Hey everyone, and welcome! Ever looked at a massive skyscraper or a super tall tree and wondered, “Just how tall is that?” You can’t exactly whip out a measuring tape and climb to the top. This isn’t just a random thought experiment; it’s a real-world problem that architects, surveyors, and urban planners face all the time. Imagine you’re in charge of a city park, and a developer wants to build a new tower next door. You have to make sure its shadow won’t block the sun from the park. To figure that out, you first need the height of the existing building next to it. You can measure your distance from the building on the ground and the angle from your eyes to its roof, but that’s it. How do you find the height?
This is the puzzle we’re going to solve today. It’s a classic case of indirect measurement, and the keys to unlocking it lie in the magic of right-angled triangles. We’re going to dust off the Pythagorean Theorem, become best friends with the legendary SOH CAH TOA, and work through some IB-style challenges. By the end of this, you’ll be able to solve our urban planner’s dilemma and feel like a math wizard. Let’s get started!
Table of Contents
- Pythagoras Revisited: The OG of Triangles
- Pythagoras in Action
- The Converse: Is It a Right-Angled Triangle?
- Enter SOH CAH TOA: Our Angle-Finding Superpower
- SOH CAH TOA in Action
- The ‘One Side, One Angle’ Golden Rule
- IB-Style Problems: Putting It All Together
- Solving the Urban Planner’s Challenge
- Conclusion: You’ve Got This!
- Additional Practice Problems
Pythagoras Revisited: The OG of Triangles
Before we can tackle angles, let’s start with a classic tool you’ve known for years: the Pythagorean Theorem. It’s the bedrock of right-angled triangle math, but it’s so important it’s worth a quick, solid review.
Rule: The Pythagorean Theorem
For any right-angled triangle, the square of the two shorter sides (the legs) add up to the square of the longest side (the hypotenuse).
The formula is:
$$ a^2 + b^2 = c^2 $$
Where ‘a’ and ‘b’ are the legs, and ‘c’ is always the hypotenuse (the side opposite the right angle).
Pythagoras in Action
Example 1: Finding the Hypotenuse
Example: Finding the Hypotenuse
You’re building a ramp. The base of the ramp is 12 feet long, and it needs to reach a height of 5 feet. How long does the ramp’s surface (the hypotenuse) need to be?
Let $a = 5$ ft and $b = 12$ ft. We need to find $c$.
1. Formula: $a^2 + b^2 = c^2$
2. Substitute: $5^2 + 12^2 = c^2$
3. Calculate: $25 + 144 = c^2$
4. Simplify: $169 = c^2$
5. Solve for c: $c = \sqrt{169} = 13$
The ramp surface needs to be 13 feet long. This is a classic 5-12-13 Pythagorean triple!
Example 2: Finding a Missing Leg
Example: Finding a Missing Leg
A 25-foot ladder is leaning against a wall. The top of the ladder touches the wall 24 feet above the ground. How far is the base of the ladder from the wall?
Here, the ladder is the hypotenuse ($c = 25$ ft) and the wall height is one leg ($a = 24$ ft). We need to find the other leg, $b$.
1. Formula: $a^2 + b^2 = c^2$
2. Substitute: $24^2 + b^2 = 25^2$
3. Calculate: $576 + b^2 = 625$
4. Rearrange: $b^2 = 625 – 576$
5. Simplify: $b^2 = 49$
6. Solve for b: $b = \sqrt{49} = 7$
The base of the ladder is 7 feet from the wall. (Another famous triple: 7-24-25).
The Converse: Is It a Right-Angled Triangle?
This is a neat trick. The Converse of the Pythagorean Theorem lets us work backward. If you know three side lengths, you can test if they form a right-angled triangle.
Example: Checking the Corners
You’re marking out a rectangular foundation for a shed. You measure the sides as 8 meters and 11 meters. To make sure your corner is a perfect 90 degrees, you measure the diagonal and find it is 13 meters. Is your corner square?
We need to check if $a^2 + b^2 = c^2$. Crucially, $c$ must be the longest side. So, $a=8, b=11, c=13$.
Does $8^2 + 11^2 = 13^2$?
Let’s check the left side (LHS): $8^2 + 11^2 = 64 + 121 = 185$.
