Geometry

Measuring what you cannot reach

A tree is far too tall to reach with a tape measure. How can its shadow, a mirror, or one angle tell you how tall it is?

You need

Angles of elevation and depressionGeometry Measuring and drawing angles with a protractorGrade 4

Where you are 3 in Scale and size

Measurement12

Scale and size3

Measuring what you cannot reach

Scale factor multiplies every length Scale factor, area and volume

When you reach for it

Some heights cannot be reached with a tape measure: a tree, a flagpole, the roof of a building, a monument. What you can still measure is on the ground near them: the length of a shadow, a distance you pace out, an angle read off a protractor. Finding a length from other measurements, without ever touching the length itself, is called indirect measurement.

Every method in this lesson rests on one fact about similar figures: two triangles with the same angles have the same shape, so their sides are in the same ratio. Each method finds a small triangle you can measure completely, and a large triangle with the same angles that contains the height you want. The small triangle gives the ratio, and the ratio carries over to the large one.

Reach for a shadow on a sunny day, when the shadow falls on level ground and you can measure it from the point under the top of the object. Reach for a mirror when there is no sun. Reach for an angle and a distance when you can stand a measured distance from the foot of the object and see its top.

The procedure

Why the shadow triangles have the same shape

Stand a post upright on level ground in sunshine, near a tree. Each casts a shadow. The object, its shadow, and the ray of sunlight that just passes its top make a triangle.

Figure 1The post and the tree with their shadows make two triangles with the same angles, so \displaystyle \frac{a}{b} = \frac{h}{s}.

The two triangles have the same three angles.

A right angle at each foot. The post and the tree are upright and the ground is level, so each meets the ground at 90^\circ.

The same angle at each shadow tip. The Sun is so far away that the rays reaching the post and the tree are parallel. The angle at each shadow tip, between the ground and the ray, is the Sun’s angle of elevation. With both shadows along one line of level ground, as in the figure, the two parallel rays are crossed by the ground, and the two angles are corresponding angles, so they are equal. Call that angle \theta.

The third angles match too. The angles of a triangle add up to 180^\circ, so each third angle is 180^\circ - 90^\circ - \theta.

Triangles with the same angles are similar (two pairs of equal angles are already enough), and the sides of similar triangles are in proportion (proportions in similar triangles).

Setting up the proportion

There are two ways to write it, and they are the same equation rearranged.

Within each triangle. Height divided by shadow is the same in both triangles. With the letters of the figure, the post’s height a over its shadow b equals the tree’s height h over its shadow s:

\frac{a}{b} = \frac{h}{s}

That ratio belongs to the moment, not to the object. At one moment, every upright thing on level ground has the same height-to-shadow ratio. Measure it once on something you can reach, and any shadow times that ratio is a height.

Between the triangles. The tree’s shadow is some number of times the post’s shadow, and since the triangles are similar, the tree’s height is the same number of times the post’s height. That number is the scale factor, and a scale factor multiplies every length.

Either way, the two lengths in one ratio must be in the same unit. Then the units cancel, and the ratio is a plain number that works for any unit of the shadow you multiply. Solving the proportion is an ordinary proportion equation.

Measure from the point under the top

The shadow side of each triangle runs from the point directly under the top out to the shadow tip. For a thin pole that point is the foot. For a building with upright walls and a flat roof, the roof edge is directly above the wall, so the shadow is measured from the wall. For a cone or a pyramid the top is above the center of the base, so part of the triangle’s base is hidden under the object and has to be added to the visible shadow (Example 4).

A mirror instead of a shadow

Lay a small flat mirror on level ground, and stand back from it until you see the top of the building in the middle of the mirror. Light from the top bounces off the mirror into your eye. A flat mirror sends light away at the same angle to the mirror as it arrived; that is the law of reflection, a fact about light that physics measures rather than a theorem of geometry.

That gives two triangles: one from your eye down to your feet and across to the mirror, and one from the top of the building down to its foot and across to the mirror. Each has a right angle at the foot, and the two angles at the mirror are equal, so the triangles are similar. Your eye height over your distance to the mirror equals the building’s height over its distance to the mirror. With e for the eye height, p for your distance, h for the height and m for the building’s distance,

\frac{e}{p} = \frac{h}{m}

It is your eye height that goes in, not your full height, because the light ends at your eye.

