Some Applications of Trigonometry
Why This Matters
How tall is the Qutub Minar? How wide is a river you can’t cross? How high is a kite, or a cloud, or a mountain peak? You can’t run a measuring tape up a 70-metre tower or across flowing water — but you don’t have to. Stand at a known distance, point at the top, and measure the angle your line of sight makes with the ground. That single angle, together with one distance you can measure, is enough to pin down the height exactly.
This is trigonometry stepping out of the textbook and into the world. In the last chapter you learned the ratios sin, cos and tan for an angle in a right triangle. Here you put them to work: every “how tall / how far / how high” problem becomes a right triangle, where one side and one angle are known and the side you want is just a ratio away.
The beautiful part is that you never need a special angle-measuring gadget on paper — the questions are built around the standard angles 30°, 45° and 60°, whose ratios you already know by heart. Master the small habit of drawing the right triangle first, and these problems become almost mechanical.
The Big Idea
To find a height or distance you can’t reach, turn the situation into a right triangle. The line of sight is the straight line from the observer’s eye to the object. The angle this line makes with the horizontal is the angle of elevation when you look up (object above eye level) and the angle of depression when you look down (object below eye level). Once one side and one of the standard angles 30°/45°/60° are known, a single trig ratio — usually tan — gives the unknown side.
Let’s Break It Down
The line of sight and the angle of elevation
Imagine standing on the ground, looking up at the top of a tower. The straight line from your eye to the top of the tower is the line of sight. The flat, level direction you’d look if you stared straight ahead is the horizontal.
When the object is above your eye level, you raise your head — and the angle between your line of sight and the horizontal is the angle of elevation.
To turn this into a right triangle: the vertical is the height of the object (above eye level), the horizontal is the distance to its foot, and the line of sight is the hypotenuse. The right angle is at the foot. The angle of elevation sits at the observer’s eye, with the height opposite it and the distance adjacent — so tan θ = height / distance.
The angle of depression
Now climb to the top of a building and look down at something on the ground — a car, a boat, a flower pot. Your line of sight now slopes downward. The horizontal is still the level direction at your eye. The angle between the horizontal and your downward line of sight is the angle of depression.
Here is the single most useful fact about depression problems. The horizontal at the observer’s eye and the horizontal ground are parallel. The line of sight is a transversal cutting them. So the angle of depression (at the top) equals the angle of elevation of the observer as seen from the object on the ground — they are alternate angles.
The depression angle “drops down” to become an equal angle at the bottom of your triangle. This lets you put the known angle inside the right triangle where it’s easy to use.
An angle of depression of 40° is measured from the top of a cliff to a boat. What is the angle of elevation of the cliff-top from the boat?
Choosing the right ratio
In almost every problem you’ll know (or want) two of these three: the height (vertical, opposite the angle), the distance to the foot (horizontal, adjacent), and the line of sight (hypotenuse). Pick the ratio that links what you know to what you want:
| You know / want | Use | Formula (angle θ at the observer) |
|---|---|---|
| height & distance | tan θ | tan θ = height / distance |
| height & line of sight | sin θ | sin θ = height / hypotenuse |
| distance & line of sight | cos θ | cos θ = distance / hypotenuse |
And the standard-angle values you’ll lean on constantly:
| θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
Worked problems
Start with the cleanest possible case — a tower, a known distance, find the height.
A tower stands vertically on the ground. From a point on the ground 15 m away from its foot, the angle of elevation of the top is 60°. Find the height of the tower.
- Draw the right triangle ABC: AB is the tower (vertical), B is the foot (right angle), and C is the point on the ground with CB = 15 m. The angle of elevation at C is 60°.
- We know the distance (adjacent = 15 m) and want the height (opposite = AB). The ratio linking opposite and adjacent is the tangent: tan 60° = AB / CB.
- So AB = CB × tan 60° = 15 × √3.
- Therefore the height of the tower is 15√3 m ≈ 25.98 m.
Next, a case where the line of sight itself (the hypotenuse) is the unknown — so we reach for sin.
