Math by First Principles

Trigonometry in Measurement

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Learning goals

  • Identify elevation and depression angles from horizontal references
  • Build height and distance models that include observer height and ground levels
  • Relate ramp inclination to rise, horizontal run, sloped length, and percent gradient
  • Combine two compatible observations to recover inaccessible measurements
  • Check assumptions, units, and rounding before reporting a measurement

Start from a horizontal reference

An angle of elevation is measured upward from a horizontal ray through the observer’s eye to the line of sight. An angle of depression is measured downward from that horizontal ray. A vertical wall or sloping ground is a different reference.

For example, a sight line 20∘20^\circ below the horizontal makes a depression angle of 20∘20^\circ. It makes 70∘70^\circ with the downward vertical. Naming the reference prevents those angles from being exchanged.

Measurements in a vertical planeEye E is above the base B by drop. Target P is 0 m above that same horizontal base level. Two dashed horizontal references are parallel. The line EP slopes down at depression angle 28° from the upper horizontal. The horizontal run is labeled horizontal run.EBP28°horizontalhorizontal rundropTarget above base level: 0 m
The upper and lower horizontal references are parallel.

For two points at different heights in one vertical plane, draw a horizontal through each point. These lines are parallel. The common sight line cuts them, giving equal alternate interior angles. Thus the angle of depression from the higher point equals the angle of elevation from the lower point. Neither is generally the angle between the sight line and a vertical side.

Why the two sight angles match#

For a depression angle of 28∘28^\circ, the parallel horizontals and the sight-line transversal give an alternate interior angle of 28∘28^\circ at the lower point. The angle between that sight line and the vertical at the higher point is instead

90∘−28∘=62∘.90^\circ-28^\circ=62^\circ.

The same parallel-line argument works for any acute depression angle θ\theta. The lower horizontal points toward the higher point, and the upper horizontal points toward the lower point. Their alternate interior angles both have measure θ\theta. This uses parallel lines, not a guess that the sloping sight line is horizontal.

Measurements in a vertical planeEye E is above the base B by drop. Target P is 0 m above that same horizontal base level. Two dashed horizontal references are parallel. The line EP slopes down at depression angle 28° from the upper horizontal. The horizontal run is labeled run.28°EBP28°horizontalrundropTarget above base level: 0 m
Matching angles follow from the marked parallel horizontals.

Check your understanding

A sight line is 34 degrees below a horizontal through an observer's eye. What is the angle of depression?

Height above the eye and height above the ground

Suppose a vertical post stands on the same level ground as an observer. The horizontal separation is 1010 m, the elevation angle is 45∘45^\circ, and the eye is 1.61.6 m above ground. Tangent gives a rise of 1010 m above the eye. The whole post is therefore 11.611.6 m tall.

For the general model, let dd be the positive horizontal separation, ee the eye height, HH the object height, and θ\theta an acute elevation angle. The opposite leg is the height difference, so

tan⁡θ=H−ed.H=e+dtan⁡θ.\begin{aligned} \tan\theta&=\frac{H-e}{d}.\\ H&=e+d\tan\theta. \end{aligned}

This form assumes a vertical object, level ground at the same elevation at both bases, and a shared vertical sight plane. If the observer’s ground is higher than the object base by a known vertical amount gg, the eye is e+ge+g above that base instead. If it is lower by gg, subtract that ground difference. Ground slope cannot silently be treated as horizontal distance.

Worked example 1 A height observation

Problem. A vertical mast and an observer stand on the same level ground in one vertical plane. The measurements are

d=28.0 m horizontally.e=1.65 m.θ=37.0∘.\begin{aligned} d&=28.0\text{ m horizontally}.\\ e&=1.65\text{ m}.\\ \theta&=37.0^\circ. \end{aligned}

Estimate the mast height to the nearest tenth of a meter.

Measurements in a vertical planeE is the eye above ground point F. The object BT is vertical. F and B have the same ground elevation. The dashed horizontal through E meets BT and forms a right angle. The line ET rises at 37° above that horizontal. The horizontal run is 28 m, eye height is 1.65 m, and object height is labeled H.EFBCThorizontal28 mH37°Eye height EF: 1.65 m
The dashed reference begins at the eye, above the ground.

Solution. The right triangle begins at the eye. Its rise is the mast height minus eye height:

tan⁡37∘=H−1.6528.\tan37^\circ=\frac{H-1.65}{28}.

Solve without rounding the tangent:

H=1.65+28tan⁡37∘.H=1.65+28\tan37^\circ.

The model gives

H≈22.7 m.H\approx22.7\text{ m}.

The height difference above the eye is about 21.121.1 m. That is only part of the whole mast.

Check your understanding

For a vertical tree on the same level ground as an observer, the eye is 1.8 m high, the horizontal separation is 15 m, and the elevation angle is 40 degrees. Which expression gives the whole tree height in meters?

Downward sight lines

For a lookout eye 1212 m above a horizontal water surface and a target on that surface, a 30∘30^\circ depression angle gives

tan⁡30∘=12d.\tan30^\circ=\frac{12}{d}.