An angle and a distance

With no shadow and no mirror, stand on level ground a measured distance from the foot of the object and measure the angle of elevation of its top. A clinometer does this: a protractor held with its straight edge pointing at the top, with a weighted string hanging from the center. The string hangs straight down, and the angle between the string and the protractor’s 90^\circ mark is the angle of elevation.

The right triangle now has its corners at your eye, at the point on the object level with your eye, and at the top. You know one acute angle and the side next to it, and that fixes the triangle’s shape and size. There are two ways to read off the height.

A scale drawing. Draw the triangle small: a base of the right length at a chosen scale, the angle at its left end with a protractor, an upright line at its right end. The drawing and the real triangle both have a right angle and the angle \theta, so they are similar. Measure the drawing’s upright side and scale it back up (scale drawings).

The tangent. All right triangles with the same acute angle \theta are similar, so the ratio of the side opposite \theta to the side next to it depends only on \theta. That ratio is the tangent of \theta, written \tan\theta (right triangle trigonometry):

\text{height above the eye} = \text{distance} \times \tan\theta

The tangent is the scale drawing done once and for all, to many more decimal places than a pencil can manage. Either way, the triangle stops at eye level, so add the eye height at the end.

How exact the answer is

Every input is measured, so the answer can be no more exact than the measurements. Keep full values through the working and round once, at the end. Where a tangent is used, lengths below are given to two decimal places.

Worked examples

Example 1

A flagpole from a post’s shadow

A post 1.2 m tall casts a shadow 0.8 m long. At the same moment a flagpole’s shadow is 6.4 m long. How tall is the flagpole?

Solution

Both triangles have a right angle at the foot and the Sun’s angle at the shadow tip, so they are similar, and height over shadow is the same for both. Call the flagpole’s height h:

\frac{1.2}{0.8} = \frac{h}{6.4}

The left side is 1.5: at this moment every upright object is 1.5 times as tall as its shadow is long. So h = 6.4 \times 1.5 = 9.6.

The scale factor gives the same answer by the other route. The flagpole’s shadow is \displaystyle \frac{6.4}{0.8} = 8 times the post’s, so its height is 8 times the post’s: 8 \times 1.2 = 9.6.

The flagpole is 9.6 m tall. As a check, it is taller than its shadow, just as the post is.

Example 2

Two units in one problem

Ines is 160 cm tall, and her shadow is 2.4 m long. At the same moment a tree’s shadow is 15 m long. How tall is the tree?

Solution

Within one ratio both lengths must be in the same unit, so write Ines’s height as 1.6 m. Her height-to-shadow ratio is

\frac{1.6}{2.4} = \frac{2}{3}

The tree’s height is two thirds of its shadow: \displaystyle 15 \times \frac{2}{3} = 10.

The tree is 10 m tall. Writing \displaystyle \frac{160}{240} with both lengths in centimeters gives the same ratio \displaystyle \frac{2}{3}, which is the point: once the units match, they cancel.

Example 3

From a height to a shadow

At a moment when a 2 m post casts a shadow 3.5 m long, how long is the shadow of a 28 m office building with upright walls and a flat roof? A tennis court begins 45 m from the building’s wall, on the shadow side. Is any of the court in shadow?

Solution

The proportion runs either way. This time the height is known and the shadow is wanted, so use shadow over height, which is \displaystyle \frac{3.5}{2} = 1.75:

\text{shadow} = 28 \times 1.75 = 49 \text{ m}

The roof edge is directly above the wall, so the shadow is measured from the wall. It reaches 49 m out, and the court starts at 45 m, so the first 49 - 45 = 4 m of the court is in shadow.

Example 4

A shadow that starts under the object

A monument is a pyramid on a square base 30 m along each side, with its top directly above the center of the base. At one moment the Sun shines square onto one side, and the tip of the shadow lies 12 m out from the middle of that side. At the same moment a 2 m stick casts a shadow 1.5 m long. How tall is the monument?

Figure 2Example 4. The triangle that matters runs from the point under the top, so its base is 15 + 12 = 27 m.