An electrician must reach a point 1.3 m below the top of a 5 m pole. Her ladder leans at 60° to the horizontal. How long must the ladder be, and how far from the pole's foot should she place it? (Take √3 = 1.73.)
- The point she must reach is at height BD = 5 − 1.3 = 3.7 m up the pole. The ladder BC is the hypotenuse of a right triangle, with the 60° angle at the ground point C and the right angle at the pole’s foot D.
- Height (opposite) and ladder (hypotenuse) are linked by sine: sin 60° = BD / BC, so BC = BD / sin 60° = 3.7 / (√3/2) = (3.7 × 2)/√3.
- BC = 7.4 / 1.73 ≈ 4.28 m. So the ladder must be about 4.28 m long.
- For the distance DC of the foot from the pole, use the tangent: tan 60° = BD / DC, so DC = BD / tan 60° = 3.7 / √3 = 3.7 / 1.73 ≈ 2.14 m from the pole’s foot.
When the observer has a height of their own (eyes are not on the ground), the triangle gives you the height above eye level — you add the observer’s height at the end.
An observer 1.5 m tall stands 28.5 m from a chimney. The angle of elevation of the chimney's top from her eyes is 45°. Find the height of the chimney.
- Her eyes are 1.5 m up. Let the chimney’s top be AE above her eye level, where the right triangle has the horizontal eye-level distance DE = 28.5 m and the 45° angle at her eye.
- tan 45° = AE / DE → AE = DE × tan 45° = 28.5 × 1 = 28.5 m. This is the part of the chimney above her eyes.
- The chimney’s full height = (part above eyes) + (her eye height) = AE + 1.5.
- Height = 28.5 + 1.5 = 30 m.
Many exam problems give two angles for the same object — handle them as two right triangles sharing a side, and solve them together.
From a point P on the ground the angle of elevation of the top of a 10 m building is 30°. A flag is hoisted on top, and the elevation of the flag's top from P is 45°. Find the length of the flagstaff and the distance of P from the building. (Take √3 = 1.732.)
- Let the foot of the building be A, its top B, the flag’s top D, and PA = the horizontal distance. First triangle PAB: tan 30° = AB / PA = 10 / PA.
- So 1/√3 = 10 / PA → PA = 10√3 = 10 × 1.732 = 17.32 m (the distance of P from the building).
- Let the flagstaff DB = x, so the whole height AD = 10 + x. Second triangle PAD: tan 45° = AD / PA → 1 = (10 + x) / (10√3).
- So 10 + x = 10√3 = 17.32 → x = 17.32 − 10 = 7.32 m, the length of the flagstaff.
A classic “difference of two shadows” problem — the unknowns appear in both triangles, so set up two equations and combine.
The shadow of a tower on level ground is 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower.
- Let the height be h and the shorter shadow (at 60°) be x. The longer shadow (at 30°) is then x + 40.
- At 60°: tan 60° = h / x → √3 = h / x → h = x√3. … (1)
- At 30°: tan 30° = h / (x + 40) → 1/√3 = h / (x + 40) → h = (x + 40)/√3. … (2)
- Set (1) = (2): x√3 = (x + 40)/√3. Multiply both sides by √3: 3x = x + 40 → 2x = 40 → x = 20. Then h = x√3 = 20√3, so the tower is 20√3 m ≈ 34.64 m tall.
Now a depression problem with two angles — watch how each depression angle drops into the triangle as an equal angle.
From the top of a multi-storeyed building, the angles of depression of the top and bottom of an 8 m tall building are 30° and 45°. Find the height of the multi-storeyed building and the distance between the two buildings.
- Let P be the top of the tall building, C its foot; A the foot and B the top of the 8 m building. The horizontal at P is parallel to the ground, so by alternate angles the depression of 30° to B becomes 30° at B, and 45° to A becomes 45° at A. Let PD be the part of the tall building above B’s level, and the gap between buildings be AC = BD = d.
