E7.2 - Elevation, depression and shortest distance (Extended)

E7.2 - Elevation, depression and shortest distance (Extended)

An observer is 18 m above level ground, 30 m horizontally from a tower, and sees its top at an angle of elevation of 34∘34^\circ. Is the tower's height 30tan⁡34∘30\tan34^\circ, or is something missing? E7.2 connects horizontal sight lines, perpendicular distances and linked right triangles.

Elevation and depression

Draw a horizontal line through the observer's eye. It is the reference line for both angles: looking upwards gives elevation; looking downwards gives depression. A vertical line is a different reference and produces the complementary angle.

Elevation and depression

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

If a person looks up at a roof at 28∘28^\circ, the angle between their sight line and an upward vertical is 90∘−28∘=62∘90^\circ-28^\circ=62^\circ. The elevation is still 28∘28^\circ, not 62∘62^\circ.

[DIAGRAM: asset_name: Horizontal references for elevation and depression; asset_slug: elevation_depression_horizontal; caption: Elevation is above the observer's horizontal and depression is below it.; description: Two exact side-view panels. In the left, a sight line rises 28 degrees from an observer's dashed horizontal to a target. In the right, a sight line falls 35 degrees below a dashed horizontal to a lower target. The lower target also has a parallel horizontal line, and the equal 35 degree elevation looking back is marked. Vertical directions are shown distinctly from the horizontal references.]
Elevation is above the observer's horizontal and depression is below it.

When two observers look directly at each other, their horizontal lines are parallel. A depression of 35∘35^\circ from the upper observer equals an elevation of 35∘35^\circ from the lower observer, by alternate angles. This equality concerns the same sight line, not a different target.

Watch out

A right triangle drawn beside a vertical wall may show the angle at the top beside the vertical. Subtract that angle from 90∘90^\circ to obtain the depression from horizontal.

Identify both the reference and the angle.

Shortest distance

The shortest distance from a point to a straight line meets that line perpendicularly. To see why, let HH be the perpendicular foot from point PP, and choose another point XX on the line. Pythagoras gives PX2=PH2+HX2PX^2=PH^2+HX^2. For X≠HX\ne H, the added square is positive, so PX>PHPX>PH.

Perpendicular distance

The perpendicular distance from a point to a line is the length of the segment meeting the line at 90∘90^\circ. It is the shortest distance to that line.

The foot can lie on an extension of a drawn segment. A question asking for distance to a finite segment needs its endpoints considered if the foot falls outside it; the shortest-distance-to-a-line result refers to the complete straight line.

[DIAGRAM: asset_name: Why the perpendicular is shortest; asset_slug: perpendicular_shortest_distance; caption: Every slanted route PX adds a positive horizontal square to the perpendicular distance PH.; description: Exact straight horizontal line with foot H, point X further along it, and point P vertically above H. PH is highlighted and has a right-angle square at H. PX is a slanted segment, HX is dashed, and the relationship PX squared equals PH squared plus HX squared is displayed in a separate uncluttered label.]
Every slanted route PX adds a positive horizontal square to the perpendicular distance PH.

Worked example 1: distance to a straight road

Question. Point PP is 24 m from a point AA on a straight road. The angle between APAP and the road is 32∘32^\circ. Calculate the shortest distance from PP to the road.

Plan. Drop a perpendicular from PP to the road; use sine in the resulting right triangle.

sin⁡32∘=d24perpendicular side opposite the angled=24sin⁡32∘=12.718062… m\begin{aligned} \sin32^\circ&=\frac{d}{24} &&\text{perpendicular side opposite the angle}\\ d&=24\sin32^\circ=12.718062\ldots\,\mathrm{m} \end{aligned}

Answer. 12.7 m12.7\,\mathrm{m} to three significant figures.

Check. The along-road distance is 24cos⁡32∘24\cos32^\circ. The two unrounded component squares sum to 24224^2, and the perpendicular is shorter than the 24 m slanted connection.

Keep the 24 m slanted connection but increase the angle it makes with the road. The perpendicular distance should increase.

Worked example 2: a steeper connection

Question. Another point QQ is 24 m from AA, and AQAQ makes 58∘58^\circ with the same straight road. Calculate the shortest distance from QQ to the road.

Plan. Use the new acute angle in the same perpendicular triangle structure.

sin⁡58∘=d24d=24sin⁡58∘=20.353154… m\begin{aligned} \sin58^\circ&=\frac{d}{24}\\ d&=24\sin58^\circ=20.353154\ldots\,\mathrm{m} \end{aligned}

Answer. 20.4 m20.4\,\mathrm{m} to three significant figures.

Check. This lies between the previous perpendicular distance and 24 m, as the steeper connection predicts. Since 58∘=90∘−32∘58^\circ=90^\circ-32^\circ, the new perpendicular equals the previous along-road component.

Continue this lesson.

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