The horizontal distance is therefore 12/tan⁡30∘12/\tan30^\circ meters. If the target were on a deck above the water, the numerator would be the vertical drop from the eye to that deck, rather than the whole eye height above water.

Worked example 2 A lookout and a marker

Problem. A lookout’s eye is 30.030.0 m above a horizontal water surface. A floating marker is modeled as a point on that surface, in the same vertical sight plane. The depression angle is 25.0∘25.0^\circ from the horizontal. Find the horizontal distance to the marker and the straight sight-line length, each to the nearest tenth of a meter.

Measurements in a vertical planeEye E is above the base B by 30 m. Target P is 0 m above that same horizontal base level. Two dashed horizontal references are parallel. The line EP slopes down at depression angle 25° from the upper horizontal. The horizontal run is labeled d.LEBP25°horizontald30 mTarget above base level: 0 m
The horizontal run differs from the sloping sight line.

Solution. The vertical drop is 3030 m. The depression angle transfers to the marker’s elevation angle because the two horizontal references are parallel. Tangent relates the drop and horizontal run:

tan⁡25∘=30d.d=30tan⁡25∘.d≈64.3 m.\begin{aligned} \tan25^\circ&=\frac{30}{d}.\\ d&=\frac{30}{\tan25^\circ}.\\ d&\approx64.3\text{ m}. \end{aligned}

The sight line is the hypotenuse LL. Sine relates it to the same drop:

sin⁡25∘=30L.L=30sin⁡25∘.L≈71.0 m.\begin{aligned} \sin25^\circ&=\frac{30}{L}.\\ L&=\frac{30}{\sin25^\circ}.\\ L&\approx71.0\text{ m}. \end{aligned}

The longer value is the sloping sight line. Replacing horizontal run by that length would change the model.

Ramps, inclination, and gradient

A ramp’s rise is its vertical height change. Its run is its horizontal distance, not its sloped surface length. For a straight ramp that rises 11 m over a horizontal run of 1010 m, the rise/run ratio is 0.10.1. Its percent gradient is 100100 times that ratio, or 10%10\%.

If rr is a positive rise, dd a positive run, LL the sloped length, and θ\theta the acute inclination from horizontal, the right triangle gives

tan⁡θ=rd.sin⁡θ=rL.cos⁡θ=dL.\begin{aligned} \tan\theta&=\frac rd.\\ \sin\theta&=\frac rL.\\ \cos\theta&=\frac dL. \end{aligned}

Thus a gradient percentage is not an angle in degrees. A 10%10\% gradient has inclination tan⁡−1(0.1)\tan^{-1}(0.1), approximately 5.7∘5.7^\circ. A rise:run label of 1:101:10 means ten horizontal units for each vertical unit.

Worked example 3 A ramp profile

Problem. A straight ramp rises 0.750.75 m over a horizontal run of 9.009.00 m. Find its percent gradient to the nearest tenth of a percent, inclination to the nearest tenth of a degree, and sloped length to the nearest hundredth of a meter.

Measurements in a vertical planeA straight ramp is the sloping hypotenuse, labeled L. The horizontal run is 9.00 m and the perpendicular vertical rise is 0.75 m. The angle with the horizontal is labeled θ.9.00 m0.75 mLθHorizontal run and vertical rise
The profile distinguishes all three lengths.

Solution. Compare rise with horizontal run:

G=100(0.759).G≈8.3%.\begin{aligned} G&=100\left(\frac{0.75}{9}\right).\\ G&\approx8.3\%. \end{aligned}

Use the acute inverse tangent in degree mode for the inclination:

θ=tan⁡−1(0.759).θ≈4.8∘.\begin{aligned} \theta&=\tan^{-1}\left(\frac{0.75}{9}\right).\\ \theta&\approx4.8^\circ. \end{aligned}

Pythagoras gives the sloped length:

L=92+0.752.L≈9.03 m.\begin{aligned} L&=\sqrt{9^2+0.75^2}.\\ L&\approx9.03\text{ m}. \end{aligned}

The slope length is slightly greater than the horizontal run. These are measurements of three different features of one profile.

Check your understanding

A ramp rises 0.60 m over a horizontal run of 7.50 m. What is its percent gradient?

Combine two compatible observations

Two views can supply a length that neither view supplies alone. The station positions must belong to one stated configuration. Suppose two observers have equal eye heights on the same level ground, one 1010 m farther from a vertical pole along the same straight line through its base. If the top is 1010 m above their eyes and the nearer run is 1010 m, the nearer elevation is 45∘45^\circ. The farther run is 2020 m, and its elevation is tan⁡−1(10/20)\tan^{-1}(10/20), about 26.6∘26.6^\circ. Moving farther away makes the angle smaller under these conditions.