Now the right side (RHS): $13^2 = 169$.
Since $185 \neq 169$, this is not a right-angled triangle. Your corner isn’t square!
Enter SOH CAH TOA: Our Angle-Finding Superpower
Pythagoras is awesome for sides, but it can’t help us with angles. For that, we need trigonometry. Specifically, three little ratios that form the foundation of trig: Sine, Cosine, and Tangent. The best way to remember them is with the magical mnemonic: SOH CAH TOA.
First, Let’s Label Things Correctly
Before we can use SOH CAH TOA, we have to label the sides of our triangle from the perspective of a specific angle (we’ll call it theta, or $\theta$).
- Hypotenuse (H): Always the longest side, always opposite the right angle. This one never changes.
- Opposite (O): The side directly across from our angle $\theta$.
- Adjacent (A): The side next to our angle $\theta$ that is NOT the hypotenuse.
Rule: The Trigonometric Ratios
SOH: Sine of an angle is the ratio of the Opposite side to the Hypotenuse.
$$ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{O}{H} $$$$
CAH: Cosine of an angle is the ratio of the Adjacent side to the Hypotenuse.
$$ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{A}{H} $$$$
TOA: Tangent of an angle is the ratio of the Opposite side to the Adjacent side.
$$ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{O}{A} $$
SOH CAH TOA in Action
Example 3: Finding a Missing Angle
Example: Finding an Angle
A pilot is flying and sees an airport runway. The plane is at an altitude of 2000 meters (Opposite side) and is a horizontal distance of 5000 meters from the start of the runway (Adjacent side). What is the angle of depression from the plane to the runway?
We have the Opposite (2000 m) and the Adjacent (5000 m) sides relative to the angle of depression, $\theta$.
1. Choose the ratio: O and A point us to TOA.
2. Formula: $\tan(\theta) = \frac{O}{A}$
3. Substitute: $\tan(\theta) = \frac{2000}{5000} = 0.4$
4. Solve for $\theta$: To get the angle, we need to use the inverse tangent function on our calculator (often labeled as $\arctan$ or $\tan^{-1}$).
$$ \theta = \arctan(0.4) $$
5. Calculate: $\theta \approx 21.8^\circ$
The angle of depression is about 21.8 degrees.
Example 4: Finding a Missing Side (Given Hypotenuse)
Example: Finding a Side
A kite is flying on a 50-meter string. The string makes an angle of $40^\circ$ with the ground. How high is the kite above the ground (assuming the string is taut)?
We know the Hypotenuse (50 m) and an angle ($40^\circ$). We want to find the Opposite side (the kite’s height, let’s call it $h$).
1. Choose the ratio: O and H point us to SOH.
2. Formula: $\sin(\theta) = \frac{O}{H}$
3. Substitute: $\sin(40^\circ) = \frac{h}{50}$
4. Rearrange: $h = 50 \times \sin(40^\circ)$
5. Calculate: $h \approx 32.14$ meters.
The kite is approximately 32.14 meters high.
Example 5: Finding a Missing Side (Given a Leg)
Example: Finding Another Side
You are standing 30 meters from the base of a flagpole. You look up to the top of the pole at an angle of elevation of $25^\circ$. How tall is the flagpole?
We know the Adjacent side (30 m) and an angle ($25^\circ$). We want to find the Opposite side (the flagpole’s height, let’s call it $y$).
1. Choose the ratio: O and A point us to TOA.
2. Formula: $\tan(\theta) = \frac{O}{A}$
3. Substitute: $\tan(25^\circ) = \frac{y}{30}$
4. Rearrange: $y = 30 \times \tan(25^\circ)$
5. Calculate: $y \approx 13.99$ meters.
The flagpole is approximately 13.99 meters tall. (This assumes you are measuring from ground level! More on that later…)
The ‘One Side, One Angle’ Golden Rule
Pro Tip: The Golden Rule
This is a game-changer for IB exams. In any right-angled triangle, if you know:
- at least one side length, and
- at least one non-right angle
…you can find out everything else about that triangle. You can find the third angle (since they add to $180^\circ$), and you can use SOH CAH TOA to find the other two sides. You have the power to completely solve the triangle.
IB-Style Problems: Putting It All Together
Let’s look at how the IB might frame these questions. They love adding a little context and sometimes a tricky extra step.