Solution

The triangle that is similar to the stick’s runs from the point directly under the top, which is the center of the base, out to the shadow tip. The center is half of 30, that is 15 m, in from the middle of the side, so the triangle’s base is

15 + 12 = 27 \text{ m}

and not just the 12 m that can be seen. The stick’s height-to-shadow ratio is \displaystyle \frac{2}{1.5} = \frac{4}{3}, so the monument’s height is

h = 27 \times \frac{4}{3} = 36 \text{ m}

Using only the visible 12 m would have given 16 m, less than half the true height. The shadow shows beyond the base at all only because the Sun is lower than the slope of the monument’s face.

Example 5

Using a mirror

Joel lays a mirror flat on the ground 14 m from the foot of a building and walks back from it, away from the building, until he sees the top of the building in the mirror. He is then 2 m from the mirror, and his eyes are 1.5 m above the ground. How tall is the building?

Figure 3Example 5. Light leaves the mirror at the same angle it arrives, so the small triangle and the large one have the same angles.

Solution

The light from the top of the building leaves the mirror at the same angle it arrived, so the two angles at the mirror are equal. Joel and the building both stand upright on level ground, so each triangle has a right angle at its foot. Two pairs of equal angles make the triangles similar, and

\frac{1.5}{2} = \frac{h}{14}

So h = 14 \times 0.75 = 10.5. The building is 10.5 m tall.

His full height played no part: the light ends at his eye, so the eye height is the side of his triangle.

Example 6

An angle and a distance, two ways

Priya stands on level ground 30 m from the foot of a tree. With a clinometer she measures the angle of elevation of the top as 35^\circ. Her eye is 1.5 m above the ground. Find the height of the tree (a) with a scale drawing and (b) with the tangent.

Figure 4Example 6. The right triangle sits above eye level, so the height it gives is the height above the eye.

Solution

The right triangle runs from her eye across to the tree at eye level, then up to the top. Its base is 30 m and its angle at her eye is 35^\circ.

(a) Choose a scale of 1 cm to 5 m. The 30 m base becomes 6 cm. Draw it, mark 35^\circ at the left end with a protractor, and raise a line at right angles from the right end. Where the slanted line meets the upright one, the upright side measures about 4.2 cm. That stands for

4.2 \times 5 = 21 \text{ m}

above her eye, so the tree is about 21 + 1.5 = 22.5 m tall.

(b) The side opposite the 35^\circ angle is the height above her eye, and the side next to it is the 30 m:

\text{height above the eye} = 30\tan 35^\circ \approx 21.01 \text{ m}

Adding her eye height, the tree is about 22.51 m tall.

The two answers agree to the precision of the drawing. The drawing could not have done better: one millimeter on it is half a meter on the tree.

Example 7

The Sun’s angle, read from a shadow

A 1.8 m post casts a shadow 2.4 m long.

(a) What is the Sun’s angle of elevation?

(b) At the same moment a tower’s shadow is 40 m long. How tall is the tower?

Solution

(a) The post’s triangle has the Sun’s angle at the shadow tip, with the post opposite it and the shadow next to it. So the height-to-shadow ratio is the tangent of the Sun’s angle:

\tan\theta = \frac{1.8}{2.4} = 0.75

The angle with that tangent is found with the inverse tangent: \theta = \tan^{-1}(0.75) \approx 36.87^\circ. A scale drawing of the post and its shadow, measured with a protractor, shows about 37^\circ.

(b) Same moment, same ratio: 40 \times 0.75 = 30. The tower is 30 m tall.

(c) The tower is still 30 m tall, but the ratio has changed with the Sun. Now \displaystyle \tan 50^\circ = \frac{30}{\text{shadow}}, so

\text{shadow} = \frac{30}{\tan 50^\circ} \approx 25.17 \text{ m}

A higher Sun gives a shorter shadow. This example joins the two methods: the shadow ratio of a moment is the tangent of the Sun’s angle at that moment.

Boundaries

The ground must be level, and the object upright. Both are what put a right angle at the foot. With the tree on a slope and the post on level ground, or with either one leaning, the angles at the two feet differ, the triangles no longer have the same angles, and the ratio does not carry over.

Both shadows must be measured at the same moment. The ratio of height to shadow changes as the Sun moves across the sky. A shadow measured an hour later belongs to a different triangle.

The shadow is measured from the point under the top. For a pole that is the foot. For anything wider, measure from the point directly beneath the highest point (Example 4).

