is the difference between these two bearings: $\angle HCD = 220^\circ – 130^\circ = 90^\circ$.
- Since $\triangle HCD$ is a right-angled triangle, use Pythagoras theorem:
- $HD^2 = HC^2 + CD^2$
- $HD^2 = 12^2 + 15^2 = 144 + 225 = 369$
- $HD = \sqrt{369} \approx 19.209 \text{ km}$
- Distance HD: $19.2 \text{ km}$ (3 s.f.)
Question 6: From the solution to Question 5, find the bearing of D from H.
Show Answer
Solution:
- We have $\triangle HCD$ with $HC = 12$ km, $CD = 15$ km, $HD = 19.209$ km, and $\angle HCD = 90^\circ$.
- We need to find $\angle CHD$ (the angle at H in the triangle) to add to the initial bearing.
- Using the Sine Rule: $$\frac{\sin(\angle CHD)}{CD} = \frac{\sin(\angle HCD)}{HD}$$
- $$\frac{\sin(\angle CHD)}{15} = \frac{\sin(90^\circ)}{19.209}$$
- $$\sin(\angle CHD) = \frac{15 \cdot \sin(90^\circ)}{19.209} = \frac{15 \cdot 1}{19.209} \approx 0.78093$$
- $$\angle CHD = \sin^{-1}(0.78093) \approx 51.34^\circ$$
- The bearing of C from H is $040^\circ$. The angle $\angle CHD$ is clockwise from the line HC to HD.
- Bearing of D from H = (Bearing of C from H) + $\angle CHD$
- Bearing of D from H = $040^\circ + 51.34^\circ = 091.34^\circ$
- Bearing of D from H: $091.3^\circ$ (3 s.f.)
Question 7: A surveyor at point A measures the bearing to point B as $030^\circ$ and the distance as 7 km. From point A, the bearing to point C is $080^\circ$ and the distance is 9 km. Calculate the distance BC.
Show Answer
Solution:
- Draw A as origin. North line from A. Plot B (7 km, $030^\circ$) and C (9 km, $080^\circ$). This forms $\triangle ABC$.
- Find the included angle $\angle BAC = 080^\circ – 030^\circ = 50^\circ$.
- Use Cosine Rule to find BC:
- $BC^2 = AB^2 + AC^2 – 2(AB)(AC)\cos(\angle BAC)$
- $BC^2 = 7^2 + 9^2 – 2(7)(9)\cos(50^\circ)$
- $BC^2 = 49 + 81 – 126(0.64278)$
- $BC^2 = 130 – 80.99 \approx 49.01$
- $BC = \sqrt{49.01} \approx 7.0007 \text{ km}$
- Distance BC: $7.00 \text{ km}$ (3 s.f.)
Question 8: From the solution to Question 7, calculate the bearing of C from B.
Show Answer
Solution:
- From Q7, $AB=7$ km, $AC=9$ km, $BC=7$ km. $\angle BAC = 50^\circ$.
- Since $AB = BC = 7$ km, $\triangle ABC$ is an isosceles triangle with base AC. This means the angles opposite the equal sides are equal: $\angle BCA$ (opposite AB) is equal to $\angle BAC$ (opposite BC). So, $\angle BCA = 50^\circ$.
- Therefore, the third angle $\angle ABC = 180^\circ – (50^\circ + 50^\circ) = 180^\circ – 100^\circ = 80^\circ$.
- Now, to find the bearing of C from B:
- The bearing of B from A is $030^\circ$. The back bearing of A from B (angle from North at B clockwise to BA) is $030^\circ + 180^\circ = 210^\circ$.
- The angle $\angle ABC = 80^\circ$. From the diagram, the line BC is positioned further clockwise from the line BA.
- Bearing of C from B = (Bearing of A from B) + $\angle ABC$
- Bearing of C from B = $210^\circ + 80^\circ = 290^\circ$.
- Bearing of C from B: $290^\circ$.
Question 9: A hot air balloon (B) is observed from two ground stations, X and Y. Station X is 15 km due South of Station Y. The bearing of the balloon from X is $045^\circ$. The bearing of the balloon from Y is $120^\circ$. Calculate the distance of the balloon from Station Y (BY).
Show Answer
Solution:
- Draw diagram: Place Y at the origin. Draw a North line from Y. Station X is 15 km due South of Y, so YX = 15 km, and the bearing of X from Y is $180^\circ$.
- Draw North line from X (parallel to North line from Y).