IB Question 1: The Surveyor’s Plot
IB Exam Question
A surveyor is mapping a triangular plot of land. The survey map shows the side lengths as 20 m, 21 m, and 29 m. The buyer is told it is a perfect right-angled plot.
(a) Show that the plot of land is a right-angled triangle.
To show this, we use the Converse of the Pythagorean Theorem. The longest side must be the hypotenuse, so $c = 29$.
We test if $a^2 + b^2 = c^2$.
LHS: $20^2 + 21^2 = 400 + 441 = 841$.
RHS: $29^2 = 841$.
Since LHS = RHS, the theorem holds true, and the plot is indeed a right-angled triangle.
IB Question 2: The Botanist’s Dilemma
IB Exam Question
A botanist, Maria, stands 25 meters from the base of a vertical cliff. She spots a rare plant, P, growing on the cliff face. Using a clinometer, she measures the angle of elevation from her eye level to the plant as $32^\circ$. The triangle formed is a right-angled triangle.
(a) Calculate the vertical height of the plant above her eye level.
Let this height be $h$. We know the Adjacent side (25 m) and the angle ($32^\circ$), and we want the Opposite side ($h$). We use TOA.
$\tan(32^\circ) = \frac{h}{25}$
$h = 25 \times \tan(32^\circ) \approx 15.62$ m.
So, the plant is about 15.62 meters above her eye level.
(b) State what other information Maria needs to find the total height of the plant from the ground.
This is the classic IB trick! The height we found is only from her eye level up. To find the total height from the ground, Maria needs to know her own eye height (the height from the ground to her eyes/clinometer).
Solving the Urban Planner’s Challenge
Alright, it’s time. Let’s go back to our intro problem and solve it for Alex, the urban planner.
Challenge Solved: The Skyscraper’s Height
The Data:
- Alex stands 40 meters from the base of the building.
- The angle of elevation from Alex’s eyes to the top is 70 degrees.
- Alex’s eye height is 1.5 meters above the ground.
The Goal: Find the total height of the building.
Step 1: Find the height of the building above eye level.
Let’s call this partial height $h_1$. This is the Opposite side to the $70^\circ$ angle. The distance from the building, 40 m, is the Adjacent side.
We use TOA: $\tan(\theta) = \frac{O}{A}$
$\tan(70^\circ) = \frac{h_1}{40}$
$h_1 = 40 \times \tan(70^\circ) \approx 109.89$ m.
Step 2: Add Alex’s eye height.
The total height of the building is this partial height plus the height from the ground to Alex’s eyes.
Total Height = $h_1$ + Eye Height
Total Height = $109.89 + 1.5 = 111.39$ m.
Result: The building is approximately 111.39 meters tall. Alex now has the data needed to calculate the shadow’s length and protect the park. Problem solved!
Conclusion: You’ve Got This!
And there you have it! We started with a real-world puzzle and used our mathematical toolkit to find the solution. We’ve seen that the Pythagorean Theorem is our go-to for problems involving only side lengths, while SOH CAH TOA is our champion for any problem that involves angles. The most powerful idea is that indirect measurement, powered by these simple rules, lets us measure the unmeasurable.
This skill isn’t just for passing your IB Math exam; it’s a way of thinking and problem-solving that applies everywhere. Don’t just memorize the formulas—understand *when* and *why* you use them. 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: A right-angled triangle has legs of length 10 cm and 24 cm. What is the length of the hypotenuse?
Show Answer
Solution: Use Pythagoras: $a^2 + b^2 = c^2$.
$10^2 + 24^2 = c^2$
$100 + 576 = c^2$
$676 = c^2$
$c = \sqrt{676} = 26$ cm.
Question 2: A right-angled triangle has a hypotenuse of 61 m and one leg of 60 m. What is the length of the other leg?
Show Answer
Solution: Use Pythagoras: $a^2 + b^2 = c^2$.
$a^2 + 60^2 = 61^2$
$a^2 + 3600 = 3721$
$a^2 = 3721 – 3600 = 121$
$a = \sqrt{121} = 11$ m.
Question 3: A triangle has side lengths of 9 inches, 12 inches, and 16 inches. Is it a right-angled triangle? Show your work.
