Big picture
Applications of trigonometry uses right triangles to calculate heights and distances that are difficult to measure directly. A person can estimate a building's height from its viewing angle and horizontal distance, while two observations can establish an unknown distance without approaching an object.
The crucial work happens before calculation. Identify the observer's eye level, draw a horizontal reference and distinguish horizontal distance from sloping line of sight. Most errors arise from modelling the wrong triangle, adding the wrong distances or forgetting that a calculated height may begin at the observer's eyes rather than at ground level.
Learning goals
Identify line of sight and angles of elevation and depression.
Translate a verbal situation into labelled right triangles.
Select sine, cosine or tangent from the required sides.
Solve height problems with and without an observer's eye height.
Combine two observations using clear distance equations.
Check units, positivity and the relationship between angle and distance.
Core ideas
1. Line of sight and horizontal reference
The line of sight joins the observer's eye to the observed point. An angle of elevation is measured upward from a horizontal through the eye. An angle of depression is measured downward from that horizontal. Neither is normally measured from the vertical wall.
Picture a student at E looking towards a tower top T. Draw horizontal EH to meet the vertical tower at H. The angle TEH is the elevation angle, and triangle EHT is right-angled at H. Its vertical side HT measures the height above eye level, not automatically the complete tower height.
A tower's top may be observed at elevation 30°, while the angle between the line of sight and a vertical is 60°. These are complementary because horizontal and vertical are perpendicular. Read how the angle is defined rather than transferring the given number to the most convenient corner.
2. Choose the appropriate trigonometric ratio
Relative to the chosen acute angle θ, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, and tan θ = opposite/adjacent. The hypotenuse is opposite the right angle and is the longest side. Here adjacent normally means the horizontal leg, not the sloping sight line.
θ | sin θ | cos θ | tan θ |
|---|---|---|---|
30° | ½ | √3/2 | 1/√3 |
45° | 1/√2 | 1/√2 | 1 |
60° | √3/2 | ½ | √3 |
Choose tangent for vertical height and horizontal distance. Use sine if vertical height and line of sight are involved; cosine links horizontal distance and line of sight. Avoid using all three ratios when one directly connects the known and unknown quantities.
Worked example. A cable of length 20 m rises at 30° to horizontal ground. Vertical rise = 20 sin 30° = 20 × ½ = 10 m. Horizontal reach = 20 cos 30° = 10√3 m. Using 20 as a horizontal distance would describe a different triangle.
3. One observation from ground level
For an ideal observation point at the same ground level as a vertical object's base, let height be h and horizontal distance d. Then tan θ = h/d, so h = d tan θ or d = h/tan θ.
Worked example. The top of a tower is seen at 30° from a point 24√3 m from its foot. Then tan 30° = h/(24√3). Thus h = 24√3 × 1/√3 = 24 m. If required, the line of sight is h/sin 30° = 24 ÷ ½ = 48 m.
Inverse example. A 16 m pole subtends elevation 45°. Then 1 = 16/d, giving d = 16 m. The 45° right triangle has equal legs; this also checks the result without further computation. State that ground is level and the pole vertical where the model assumes those conditions.
4. Observer's eye height
If an observer's eyes are e metres above the same level ground, the right triangle's vertical side is h − e. Hence tan θ = (h − e)/d, giving h = e + d tan θ. Adding eye height is necessary only when the required height is measured from ground and the triangle starts above it.
Worked example. An observer's eyes are 1.6 m high. From 18 m away, a roof is seen at elevation 45°. Height above eye level = 18 × 1 = 18 m. Complete height = 18 + 1.6 = 19.6 m. Do not add 1.6 m to the horizontal distance or line of sight.
If an instrument stands on a raised platform, identify all relevant vertical levels. An eye height measured above the platform and the platform's height above the object's base may both contribute. Conversely, a question asking only how far the roof is above the instrument does not require those additions.
5. Angles of depression
Draw the horizontal through the higher observer and another through the lower observation point. These horizontals are parallel. The line of sight is a transversal, so the angle of depression equals the corresponding angle of elevation by alternate interior angles.
Worked example. From a platform 36 m above level ground, the angle of depression to a cart is 60°. Let horizontal distance be d. Using the lower elevation angle, tan 60° = 36/d. Therefore d = 36/√3 = 12√3 m. The line of sight is 36/sin 60° = 36 ÷ (√3/2) = 24√3 m.