Worked example 4 Two survey stations

Problem. Stations FF and NN and the base BB of a vertical tower lie in that order on one straight horizontal line. The nearer station is NN. Both sighting instruments are 1.501.50 m above this ground and point at the same tower top in their common vertical plane. The data are

FN=18.0 m.αF=31.0∘.βN=52.0∘.\begin{aligned} FN&=18.0\text{ m}.\\ \alpha_F&=31.0^\circ.\\ \beta_N&=52.0^\circ. \end{aligned}

Find the horizontal distance NBNB and the tower height, each to the nearest tenth of a meter.

Measurements in a vertical planeAll bases lie on the same horizontal line and both eyes are 1.5 meters above it. The station order is F, N, B, with N nearer the tower. Blue and dashed red sight lines reach the same tower top T. Elevation angles are 31 and 52 degrees. The marked separation is 18 m. Object height is H.FNBT31°52°HBoth eye heights: 1.5 m18 mx
The stations and base follow the stated order F, N, B.

Solution. Let xx be NBNB and uu the tower height above instrument level. The far run is x+18x+18, so

u=xtan⁡52∘.u=(x+18)tan⁡31∘.\begin{aligned} u&=x\tan52^\circ.\\ u&=(x+18)\tan31^\circ. \end{aligned}

Equate the two expressions for the same vertical rise:

xtan⁡52∘=(x+18)tan⁡31∘.x\tan52^\circ=(x+18)\tan31^\circ.

Collect the terms containing xx:

x=18tan⁡31∘tan⁡52∘−tan⁡31∘.x=\frac{18\tan31^\circ}{\tan52^\circ-\tan31^\circ}.

This gives

x≈15.9 m.x\approx15.9\text{ m}.

Use the unrounded xx in either original equation, then add the instrument height:

H=1.50+xtan⁡52∘.H≈21.9 m.\begin{aligned} H&=1.50+x\tan52^\circ.\\ H&\approx21.9\text{ m}. \end{aligned}

Both equations return the same rise before rounding. The positive distance and larger near angle agree with the stated station order.

With stations on opposite sides of a tower, the two runs add to the station separation instead. Do not use x+18x+18 for that arrangement. Draw the station order before deciding whether to add or subtract lengths.

Worked example 5 A window observation

Problem. From eye point EE, the top of a vertical mast is above the eye’s horizontal and the base is below it. All three points lie in one vertical plane. The horizontal distance from EE to the mast is 44.044.0 m. The elevation angle to the top is 29.0∘29.0^\circ; the depression angle to the base is 17.0∘17.0^\circ. Find the whole mast height to the nearest tenth of a meter.

Measurements in a vertical planeEye E sees the top T above its horizontal at elevation 29 degrees and the base B below its horizontal at depression 17 degrees. BT is vertical, and the horizontal run from E to BT is 44 m. The total vertical height BT is H.EBT29°17°horizontal44 mH
The upper and lower sight lines share the horizontal run.

Solution. The horizontal through EE divides the mast into an upper rise and a lower drop. Both right triangles have horizontal run 4444 m.

u=44tan⁡29∘.v=44tan⁡17∘.\begin{aligned} u&=44\tan29^\circ.\\ v&=44\tan17^\circ. \end{aligned}

The two vertical parts lie on opposite sides of eye level, so add them:

H=44(tan⁡29∘+tan⁡17∘).H≈37.8 m.\begin{aligned} H&=44(\tan29^\circ+\tan17^\circ).\\ H&\approx37.8\text{ m}. \end{aligned}

The full height is the vertical separation from base to top. No unstated building height is needed: the depression observation determines the part below eye level.

Check your understanding

Two equal-height instruments on the same level ground stand on the same side of a tower base. Along one straight line, the order is far station, near station, tower base. Their horizontal separation is 12 m. If x is the near run, which run belongs to the far observation?

Check the model before reporting

Use degree mode and keep full calculator precision through the final step. A rounded intermediate angle or tangent can move the last reported digit. Report a sensible final unit and precision; extra calculator digits do not improve a rough field measurement.

Before accepting an estimate, check the physical roles. Is the run horizontal? Are the object and height directions vertical? Does the model use the eye’s actual elevation relative to the object’s base? Do both observations concern the same point, vertical plane, and station order? A scale-looking picture alone establishes none of those facts.

An independent measurement can test consistency, but agreement cannot prove every assumption. If two instruments share an incorrect reference height, their angle calculations can agree while both whole-height estimates share the same offset.

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Practice

Multiple Choice Questions (MCQ)

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More resources (optional)

Other explanations of this lesson, if you want a second take.

A bit of history (optional)

Repairs after the 2011 Virginia earthquake surrounded the Washington Monument with scaffolding. That gave NOAA surveyors a rare opportunity in 2013 and 2014: they could put instruments at the peak itself, as well as observe it from below.

The survey produced an architectural height slightly shorter than the familiar historical figure. That did not simply mean the monument had shrunk by the difference. A height needs a starting point, and the modern measurement followed a stated standard for that reference. NOAA’s explanation says the exact point treated as zero in the nineteenth-century survey remains unknown; its comparison with the older location is therefore qualified.

The team recorded both its methods and reference conventions so later surveys could make clearer comparisons. The National Geodetic Survey’s account and explanation of the differing heights describe that work.

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