What’s up, IB Math legends! Ever felt like the math you learn is just a bunch of abstract numbers? Well, get ready to blow that idea out of the water. Imagine you’re a lighthouse keeper, 80 meters up, staring out at a calm sea. You spot a fishing boat and measure its angle of depression as 20°. A little while later, you look again, and the boat is closer—the angle of depression is now 50°. Your mission, should you choose to accept it, is to figure out exactly how far that boat traveled between your two sightings. Sound like a movie plot? Nope, it’s just a classic IB Math problem, and by the end of this post, you’ll be able to solve it like a pro.
We’re diving deep into the world of right-angled trigonometry, specifically focusing on the super-useful concepts of angles of elevation and depression. We’ll break down the definitions, work through some classic IB-style examples, cover some must-know exam tips, and then, yes, we’ll solve our lighthouse mystery. Let’s get started!
Table of Contents
Up or Down? The Basics of Elevation & Depression
Before we can solve anything, we need to speak the same language. It all starts from one simple, but crucial, line: the horizontal. Imagine a laser beam shooting straight out from your eyes, perfectly parallel to the ground. This is your starting point for everything.
Core Definitions
- Angle of Elevation: If you have to look UP from the horizontal to see an object (like a drone in the sky), the angle your line of sight makes with the horizontal is the angle of elevation. You are elevating your gaze.
- Angle of Depression: If you have to look DOWN from the horizontal to see an object (like a dog on the ground), the angle your line of sight makes with the horizontal is the angle of depression. You are depressing your gaze.
The single most common mistake is putting these angles inside the triangle at the wrong vertex. Always, always, always draw the horizontal line first! The angle must touch this line.
The Z-Angle Trick: Your New Best Friend
Here’s a geometry hack that will save you tons of time and prevent major headaches. Because your horizontal line of sight is parallel to the ground (or sea level), we can use a cool property of parallel lines.
Rule: The “Z-Angle” Shortcut
The angle of depression from an observer to an object is always equal to the angle of elevation from the object back to the observer. These are called alternate interior angles. If you draw the horizontal line from the observer, the line of sight, and the ground, you’ll see they form a ‘Z’ shape. The angles in the corners of the ‘Z’ are equal. This is amazing because it lets us move the angle of depression down into our right-angled triangle, making it much easier to work with.
Trig in Action: The Skate Ramp
Let’s put the theory to work with a classic SOH CAH TOA problem. Remember: SOH (Sine = Opp/Hyp), CAH (Cosine = Adj/Hyp), TOA (Tangent = Opp/Adj).
Example: The Skate Ramp
A city is building a new competition skate ramp. The take-off point needs to be 2.5 metres vertically above the ground. For a smooth transition, the angle of elevation of the ramp must be exactly 12°.
a) What is the length of the ramp’s surface?
b) How much horizontal floor space will the ramp take up?
Step 1: Draw the Diagram! This is non-negotiable. We have a right-angled triangle. The vertical height (the Opposite side to our angle) is 2.5 m. The ramp itself is the Hypotenuse. The floor space is the Adjacent side.
a) Finding the ramp length (Hypotenuse):
We know the Opposite (O) and want the Hypotenuse (H). That’s SOH!
$$ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} $$
$$ \sin(12^{\circ}) = \frac{2.5}{\text{Hypotenuse}} $$
Rearranging to solve for the hypotenuse:
$$ \text{Hypotenuse} = \frac{2.5}{\sin(12^{\circ})} = 12.025… \text{ m} $$
So, the length of the ramp is 12.0 m (to 3 s.f.).
b) Finding the floor space (Adjacent):
We know the Opposite (O) and want the Adjacent (A). That’s TOA!
$$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} $$
$$ \tan(12^{\circ}) = \frac{2.5}{\text{Adjacent}} $$
Rearranging to solve for the adjacent side:
$$ \text{Adjacent} = \frac{2.5}{\tan(12^{\circ})} = 11.76… \text{ m} $$
The ramp will take up 11.8 m of horizontal floor space (to 3 s.f.).
Next-Level Trig: The Canyon View
Ready to level up? Many IB problems involve not one, but two triangles stacked together. The secret is to find the side they share to link them.
Example: The Canyon View
A tourist stands at the edge of a canyon lookout point. Her eye level is 20 metres above the canyon floor. She looks across to a magnificent rock pinnacle on the other side. The angle of elevation to the top of the pinnacle is 48°, and the angle of depression to the base of the pinnacle (at the same level as the canyon floor) is 15°. How tall is the rock pinnacle?
