Applications — heights and distances problems

Heights and distances problems use trigonometric ratios to find unknown lengths or angles when direct measurement is impossible. These are the why trigonometry exists — ancient surveyors needed to measure pyramid heights, sailors needed to navigate by stars, and engineers needed to calculate bridge spans without crossing rivers.

[!intuition] The Core Idea

You're standing on the ground looking up at a building. You can measure:

• Your distance from the building (horizontal)
• The angle your eyes make with the horizontal (angle of elevation)

You cannot climb up to measure the building's height directly.

Trigonometry lets you turn angles into lengths. If you know one side of a right triangle and one angle, the ratios (tan, sin, cos) unlock the other sides.

WHY this works: The angle of elevation determines the shape of the triangle. Once shape is fixed, if you scale one side (the distance you measured), all other sides scale proportionally. Tan, sin, cos are just those scale factors.

[!definition] Key Terms

Angle of elevation: The angle between the horizontal line (from your eye level) and the line of sight upward to an object above you.

Angle of depression: The angle between the horizontal line and the line of sight downward to an object below you (like from the top of a cliff looking down at a boat).

WHY these are equal from opposite viewpoints: If you're on the ground looking up at angle α, someone at the top looking down at you sees angle α too — they're alternate interior angles. The horizontal line at your eye level and the horizontal line at the top are the two parallel lines, and the line of sight is the transversal that cuts across both. The vertical height is just the segment between the two lines, not the transversal.

Line of sight: The straight line from your eye to the object. This becomes the hypotenuse of your right triangle.

Horizontal distance: The ground-level distance from you to the base of the object. Always perpendicular to the vertical height.

[!formula] Deriving the Basic Setup

GIVEN: You stand at point A, looking at the top B of a tower. The tower's base is C. You measure:

• Horizontal distance AC = $d$
• Angle of elevation ∠BAC = $\theta$

FIND: Height of tower BC = $h$

Step-by-step derivation:

Step 1: Identify the right triangle.

• The tower is vertical → ∠ACB = 90°
• Triangle ABC is right-angled at C

Step 2: Choose the correct trig ratio. We have the adjacent side (AC = $d$) and want the opposite side (BC = $h$). $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d}$

WHY tan? Because tan relates the two legs of the triangle. Sin and cos involve the hypotenuse (line of sight), which we don't know and don't need.

Step 3: Solve for the unknown. $h = d \tan \theta$

Physical meaning: If the angle doubles, does height double? NO — tan is nonlinear. $\tan 60° = \sqrt{3} \approx 1.73$, but $\tan 30° = 1/\sqrt{3} \approx 0.58$. As $\theta \to 90°$, $\tan\theta$ grows without bound (it has a vertical asymptote), so the calculated height shoots off to infinity. Note this growth is not exponential ($e^{k\theta}$) — it is the specific unbounded growth of the tangent function toward its asymptote at 90°.

[!example] Example 1: Basic Tower

Problem: A student stands 50 m from a tower. The angle of elevation to the top is 30°. Find the tower's height.

Solution:

Given: d = 50 m, θ = 30°
Find: h

$h = d \tan \theta = 50 \times \tan 30°$

WHY this step? Tan connects the known adjacent (50 m) to the unknown opposite (h).

$h = 50 \times \frac{1}{\sqrt{3}} = \frac{50}{\sqrt{3}} = \frac{50\sqrt{3}}{3}$

WHY rationalize? Standard form. Numerically: $h \approx 28.87$ m.

Verification instinct: Does this make sense? The angle is small (30°), so the tower should be shorter than the base distance (50 m). ✓ Check: 28.87 < 50.

[!example] Example 2: Two Observers (Distance Between)

Problem: From the top of a 75 m cliff, the angle of depression to a boat is 30°. How far is the boat from the base of the cliff?

Solution:

Given: h = 75 m (cliff height), θ = 30° (depression angle)
Find: d (horizontal distance)

CRITICAL INSIGHT: Draw it! The angle of depression from the top equals the angle of elevation from the boat (alternate angles).

From the cliff-top looking down: $\tan 30° = \frac{h}{d} = \frac{75}{d}$

WHY this setup? The cliff height (75 m) is opposite to the angle of depression, the boat distance is adjacent.

$d = \frac{75}{\tan 30°} = 75 \times \sqrt{3} = 75\sqrt{3}$

Numerically: $d \approx 129.9$ m.

Verification: Angle is 30° (small) → distance should be longer than height. ✓ Check: 129.9 > 75.

[!example] Example 3: Change in Angle (Moving Observer)

Problem: A man observes the top of a tower at an angle of 30°. He walks 20 m closer; now the angle is 60°. Find the tower's height.

Solution:

Let tower height = h
Initial distance from tower = x
After walking closer: distance = x - 20

From the first position: $\tan 30° = \frac{h}{x} \implies \frac{1}{\sqrt{3}} = \frac{h}{x} \implies x = h\sqrt{3} \quad \text{..(1)}$

WHY this step? We express the unknown distance $x$ in terms of $h$, which we want to find.

From the second position: $\tan 60° = \frac{h}{x - 20} \implies \sqrt{3} = \frac{h}{x - 20} \implies x - 20 = \frac{h}{\sqrt{3}} \quad \text{..(2)}$

Subtract equation (2) from equation (1): $x - (x - 20) = h\sqrt{3} - \frac{h}{\sqrt{3}}$

WHY subtract? Eliminates $x$, leaving only $h$.

$20 = h\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = h\left(\frac{3 - 1}{\sqrt{3}}\right) = h \cdot \frac{2}{\sqrt{3}}$

$h = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32 \text{ m}$

Physical intuition check: He walked 20 m closer and the angle increased (30° → 60°), which makes sense — moving toward a tower makes it look taller (steeper line of sight). The height (17.32 m) is comparable to how far he walked (20 m), which is reasonable.

