E6.2 - Elevation, depression and shortest distance

E6.2 - Elevation, depression and shortest distance

A sight line can tell you the height of an inaccessible object, provided you measure its angle from the right reference. You will interpret elevation and depression, connect right-angled triangles, and find the shortest distance from a point to a straight line. This lesson is for Extended.

Angles from the horizontal

Two people look along the same sight line in opposite directions. Why can one person's angle of elevation equal the other person's angle of depression? The reference lines through their eyes are different lines, but both are horizontal.

Imagine looking straight ahead, then lifting your gaze towards the top of a tower. The turn upwards starts at the horizontal through your eye, not at the vertical tower or the ground below your feet.

Elevation and depression

An angle of elevation is the angle measured upwards from the horizontal to a sight line. An angle of depression is the angle measured downwards from the horizontal to a sight line.

The horizontal is drawn through the observer. An angle measured from a vertical is not the angle of elevation or depression.

Horizontal lines are parallel. A sight line crossing two of them forms equal alternate angles, so reversing the view changes elevation to depression without changing the angle's size.

[DIAGRAM: asset_name: Elevation and depression share a sight line; asset_slug: elevation_depression; caption: Parallel horizontals give equal elevation and depression angles.; description: An exact rising sight line joins a lower observer A to a higher point B. Dashed horizontal lines through A and B are parallel. The 37 degree angle above the lower horizontal and the 37 degree angle below the upper horizontal are marked, with a separate vertical reference at A.]
Parallel horizontals give equal elevation and depression angles.

Worked example 1: reversing an upward view

Question. From a point AA, the angle of elevation of BB is 37∘37^\circ. Find the angle of depression of AA from BB, and the angle between ABAB and the upward vertical at AA.

Plan. Use parallel horizontals for the reversed view, then a right angle for the vertical reference.

depression from B=37∘alternate anglesangle with vertical=90∘−37∘vertical is perpendicular to horizontal=53∘\begin{aligned} \text{depression from }B&=37^\circ && \text{alternate angles}\\ \text{angle with vertical}&=90^\circ-37^\circ && \text{vertical is perpendicular to horizontal}\\ &=53^\circ \end{aligned}

Answer. The depression is 37∘37^\circ; the angle with the upward vertical is 53∘53^\circ.

Check. The two angles at AA add to 90∘90^\circ. A line closer to horizontal than vertical should make the smaller angle with horizontal.

Now the first observer looks downwards. Only the direction of the view changes; the horizontal reference remains essential.

Worked example 2: reversing a downward view

Question. From PP, the angle of depression of QQ is 28∘28^\circ. Find the angle of elevation of PP from QQ, and the angle between PQPQ and the downward vertical at PP.

Plan. Transfer the angle to the parallel horizontal through QQ, then find its complement at PP.

elevation from Q=28∘alternate anglesangle with downward vertical=90∘−28∘=62∘complete the right angle\begin{aligned} \text{elevation from }Q&=28^\circ && \text{alternate angles}\\ \text{angle with downward vertical}&=90^\circ-28^\circ\\ &=62^\circ && \text{complete the right angle} \end{aligned}

Answer. 28∘28^\circ and 62∘62^\circ, respectively.

Check. 28∘+62∘=90∘28^\circ+62^\circ=90^\circ. Reversing the sight line again restores the original depression of 28∘28^\circ.

Watch out

“An angle of 28∘28^\circ from the vertical is a depression of 28∘28^\circ” uses the wrong starting line. The depression would be 90∘−28∘=62∘90^\circ-28^\circ=62^\circ.

Elevation and depression problems

A height problem begins with a right-angled triangle between the sight line, the horizontal through the eye and a vertical height difference. That height difference may be less than the object's total height.

You already know

In a right-angled triangle, tan⁡θ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}. Here the opposite side is the vertical difference between eye and target; the adjacent side is the horizontal separation.

Sketch the observer's eye first. Draw a horizontal to the object's vertical, then label only the sides belonging to that triangle. For a downward view, the vertical difference is still a positive length. Add or subtract it from the observer's level as the context requires.

For instance, a lookout's eye is 24 m24\,\mathrm{m} above level ground. A point on that ground has an angle of depression of 32∘32^\circ. If its horizontal distance is dd, the downward triangle gives tan⁡32∘=24/d\tan32^\circ=24/d, so d=24/tan⁡32∘=38.4 md=24/\tan32^\circ=38.4\,\mathrm{m} to 33 significant figures. The positive vertical difference is 2424, even though the sight line goes downwards. As a check, a run of about 38.438.4 and a drop of 2424 give an angle close to 32∘32^\circ from the horizontal.

Worked example 3: adding the eye height

Question. An observer stands on level ground 20 m20\,\mathrm{m} from a vertical tower. Their eye is 1.5 m1.5\,\mathrm{m} above the ground. The angle of elevation of the top is 35∘35^\circ. Calculate the tower's height.

Plan. Find the height above eye level using tangent, then add the eye height. Use degree mode.

tan⁡35∘=H−1.520opposite over adjacentH−1.5=20tan⁡35∘isolate the vertical differenceH=1.5+20tan⁡35∘=15.5041…keep the calculator value\begin{aligned} \tan35^\circ&=\frac{H-1.5}{20} && \text{opposite over adjacent}\\ H-1.5&=20\tan35^\circ && \text{isolate the vertical difference}\\ H&=1.5+20\tan35^\circ\\ &=15.5041\ldots && \text{keep the calculator value} \end{aligned}

Answer. The tower is 15.5 m15.5\,\mathrm{m} high, to 33 significant figures.

Check. The height above the eye is about 14.0 m14.0\,\mathrm{m}. Its ratio to 20 m20\,\mathrm{m} is about 0.7000.700, giving an angle close to 35∘35^\circ, and adding 1.5 m1.5\,\mathrm{m} restores the total height.

What if the distance to the tower cannot be measured? Add one observation from a known distance farther away. Both triangles share the same vertical height difference, so two tangent equations can determine it.

[DIAGRAM: asset_name: Two observations of one tower; asset_slug: linked_observations; caption: Moving 12 metres away changes the angle but leaves the height above eye level unchanged.; description: To-scale side view of a vertical tower with top T. Two eyes F and N at the same height 1.5 metres are on one horizontal line, F 12 metres farther from the tower than N. Elevation angles are 35 degrees at F and 55 degrees at N. The nearer horizontal distance is d and the tower height above the eye line is y.]
Moving 12 metres away changes the angle but leaves the height above eye level unchanged.

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