Depression is not the angle between the sight line and the vertical platform. That angle would be 30° here. Either acute angle can be used correctly with suitable ratios, but mixing one angle with the side roles of the other gives an incorrect model.
6. Two observations on the same side
Suppose an observer moves towards a vertical tower on level ground. Let the nearer distance be x and the movement length be a. The farther distance is x + a, not a alone. For equal eye levels, the height above eye level is the same in both triangles.
Worked example. Tower elevation is 30° at a farther point and 60° after moving 20 m towards it. Use ground-level observations and let nearer distance be x. From the near point, h = x√3. From the far point, h = (x + 20)/√3. Equating: 3x = x + 20, so 2x = 20 and x = 10 m. Height = 10√3 m. Far distance = 30 m.
The larger elevation should occur closer to the same top. Check: h/10 = √3 and h/30 = 1/√3. If your answer places the larger angle farther away under the same conditions, revisit the labels before accepting the algebra.
7. Observers on opposite sides
If two ground points are on opposite sides of the tower foot and separated by L, their distances to the foot add to L. This differs from same-side movement, where distances differ by the movement length.
Worked example. Two points on opposite sides are 40 m apart. Elevations to the tower top are 30° and 60°. Let the distance on the 60° side be x, so the other is 40 − x. Height from the first side is x√3, and from the other it is (40 − x)/√3. Equating gives 3x = 40 − x, so 4x = 40 and x = 10 m. Other distance = 30 m and height = 10√3 m.
Both distances must be positive and their sum must be 40 m. A sketch is essential: the same two angles can produce different answers when the observation points are arranged differently.
8. Two objects, a building and a flagstaff
A viewpoint can create two right triangles sharing a horizontal distance but having different vertical differences. Do not assume the sight line reaches only the higher object's full height.
Worked example. From a building roof 12 m high, the top and foot of a pole have angles of elevation 30° and depression 45° respectively. Let the horizontal gap be d. For the foot, tan 45° = 12/d, so d = 12 m. Height of the pole above the roof = d tan 30° = 12/√3 = 4√3 m. Complete pole height = 12 + 4√3 m, approximately 18.93 m.
Flagstaff example. From a point 18 m from a building, the roof is seen at 30° and the top of a flagstaff on it at 60°. Building height = 18/√3 = 6√3 m. Combined height = 18√3 m. Flagstaff height = 18√3 − 6√3 = 12√3 m. Subtract heights, not angles: the tangent of an angle difference is not the difference of the original tangents.
9. Broken objects and practical assumptions
A broken tree can form a right triangle. The remaining vertical stump is one leg; the ground distance to the fallen top is the other; the broken portion is the hypotenuse. Original height is stump plus broken length.
Worked example. A tree breaks 9 m above ground, and its top touches the ground 12 m from the foot. Broken length = √(9² + 12²) = √225 = 15 m. Original height = 9 + 15 = 24 m. Reporting 15 m alone gives only the fallen part.
These models assume straight, vertical structures, level ground and unobstructed lines of sight unless otherwise stated. In practical surveying, uneven ground and measurement uncertainty matter. In an exam, use the stated idealisation without inventing additional slope data. Retain exact surds until a decimal is requested, then round only at the end.
A reliable approach to questions
Draw a vertical–horizontal diagram and mark the observer's eye.
Place the given angle relative to its horizontal reference.
Label known lengths and distinguish sight lines from ground distances.
Write a ratio equation for each relevant right triangle.
Connect shared heights or distances and solve before rounding.
Restore eye height or subtract component heights, then check physical plausibility.
Connections that matter
Similar triangles explain why a fixed viewing angle fixes the height-to-distance ratio.
Pythagoras finds line-of-sight lengths and broken-object parts.
Linear equations combine two observations with a common height or distance.
Common mistakes
Measuring depression from the vertical: use a horizontal reference.
Using sight-line length as ground distance: identify the hypotenuse.
Forgetting eye height: add it to a height calculated above eye level.
Adding distances for same-side movement or subtracting opposite-side distances: sketch their order first.
Revision checklist
I can draw elevation and depression angles correctly.
I can choose a ratio from the required sides.
I can handle eye height and different vertical levels.
I can combine same-side and opposite-side observations.
I can solve building, flagstaff and broken-tree applications.
I can verify exact values, units and geometric plausibility.