Show Answer
Solution: Use the Converse of Pythagoras. The longest side is the potential hypotenuse, $c = 16$.
Check: Does $9^2 + 12^2 = 16^2$?
LHS: $81 + 144 = 225$.
RHS: $16^2 = 256$.
Since $225 \neq 256$, it is not a right-angled triangle.
Question 4: In a right-angled triangle, the side opposite angle $\theta$ is 5 cm and the hypotenuse is 13 cm. What is the value of $\sin(\theta)$?
Show Answer
Solution: $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13}$.
Question 5: In a right-angled triangle, the side adjacent to a $50^\circ$ angle is 10 m long. Find the length of the hypotenuse.
Show Answer
Solution: We have Adjacent and want Hypotenuse, so we use CAH.
$\cos(50^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{10}{H}$
$H = \frac{10}{\cos(50^\circ)} \approx 15.56$ m.
Question 6: In a right-angled triangle, the side opposite an angle is 15 units and the side adjacent to it is 8 units. Find the measure of the angle.
Show Answer
Solution: We have Opposite and Adjacent, so we use TOA.
$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{15}{8}$
$\theta = \arctan(\frac{15}{8}) \approx 61.93^\circ$.
Question 7: A wheelchair ramp is 15 m long and rises to a platform that is 1.2 m high. What is the angle of inclination of the ramp?
Show Answer
Solution: The ramp length is the Hypotenuse (15 m) and the height is the Opposite side (1.2 m). We use SOH.
$\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1.2}{15}$
$\theta = \arcsin(\frac{1.2}{15}) \approx 4.59^\circ$.
Question 8: From the top of a 100 m tall lighthouse, the angle of depression to a boat at sea is $12^\circ$. How far is the boat from the base of the lighthouse?
Show Answer
Solution: The angle of depression from the top is equal to the angle of elevation from the boat. The lighthouse height is the Opposite side (100 m) and the distance to the boat is the Adjacent side (let’s call it $d$). We use TOA.
$\tan(12^\circ) = \frac{100}{d}$
$d = \frac{100}{\tan(12^\circ)} \approx 470.46$ m.
Question 9: You know a right-angled triangle has a $30^\circ$ angle and the side opposite it is 5 cm long. Find the lengths of the other two sides.
Show Answer
Solution: We have O=5 and $\theta=30^\circ$.
Find Hypotenuse (H) using SOH:
$\sin(30^\circ) = \frac{5}{H} \Rightarrow H = \frac{5}{\sin(30^\circ)} = \frac{5}{0.5} = 10$ cm.
Find Adjacent (A) using TOA:
$\tan(30^\circ) = \frac{5}{A} \Rightarrow A = \frac{5}{\tan(30^\circ)} \approx 8.66$ cm.
Question 10 (IB Exam Style Question): A communications tower is supported by two guy wires on opposite sides. Both wires are attached to the ground at the same distance from the base of the tower. Wire 1 is 80 m long and makes an angle of $60^\circ$ with the ground. Wire 2 makes an angle of $45^\circ$ with the ground.
a) Find the height, $h$, at which the wires are attached to the tower.
b) Find the length of Wire 2.
c) Find the distance from the base of the tower to where the wires are attached to the ground.
Show Answer
Solution:
a) Find the height (h): We use the information from Wire 1. We have the Hypotenuse (80 m) and the angle ($60^\circ$), and we want the Opposite side (h). Use SOH.
$\sin(60^\circ) = \frac{h}{80}$
$h = 80 \times \sin(60^\circ) \approx 69.28$ m.
b) Find the length of Wire 2 (L2): Now we look at the triangle for Wire 2. We know the height (Opposite side) is $h \approx 69.28$ m and the angle is $45^\circ$. We want the Hypotenuse (L2). Use SOH again.
$\sin(45^\circ) = \frac{69.28}{L2}$
$L2 = \frac{69.28}{\sin(45^\circ)} \approx 98.00$ m.
c) Find the distance from the base (d): We can use either triangle. Let’s use Wire 1’s triangle. We want the Adjacent side (d). We have H=80 and the angle is $60^\circ$. Use CAH.
$\cos(60^\circ) = \frac{d}{80}$
$d = 80 \times \cos(60^\circ) = 80 \times 0.5 = 40$ m.