Step 1: See the Two Triangles. Draw your diagram! You’ll see two right-angled triangles stacked on top of each other. They both share the same base: the horizontal distance across the canyon. Let’s call this distance $x$.
Step 2: Use the Bottom Triangle to Find $x$.
We’ll use the depression info. The angle inside this triangle (using our ‘Z-angle’ trick) is 15°. The opposite side is the tourist’s height above the floor, 20 m. We need the adjacent side, $x$. Time for TOA again!
$$ \tan(15^{\circ}) = \frac{20}{x} $$
$$ x = \frac{20}{\tan(15^{\circ})} = 74.641… \text{ m} $$
Pro Tip: Don’t Round Yet!
Do NOT round this value for $x$. Keep the full, glorious number in your calculator’s memory. Rounding too early will introduce errors and cost you marks!
Step 3: Use $x$ in the Top Triangle.
Now we look at the elevation triangle. The angle is 48°. The adjacent side is the value for $x$ we just found. We want to find the opposite side, which is the height of the pinnacle above the tourist’s eye level. Let’s call it $y$.
$$ \tan(48^{\circ}) = \frac{y}{x} $$
$$ y = x \times \tan(48^{\circ}) = 74.641… \times \tan(48^{\circ}) = 82.896… \text{ m} $$
Step 4: Find the Total Height.
The total height of the pinnacle is the bottom part (the 20 m from the canyon floor to the tourist’s eye level) plus the top part ($y$).
$$ \text{Total Height} = 20 + y = 20 + 82.896… = 102.896… \text{ m} $$
Rounding to 3 significant figures, the pinnacle is 103 m tall. Awesome!
Your IB Exam Survival Kit
Before we tackle the final challenge, let’s lock in some critical habits that will protect you from common mistakes on your exams.
Tip #1: Draw or Die!
Seriously. Always draw a large, clear diagram. It helps you visualize the problem and correctly identify your opposite, adjacent, and hypotenuse sides. The number one error is misplacing the angle of depression—remember, it’s from the horizontal DOWN.
Tip #2: Know Your Formula Booklet
You don’t need to memorize SOH CAH TOA. The formulas for right-angled triangle trig are in Topic 3: Geometry and Trigonometry of your IB Math AI formula booklet. Your job isn’t to memorize, it’s to know where to find the tools and how to use them efficiently.
Tip #3: Check Your Mode!
This is a classic trap. For these types of problems, you will almost always be working in Degrees (DEG). If your calculator is accidentally in Radians (RAD) mode, every single one of your calculations will be wrong. Make it a habit to check your calculator’s mode at the start of every trig question.
Tip #4: Degree of Accuracy
Read the question carefully for rounding instructions. IB exams are strict about this. Giving an answer to the wrong precision can cost you marks.
Mission Accomplished: The Lighthouse Dilemma
Alright, it’s time. Let’s put everything together and solve the problem from the introduction. Grab your calculator and let’s do this.
IB Exam Challenge: The Lighthouse Dilemma
A lighthouse keeper is in the lantern room, 80 metres above sea level. He spots a fishing boat at an angle of depression of 20°. A few minutes later, he observes the same boat, which has sailed directly towards the lighthouse, at a new angle of depression of 50°. How far did the boat travel between the two observations?
Step 1: Visualize and Diagram. This problem involves two right-angled triangles. The lighthouse height (80m) is the ‘opposite’ side for both. The ‘adjacent’ sides are the horizontal distances from the base of the lighthouse to the boat at each position.
Step 2: Calculate the Initial Distance.
Let’s look at the first observation when the boat is farther away. The angle of depression is 20°, which means the angle of elevation from the boat is also 20° (our Z-angle trick!). The opposite side is 80 m.
$$ \tan(20^{\circ}) = \frac{80}{\text{Distance}_1} $$
$$ \text{Distance}_1 = \frac{80}{\tan(20^{\circ})} = 219.799… \text{ m} $$
Keep this full value handy!
Step 3: Calculate the Final Distance.
Now for the second observation. The new angle of depression is 50°, so the angle of elevation from the boat’s new position is 50°.
$$ \tan(50^{\circ}) = \frac{80}{\text{Distance}_2} $$
$$ \text{Distance}_2 = \frac{80}{\tan(50^{\circ})} = 67.127… \text{ m} $$
Step 4: Find the Difference.
