Translating Heights and Distances into Diagrams and Trigonometric Equations

Hello! In the previous lesson, you built the trigonometric ratios $\sin \theta$, $\cos \theta$, and $\tan \theta$ from right triangles and found their exact values at $30^{\circ }$, $45^{\circ }$, and $60^{\circ }$. Now the main challenge is not memorising a new formula: it is turning a written situation into the right triangle before choosing a ratio.

By the end of this lesson, you should be able to identify the line of sight and angle of elevation or depression, draw a labelled model, and write a trigonometric equation that matches the information in the question.

---

1. The three ideas hidden inside the story

Heights-and-distances questions usually describe an observer looking at an object such as a tower, tree, building, kite, or boat. The drawing becomes a right triangle because we normally assume:

• the object is vertical;
• the ground is level and horizontal;
• the vertical object is perpendicular to the ground.

The key vocabulary is:

• Line of sight: the straight line from the observer's eye to the point being viewed.
• Angle of elevation: the angle measured upwards from a horizontal line to the line of sight.
• Angle of depression: the angle measured downwards from a horizontal line to the line of sight.

A useful mental image is to imagine a horizontal beam leaving the observer's eyes. Looking upwards creates an angle of elevation; looking downwards creates an angle of depression. The reference line is always horizontal, not the vertical object.

---

2. A reliable method for drawing the diagram

Do not begin by writing $\sin$, $\cos$, or $\tan$. First translate the story into geometry.

Step 1: Mark the observer and the target

Decide exactly:

• where the observer is;
• which point is being viewed;
• where the foot of the vertical object lies.

For example, if someone is on the ground looking at the top of a tower, mark:

• $P$: observer on the ground;
• $B$: foot of the tower;
• $T$: top of the tower.

The triangle $PBT$ is right-angled at $B$.

Step 2: Draw the physical directions correctly

Draw the ground as a horizontal line. Draw the tower vertically. Then join the observer to the top of the tower: this sloping segment is the line of sight.

Mark the right angle where the vertical object meets the horizontal ground.

Step 3: Place the given angle at its true location

If the question says “the angle of elevation of the top is $60^{\circ }$,” place $60^{\circ }$ at the observer, between:

• the horizontal ground; and
• the upward line of sight.

If it says “the angle of depression is $30^{\circ }$,” place $30^{\circ }$ at the observer below a horizontal line through the observer.

This distinction prevents one of the most common errors: putting an angle of depression directly inside the triangle at the top of a tower. It is initially outside the triangle.

Step 4: Label every known length and the unknown

Write units where supplied, such as $15\text{ m}$. Use a clear variable such as $h$, $x$, or $d$ for the unknown.

A diagram should tell you what every side means:

Part of drawing	Usual meaning
Vertical side	height, or difference in height
Horizontal side	distance along level ground
Sloping side	line of sight, ladder, rope, or string

---

3. Choose the ratio from the labelled sides

Once the angle is in the right triangle, identify the sides relative to that angle:

• opposite is directly across from the angle;
• adjacent touches the angle but is not the hypotenuse;
• hypotenuse is opposite the right angle.

Then use the ratio containing the side you know and the side you need.

Sides involved	Ratio to use
Vertical and horizontal	$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$
Vertical and line of sight	$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}$
Horizontal and line of sight	$\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$

In standard school questions, tangent appears very often because the question gives a height and a horizontal distance.

For a tower of height $h$, viewed from a point $15\text{ m}$ away at an elevation angle of $60^{\circ }$, the diagram gives:

• opposite side: $h$;
• adjacent side: $15$;
• angle: $60^{\circ }$.

So the correct equation is:

$$\tan 60^{\circ }=\frac{h}{15}$$

The important achievement is the equation. If needed, you can then use $\tan 60^{\circ }=\sqrt{3}$ to solve it.

Notice why this is not correct:

$$\tan 60^{\circ }=\frac{15}{h}$$

That fraction reverses opposite and adjacent.