- In the lower triangle PAC: tan 45° = PC / AC = 1, so PC = AC = d. The full height PC equals the distance d.
- In the upper triangle PBD: tan 30° = PD / BD → 1/√3 = PD / d → PD = d/√3. Since PC = PD + DC and DC = 8 (the short building), d = d/√3 + 8.
- So d − d/√3 = 8 → d(1 − 1/√3) = 8 → d(√3 − 1)/√3 = 8 → d = 8√3/(√3 − 1) = 4(3 + √3) = (12 + 4√3) m ≈ 18.93 m. That is both the height of the multi-storeyed building and the distance between the buildings.
One more — a width of a river measured from a bridge using two depressions.
From a point on a bridge across a river, the angles of depression of the banks on opposite sides are 30° and 45°. The bridge is 3 m above the banks. Find the width of the river.
- Let P be the point on the bridge, D the foot of the perpendicular below P (so PD = 3 m), A and B the banks on either side. By alternate angles, the depressions become the angles at A and B: angle A = 30°, angle B = 45°.
- Triangle PAD: tan 30° = PD / AD → 1/√3 = 3 / AD → AD = 3√3 m.
- Triangle PBD: tan 45° = PD / BD → 1 = 3 / BD → BD = 3 m.
- Width AB = AD + BD = 3√3 + 3 = 3(√3 + 1) m ≈ 8.20 m.
Common Mistakes
The angle of depression is measured up from the ground at the object.
In the picture the object sits on the ground, so it feels natural to mark the angle there, between the ground and the line of sight going up to the observer.
The angle of depression is measured at the OBSERVER'S eye, DOWN from the horizontal. You can move it to the bottom of the triangle — but only because it equals the angle of elevation at the object (alternate angles). Draw the horizontal at the eye and mark the angle below it.
Always use sin θ to find a height.
Height is 'vertical' and sin involves the 'opposite' side, which is often the height — so sine feels like the height ratio.
The ratio depends on which OTHER side you know. If you know the horizontal distance, use tan θ = height/distance. Use sin θ only when the hypotenuse (line of sight) is the side you know or want. Picking the wrong ratio drags an unknown hypotenuse into the problem.
When the observer has a height, the triangle gives the object's full height directly.
The trig ratio spits out a clean number, and it's tempting to call that 'the answer' without thinking about where the eye actually is.
A right triangle drawn from the EYE gives only the height ABOVE eye level. Add the observer's height (e.g. 1.5 m) at the end to get the true height of the object. Forgetting this is the single most common slip in these problems.
tan 30° = √3 and tan 60° = 1/√3.
Both 30° and 60° involve √3, so it's easy to swap them — and √3 'feels bigger', so it seems to belong to the bigger angle the wrong way round.
tan 30° = 1/√3 (small angle → small tangent) and tan 60° = √3 (large angle → large tangent). Sanity check: a steeper line of sight means a bigger angle and a bigger tan, so tan 60° must be the larger value, √3.
Quick Check
You stand on the ground and look up at the top of a tower. The angle your line of sight makes with the horizontal is the angle of…
A point is 30 m from the foot of a tower and the angle of elevation of the top is 30°. The height of the tower is…
From the top of a cliff the angle of depression of a boat is 35°. The angle of elevation of the cliff-top from the boat is…
A 1.5 m tall person finds the part of a pole above her eye level is 8 m (from her trig triangle). The pole's height is…
Practice Problems
Easy
A circus artist climbs a 20 m long rope tied from the top of a vertical pole to the ground. The rope makes 30° with the ground. Find the height of the pole.
The rope is the hypotenuse (20 m), the pole is the opposite side (height), and the angle at the ground is 30°.
sin 30° = height / 20 → height = 20 × sin 30° = 20 × (1/2) = 10 m.
A kite flies at a height of 60 m. The string from the kite is tied to a point on the ground, making 60° with the ground, with no slack. Find the length of the string.