The distance the boat traveled is simply the difference between its starting distance and its finishing distance.
$$ \text{Travelled Distance} = \text{Distance}_1 – \text{Distance}_2 $$
$$ \text{Travelled Distance} = 219.799… – 67.127… = 152.671… \text{ m} $$
Rounding to 3 significant figures, the boat traveled 153 m. Mission accomplished!
Wrap-Up and Final Thoughts
And there you have it! We’ve demystified angles of elevation and depression, showing how they’re just two sides of the same coin. The keys to success are always starting with a clear diagram, correctly identifying your horizontal line, and using the ‘Z-angle’ trick to bring depression angles into your triangle. From there, it’s just a matter of picking the right tool from your SOH CAH TOA toolbox and breaking down bigger problems into manageable pieces. These aren’t just textbook exercises; this is how we measure and map the world around us.
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
Question 1: From a point on the ground 50 metres from the base of a tree, the angle of elevation to the top of the tree is 38°. Find the height of the tree.
Show Answer
Solution: We have the Adjacent side (50m) and want the Opposite side (height, h). Use TOA.
$$ \tan(38^{\circ}) = \frac{h}{50} $$
$$ h = 50 \times \tan(38^{\circ}) = 39.06… $$
The height of the tree is 39.1 m (3 s.f.).
Question 2: A hawk flying at an altitude of 200m spots a mouse on the ground. The angle of depression from the hawk to the mouse is 28°. What is the horizontal distance between the hawk and the mouse?
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Solution: The angle of depression (28°) equals the angle of elevation from the mouse. We have the Opposite (200m) and want the Adjacent (distance, d). Use TOA.
$$ \tan(28^{\circ}) = \frac{200}{d} $$
$$ d = \frac{200}{\tan(28^{\circ})} = 376.15… $$
The horizontal distance is 376 m (3 s.f.).
Question 3: A 10-metre long ladder leans against a vertical wall. The base of the ladder is 2.5 metres from the wall. What angle does the ladder make with the ground?
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Solution: We have the Adjacent (2.5m) and the Hypotenuse (10m). Use CAH.
$$ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2.5}{10} = 0.25 $$
$$ \theta = \cos^{-1}(0.25) = 75.52…^{\circ} $$
The angle is 75.5° (3 s.f.).
Question 4: A kite is on a string that is 40m long. The string makes an angle of 55° with the horizontal ground. How high is the kite vertically above the ground?
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Solution: The string is the Hypotenuse (40m). We want the vertical height, which is the Opposite side. Use SOH.
$$ \sin(55^{\circ}) = \frac{\text{height}}{40} $$
$$ \text{height} = 40 \times \sin(55^{\circ}) = 32.766… $$
The kite is 32.8 m high (3 s.f.).
Question 5: From the top of a 60m tall building, the angle of depression to a car parked on the street below is 42°. How far is the car from the base of the building?
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Solution: The angle of elevation from the car is 42°. The Opposite side is 60m. We want the Adjacent side (distance, d). Use TOA.
$$ \tan(42^{\circ}) = \frac{60}{d} $$
$$ d = \frac{60}{\tan(42^{\circ})} = 66.63… $$
The car is 66.6 m from the building (3 s.f.).
Question 6: A flagpole’s shadow is 15 metres long when the sun’s angle of elevation is 65°. What is the height of the flagpole?
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Solution: The shadow is the Adjacent side (15m). The flagpole’s height is the Opposite side. Use TOA.
$$ \tan(65^{\circ}) = \frac{\text{height}}{15} $$
$$ \text{height} = 15 \times \tan(65^{\circ}) = 32.16… $$
The flagpole is 32.2 m tall (3 s.f.).
Question 7: An airplane is flying at an altitude of 1500m. The pilot observes the start of the runway at an angle of depression of 18°. After flying a bit further, the pilot observes the end of the runway at an angle of depression of 40°. What is the length of the runway?
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Solution: This is a two-triangle problem like our lighthouse example.
1. Horizontal distance to the start of the runway ($D_{start}$): $$\tan(18^{\circ}) = \frac{1500}{D_{start}} \implies D_{start} = \frac{1500}{\tan(18^{\circ})} = 4616.5…\text{ m}$$
2. Horizontal distance to the end of the runway ($D_{end}$): $$\tan(40^{\circ}) = \frac{1500}{D_{end}} \implies D_{end} = \frac{1500}{\tan(40^{\circ})} = 1787.6…\text{ m}$$
3. Runway length = $D_{start} – D_{end} = 4616.5… – 1787.6… = 2828.9…\text{ m}$
The runway is 2830 m long (3 s.f.).