[!example] Example 4: Two Angles from a Fixed Point

Problem: From a point on the ground, the angle of elevation to the bottom of a flagpole (mounted on a building) is 30°, and to the top of the flagpole is 45°. If the building is 20 m tall, find the height of the flagpole.

Solution:

Building height = 20 m
Let flagpole height = f
Let horizontal distance from observer to building = d

WHY this setup is consistent (unlike a badly-posed problem): Here both angles are measured from the same fixed point to two different heights on the same vertical line. This always has a clean solution.

Angle to bottom of flagpole (top of building), 30°: $\tan 30° = \frac{20}{d} \implies \frac{1}{\sqrt{3}} = \frac{20}{d} \implies d = 20\sqrt{3} \text{ m} \quad \text{..(1)}$

WHY? The building top (20 m) is opposite the 30° angle; $d$ is adjacent.

Angle to top of flagpole, 45°: $\tan 45° = \frac{20 + f}{d} \implies 1 = \frac{20 + f}{d} \implies 20 + f = d \quad \text{..(2)}$

Substitute (1) into (2): $20 + f = 20\sqrt{3}$ $f = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \approx 20 \times 0.732 = 14.64 \text{ m}$

Verification: The angle increased from 30° to 45° as we look higher, so the total height (34.64 m) must exceed the building (20 m). ✓ And $34.64 = 20\sqrt3 = d$, matching the 45° condition (at 45°, height = distance). ✓

Steel-man note: A tempting but ill-posed version says "angle of depression from a bird is 45°, then 30° after flying 100 m." If you assume the bird flies away at constant height, the two conditions force $d_2 - d_1 = 100$ but the geometry gives $d_2 - d_1 = 100(\sqrt3 - 1) \approx 73 \neq 100$ — no consistent solution. Always check that the given displacement matches what the angles imply before trusting an answer.

[!mistake] Common Mistakes

Mistake 1: Confusing adjacent and opposite

Wrong thinking: "The tower is next to me, so it's adjacent. I'll use $\sin \theta = h/d$."

WHY it feels right: The tower is physically beside you.

The fix: Adjacent/opposite are relative to the angle, not your physical position. If the angle is at the base where you stand, the tower (vertical) is opposite the angle. The ground distance is adjacent to the angle.

Memory aid: Draw the triangle. Label the angle. The side across from the angle is opposite.

Mistake 2: Using degrees instead of formula values

Wrong: $\tan 30° = 30$ (treating the angle as a number).

WHY it feels right: In simple arithmetic, we use numbers directly.

The fix: Trig ratios are ratios, not the angles themselves. $\tan 30° = 1/\sqrt{3} \approx 0.577$, not 30.

Use standard values:

• $\tan 30° = 1/\sqrt{3}$
• $\tan 45° = 1$
• $\tan 60° = \sqrt{3}$

Mistake 3: Forgetting the angle of depression = angle of elevation (alternate angles)

Wrong setup: Drawing the angle of depression from the top as the angle inside the triangle at the top.

WHY it feels right: The observer is at the top, so angle should be there.

The fix: The angle of depression is measured from the horizontal at the observer's eye level. When you transfer this to the triangle, use alternate interior angles: the two horizontal lines are parallel, the line of sight is the transversal, so the depression angle at the top equals the elevation angle at the base.

Visual check: Draw two parallel horizontal lines (one at the cliff top, one at ground). The line of sight is the transversal cutting both. Alternate interior angles are equal.

Mistake 4: Not checking units or reasonableness

Wrong: Calculate height = 500 m for angle of 10° from 50 m away, and accept it.

WHY it feels right: You trust your algebra.

The fix: $\tan 10° \approx 0.176$. Height = $50 \times 0.176 \approx 8.8$ m, not 500 m. Always estimate: small angles → height < base distance.

[!recall]- Explain to a 12-year-old

Imagine you want to know how tall a big tree is, but you can't climb it. Here's the trick:

Stand some distance away and look up at the top. Your eyes make an angle with the ground — that's the angle of elevation. It's like tilting your head up.

Now, you've created an invisible triangle: the ground is one side (you can measure this with a tape), the tree is the other side (that's what you want), and your line of sight to the treetop is the slanted side.

Here's the magic: tan of that angle is just a ratio. It tells you: "For every meter you are away, the tree is tan(angle) meters tall."

So if you're 10 meters away and the angle is 45°, tan(45°) = 1, so the tree is 10 × 1 = 10 meters tall. If the angle was 60° (steeper), tan(60°) ≈ 1.7, so the tree would be 10 × 1.7 = 17 meters!

Why does this work? Because angles lock in the shape of triangles. Once you know the shape (from the angle) and the size of one side (your distance), you can figure out all the other sides using these special ratios we call sine, cosine, and tangent.

[!mnemonic] Memory Aid

"STAND-SEE-CLIMB"

• STAND: Measure your distance (adjacent to the angle)
• SEE: Measure the angle of elevation (look up)
• CLIMB: Calculate the height (opposite side) using $h = d \cdot \tan\theta$

For angle of depression: "What I see DOWN from here = What they see UP to me" (alternate angles, line of sight is the transversal).

Connections

• 2.4.01-Trigonometric-ratios — The definitions of sin, cos, tan that power all these calculations
• 2.4.03-Complementary-angles — Why angle of depression equals angle of elevation
• 2.5.02-Surveying-and-navigation — Real-world applications of these techniques
• 3.2.04-Similar-triangles — Why angle determines shape, enabling these measurements
• 1.3.07-Pythagoras-theorem — Used when you need the hypotenuse (line of sight distance)

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