The string is the hypotenuse, the height (60 m) is opposite the 60° angle.
sin 60° = 60 / string → string = 60 / sin 60° = 60 / (√3/2) = 120/√3 = 40√3 ≈ 69.28 m.
Medium
A tree breaks in a storm; the broken top bends so it touches the ground 8 m from the foot, making 30° with the ground. Find the original height of the tree.
Let the standing (unbroken) part be h, and the broken part (now leaning to the ground) be the hypotenuse, length L. The foot-to-touch distance is 8 m (adjacent to the 30° angle).
The standing part: tan 30° = h / 8 → h = 8 × (1/√3) = 8/√3 = 8√3/3 m.
The broken (leaning) part: cos 30° = 8 / L → L = 8 / (√3/2) = 16/√3 = 16√3/3 m.
Original height = h + L = 8√3/3 + 16√3/3 = 24√3/3 = 8√3 m ≈ 13.86 m.
A 1.5 m tall boy stands away from a 30 m building. As he walks towards it, the elevation of the top rises from 30° to 60°. How far did he walk?
The top of the building is 30 − 1.5 = 28.5 m above his eyes. Let his two distances (from the building) be d₁ (at 30°) and d₂ (at 60°).
At 30°: tan 30° = 28.5 / d₁ → d₁ = 28.5 / (1/√3) = 28.5√3.
At 60°: tan 60° = 28.5 / d₂ → d₂ = 28.5 / √3 = 28.5/√3 = 9.5√3.
Distance walked = d₁ − d₂ = 28.5√3 − 9.5√3 = 19√3 ≈ 32.9 m.
Challenge
Two equal poles stand on opposite sides of an 80 m wide road. From a point between them on the road, the elevations of the two tops are 60° and 30°. Find the height of the poles and the distances of the point from each pole.
Let the common height be h, and let the point be x m from the first pole (the 60° one). Then it is (80 − x) m from the second pole (the 30° one).
First pole: tan 60° = h / x → h = x√3. … (1)
Second pole: tan 30° = h / (80 − x) → h = (80 − x)/√3. … (2)
Set (1) = (2): x√3 = (80 − x)/√3. Multiply by √3: 3x = 80 − x → 4x = 80 → x = 20.
So the point is 20 m from the first pole and 60 m from the second.
Height h = x√3 = 20√3 ≈ 34.64 m.
From the top of a 7 m building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.
Let the horizontal distance between the building and the tower be d. From the building’s top, looking at the tower’s foot, the depression is 45°; by alternate angles the bottom part of the tower below the building’s roof level equals 7 m.
Depression to foot (45°): the foot is 7 m below the roof level, so tan 45° = 7 / d → d = 7 m.
Elevation to top (60°): the part of the tower ABOVE the roof level, call it a, satisfies tan 60° = a / d → a = d × √3 = 7√3 m.
Total tower height = (part below roof level) + (part above) = 7 + 7√3 = 7(1 + √3) ≈ 19.12 m.
Summary
You should now be able to explain:
- The line of sight is the straight line from an observer’s eye to the object being viewed.
- The angle of elevation is the angle the line of sight makes with the horizontal when the object is above eye level (you look up); the angle of depression is the angle when the object is below eye level (you look down).
- A height/distance problem becomes a right triangle: vertical = height (opposite), horizontal = distance (adjacent), line of sight = hypotenuse.
- Choose the ratio that links what you know to what you want: tan (height & distance), sin (height & line of sight), cos (distance & line of sight).
- The angle of depression equals the angle of elevation from the object back to the observer (alternate angles), so you can move it into the triangle.
- If the observer has a height, the triangle gives only the height above eye level — add the observer’s height at the end.
- Problems with two angles split into two right triangles sharing a side; set up the equations and combine.
What’s Next
So far the figures have been triangles. Next, in Circles, the star shape is the round one — and the key idea is the tangent: a line that just touches a circle at one point. You’ll discover that a tangent is always perpendicular to the radius at the point of contact, and that the two tangents drawn to a circle from an outside point are exactly equal in length — clean, surprising facts you can prove and then use.