Question 8: You are standing 45m from the base of a skyscraper. The angle of elevation to the 30th floor is 35°, and the angle of elevation to the top of the building is 55°. What is the height from the 30th floor to the top of the building?
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Solution: Find the height to the top and the height to the 30th floor, then subtract.
1. Height to top ($H_{top}$): $$\tan(55^{\circ}) = \frac{H_{top}}{45} \implies H_{top} = 45 \times \tan(55^{\circ}) = 64.26…\text{ m}$$
2. Height to 30th floor ($H_{30}$): $$\tan(35^{\circ}) = \frac{H_{30}}{45} \implies H_{30} = 45 \times \tan(35^{\circ}) = 31.50…\text{ m}$$
3. Height difference = $H_{top} – H_{30} = 64.26… – 31.50… = 32.75…\text{ m}$
The height from the 30th floor to the top is 32.8 m (3 s.f.).
Question 9: A support wire is attached to the top of a 70m high mast. It is anchored to the ground 25m from the base of the mast. What is the length of the wire and the angle it makes with the ground?
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Solution: This requires two steps: Pythagoras for length, and trig for the angle.
1. Length of wire (L): Use Pythagoras’ Theorem. $$ L^2 = 70^2 + 25^2 = 4900 + 625 = 5525 $$ $$ L = \sqrt{5525} = 74.33…\text{ m} $$
2. Angle with ground ($\theta$): Use TOA. $$ \tan(\theta) = \frac{70}{25} = 2.8 $$ $$ \theta = \tan^{-1}(2.8) = 70.34…^{\circ} $$
The wire is 74.3 m long and makes an angle of 70.3° with the ground (3 s.f.).
Question 10: IB Exam Style Question. A surveyor stands at point A and measures the angle of elevation to the top of a mountain as 22°. She then walks 500 metres in a straight line on level ground towards the mountain to point B. At point B, the new angle of elevation to the top of the mountain is 36°. (a) Draw a clear diagram representing this information. (b) Find the height of the mountain, correct to the nearest metre.
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Solution: This is a challenging problem that requires setting up simultaneous equations. Let $h$ be the height of the mountain and $x$ be the horizontal distance from point B to the base of the mountain.
(a) Diagram: Your diagram should show two right-angled triangles. The larger triangle has an angle of 22° and a base of $(500+x)$. The smaller triangle has an angle of 36° and a base of $x$. Both triangles share the same height, $h$.
(b) Finding the height:
Set up two equations using TOA:
Eq 1 (from point B): $$ \tan(36^{\circ}) = \frac{h}{x} \implies x = \frac{h}{\tan(36^{\circ})} $$
Eq 2 (from point A): $$ \tan(22^{\circ}) = \frac{h}{500+x} $$
Now, substitute the expression for $x$ from Eq 1 into Eq 2:
$$ \tan(22^{\circ}) = \frac{h}{500 + \frac{h}{\tan(36^{\circ})}} $$
This looks tricky, but we just need to solve for $h$:
$$ \tan(22^{\circ}) \left(500 + \frac{h}{\tan(36^{\circ})}\right) = h $$
$$ 500\tan(22^{\circ}) + \frac{h\tan(22^{\circ})}{\tan(36^{\circ})} = h $$
Get all the terms with $h$ on one side:
$$ 500\tan(22^{\circ}) = h – \frac{h\tan(22^{\circ})}{\tan(36^{\circ})} $$
Factor out $h$:
$$ 500\tan(22^{\circ}) = h \left(1 – \frac{\tan(22^{\circ})}{\tan(36^{\circ})}\right) $$
Now isolate $h$:
$$ h = \frac{500\tan(22^{\circ})}{1 – \frac{\tan(22^{\circ})}{\tan(36^{\circ})}} $$
Plug this into your calculator carefully:
$$ h = \frac{202.01…}{1 – \frac{0.4040…}{0.7265…}} = \frac{202.01…}{1 – 0.5561…} = \frac{202.01…}{0.4438…} = 455.15…\text{ m} $$
The height of the mountain to the nearest metre is 455 m.