- Bearing of B from X is $045^\circ$. Draw line XB at this bearing.
- Bearing of B from Y is $120^\circ$. Draw line YB at this bearing. This forms $\triangle XYB$.
- Find angles within $\triangle XYB$:
- $\angle XYB$: The line YX points South from Y ($180^\circ$). The line YB is at $120^\circ$. So, $\angle XYB = 180^\circ – 120^\circ = 60^\circ$.
- $\angle YXB$: The line XQ (North from X) is parallel to the North line from Y. The bearing of B from X is $045^\circ$. This is the angle from XQ to XB. So, $\angle YXB = 045^\circ$.
- $\angle XBY = 180^\circ – (\angle XYB + \angle YXB) = 180^\circ – (60^\circ + 45^\circ) = 180^\circ – 105^\circ = 75^\circ$.
- Use Sine Rule to find BY: $$\frac{BY}{\sin(\angle YXB)} = \frac{XY}{\sin(\angle XBY)}$$
- $$\frac{BY}{\sin(45^\circ)} = \frac{15}{\sin(75^\circ)}$$
- $$BY = \frac{15 \cdot \sin(45^\circ)}{\sin(75^\circ)} = \frac{15 \cdot 0.707107}{0.965926} \approx 10.9810 \text{ km}$$
- Distance BY: $11.0 \text{ km}$ (3 s.f.)
Question 10 (IB Exam Style Question – Section A): A ship (S) sails 20 km from Port (P) on a bearing of $065^\circ$ to reach Island A. From Island A, it then sails 30 km to Island B on a bearing of $150^\circ$.
- Calculate the distance from Port P to Island B.
- Find the bearing of Island B from Port P.
- Determine the bearing of Port P from Island B.
Reveal Solution
Solution:
- Diagram: Draw P as origin. North line from P. Plot A (20 km, $065^\circ$). Draw North line from A. Plot B (30 km from A, $150^\circ$ from A). This forms $\triangle PAB$.
- Part a) Calculate the distance from Port P to Island B (PB).
- In $\triangle PAB$: PA = 20 km, AB = 30 km.
- To find $\angle PAB$: The bearing of A from P is $065^\circ$. The back bearing of P from A (angle from North at A clockwise to AP) is $065^\circ + 180^\circ = 245^\circ$.
- The bearing of B from A (angle from North at A clockwise to AB) is $150^\circ$.
- The angle $\angle PAB$ is the difference between these two bearings: $\angle PAB = 245^\circ – 150^\circ = 95^\circ$.
- Using the Cosine Rule to find PB:
- $PB^2 = PA^2 + AB^2 – 2(PA)(AB)\cos(\angle PAB)$
- $PB^2 = 20^2 + 30^2 – 2(20)(30)\cos(95^\circ)$
- $PB^2 = 400 + 900 – 1200(-0.087156)$
- $PB^2 = 1300 + 104.587 \approx 1404.587$
- $PB = \sqrt{1404.587} \approx 37.4778 \text{ km}$
- Distance PB: $37.5 \text{ km}$ (3 s.f.)
- Part b) Find the bearing of Island B from Port P.
- We need to find $\angle APB$ using the Sine Rule in $\triangle PAB$:
- $$\frac{\sin(\angle APB)}{AB} = \frac{\sin(\angle PAB)}{PB}$$
- $$\frac{\sin(\angle APB)}{30} = \frac{\sin(95^\circ)}{37.4778}$$
- $$\sin(\angle APB) = \frac{30 \cdot \sin(95^\circ)}{37.4778} \approx \frac{30 \cdot 0.99619}{37.4778} \approx 0.79778$$
- $$\angle APB = \sin^{-1}(0.79778) \approx 52.92^\circ$$
- The bearing of A from P is $065^\circ$. The angle $\angle APB$ is added clockwise to this to get the bearing of B from P.
- Bearing of B from P = (Bearing of A from P) + $\angle APB$
- Bearing of B from P = $065^\circ + 52.92^\circ = 117.92^\circ$
- Bearing of B from P: $118^\circ$ (3 s.f.)
- Part c) Determine the bearing of Port P from Island B.
- This is the back bearing of B from P.
- The bearing of B from P is $117.92^\circ$. Since this is less than $180^\circ$, we add $180^\circ$.
- Back Bearing of P from B = $117.92^\circ + 180^\circ = 297.92^\circ$
- Bearing of P from B: $298^\circ$ (3 s.f.)