---

4. Elevation problems with an observer's height

A question may say that an observer is $1.6\text{ m}$ tall and looks at the top of a building from eye level. In that case, the vertical side of the triangle is not the full building height.

Suppose:

• the building height is $H$;
• the observer's eye height is $1.6\text{ m}$;
• the horizontal distance to the building is $d$;
• the angle of elevation is $30^{\circ }$.

Draw a horizontal line from the observer's eye to the building. This reaches the building at the same height as the observer's eyes. The vertical side of the triangle is therefore:

$$H-1.6$$

The trigonometric equation is:

$$\tan 30^{\circ }=\frac{H-1.6}{d}$$

If the question asks for the total building height, the model eventually gives:

$$H=1.6+d\tan 30^{\circ }$$

The extra $1.6\text{ m}$ is added only after finding the vertical rise from the observer's eyes to the top.

A general eye-level template

For an observer with eye height $e$, looking at an object of height $H$ from horizontal distance $d$ at an elevation angle $\theta$:

$$\tan \theta =\frac{H-e}{d}$$

Use this only when the stated angle is measured from the observer's eyes, which is the usual meaning of an angle of elevation.

---

5. Translating an angle of depression

An angle of depression feels different because it is drawn outside the right triangle at first. The solution is to add a horizontal line through the observer.

Imagine a person at the top of a lighthouse of height $100\text{ m}$. They see a boat on level water at an angle of depression of $10^{\circ }$. Let $x$ be the horizontal distance from the boat to the foot of the lighthouse.

Your diagram should contain:

1. a vertical lighthouse of height $100\text{ m}$;
2. a horizontal waterline from the foot of the lighthouse to the boat;
3. a line of sight from the top to the boat;
4. a short horizontal line through the observer at the top;
5. the $10^{\circ }$ depression angle between that upper horizontal and the line of sight.

The upper horizontal and the waterline are parallel. The line of sight crosses both, so the angle inside the right triangle at the boat is also $10^{\circ }$. These are alternate interior angles.

Now, relative to the $10^{\circ }$ angle inside the triangle:

• the vertical height $100$ is opposite;
• the horizontal distance $x$ is adjacent.

Therefore:

$$\tan 10^{\circ }=\frac{100}{x}$$

Do not write the angle of depression at the bottom simply because it is more convenient. Instead, draw it where the observer sees it, then use the equal interior angle in the triangle.

More generally, if the vertical difference in height is $\Delta h$, the horizontal distance is $d$, and the angle of depression is $\theta$, then:

$$\tan \theta =\frac{\Delta h}{d}$$

The same tangent relationship appears in elevation and depression questions. What changes is where the angle is first described in the story.

---

6. A quick diagram audit before writing an equation

Before committing to an equation, check these five points:

1. Is the object vertical and the ground horizontal?
   If so, mark the right angle.

2. Did you join the observer to the exact point viewed?
   That segment is the line of sight.

3. Did you measure elevation or depression from a horizontal line?
   Never measure it from the vertical tower.

4. Does the vertical side represent the required height or only a height difference?
   Include the observer's eye height when the question provides one.

5. Does your chosen ratio contain the known and unknown sides?
   Do not choose a ratio merely because it is familiar.

A clean answer to a “form an equation” question often needs only a labelled diagram and a statement such as:

$$\tan 45^{\circ }=\frac{h-1.5}{28.5}$$

That equation communicates that you have understood the physical situation, the geometry, and the trigonometric ratio.

---

Key takeaways

Heights-and-distances questions are right-triangle questions in words. Translate them carefully before calculating.

• An angle of elevation is above the horizontal.
• An angle of depression is below the horizontal.
• The line of sight joins the observer to the viewed point.
• Use $\tan \theta$ when the vertical and horizontal sides are involved.
• If the observer has height $e$, the triangle may use $H-e$, not the total height $H$.
• For depression, draw a horizontal through the observer and transfer the angle into the triangle using parallel-line reasoning.

The next lesson shifts from trigonometry to circle geometry, beginning with the relationship between a radius and a tangent.