The Sun’s rays are treated as parallel. They are not exactly parallel, but the Sun is so far away that the difference across a playground or a city is far smaller than any tape measure can show. A nearby lamp is a different matter: its rays spread out from one point, so two objects at different distances from a street light have shadows in different ratios.

The triangle ends at the eye. With a mirror or a clinometer the right triangle starts at the eye, not the ground. The mirror needs eye height in place of full height; the clinometer needs eye height added at the end. When the angle is measured at ground level there is nothing to add.

The angle method needs the foot. One angle gives the shape of the triangle, and the distance to the foot gives its size. When the foot cannot be reached, one angle is not enough, and two sightings from two places are needed (angles of elevation and depression).

The answer is no more exact than the measurements. A shadow measured to the nearest 10 cm and an angle to the nearest degree give a height that is close, not exact.

Where this goes next

This lesson is proportions in similar triangles put to work outdoors. In geometry, the reasons used here become proofs: why two angles are enough for two triangles to be similar, and how to prove triangles similar in general.

The angle-and-distance method grows into right triangle trigonometry. The tangent is one of three ratios, and with the sine and cosine any right triangle can be solved from one side and one angle. When the foot of the object cannot be reached, two sightings from two places fix the height that one sighting cannot.

Scaling also reaches past lengths. A model twice as tall has twice every length but not twice the area; that is what a scale factor does to area and volume.

Exercises

Shadows

A 2 m pole casts a shadow 2.5 m long. At the same moment a tree’s shadow is 20 m long. How tall is the tree?
A man 6 ft tall casts a shadow 4 ft long. At the same moment a building’s shadow is 36 ft long. How tall is the building?
A girl 150 cm tall casts a shadow 2 m long. At the same moment a lamppost’s shadow is 4.8 m long. How tall is the lamppost, in meters?
A 3 m road sign casts a shadow 4 m long. At the same moment a ship’s mast on the quay casts a shadow 22 m long. How tall is the mast?
A 1.2 m post casts a shadow 0.9 m long. How long is the shadow of a 16 m building with upright walls at the same moment?
At a moment when a 1 m stick casts a shadow 1.75 m long, a 12 m tree stands 20 m from the edge of a pond, with its shadow pointing toward the pond. Does the shadow reach the pond? If so, how far past the edge?

Under the object, and mirrors

A pile of sand is shaped like a cone, 8 m across at the base, with its top above the center. Its shadow reaches 5 m out from the edge of the base, measured along a line through the center. At the same moment a 1 m stick casts a shadow 1.5 m long. How tall is the pile?
A mirror lies flat on level ground 15 m from the foot of a tree. Dev stands 2 m from the mirror, on the side away from the tree, and sees the top of the tree in it. His eyes are 1.6 m above the ground. How tall is the tree?
Lena’s eyes are 1.5 m above the ground, and she wants to stand 3 m from a mirror lying on the ground and see the top of a 20 m building in it. How far from the foot of the building should the mirror be?

An angle and a distance

From a point on level ground 25 m from the foot of a tower, the angle of elevation of the top, measured at ground level, is 40^\circ. Find the height (a) with a scale drawing at 1 cm to 5 m and (b) with the tangent.
Standing 60 m from the foot of a building, with eyes 1.7 m above the ground, Amara measures the angle of elevation of the roof as 28^\circ. How tall is the building?
The top of a water tower is 36 m above level ground. From a point on the ground the angle of elevation of the top is 50^\circ. How far is the point from the foot of the tower?
A 2 m pole casts a shadow 3 m long. What is the Sun’s angle of elevation?
When the Sun’s angle of elevation is 60^\circ, how long is the shadow of a 15 m tree? Give the exact value and a decimal.
A 1.6 m post casts a shadow 1.2 m long. What is the Sun’s angle of elevation, and how tall is a tree whose shadow at the same moment is 9 m long?

Reasoning

Explain why the triangle made by a post and its shadow and the triangle made by a tree and its shadow are similar. Name each fact you use.
Kwame measures a post’s shadow at ten in the morning and a tree’s shadow at three in the afternoon, and uses the two in one proportion. Explain why the answer cannot be trusted.
At what angle of elevation of the Sun is every shadow exactly as long as the object casting it? Explain why.
A tree stands at the top of a slope, and its shadow falls down the slope. Explain why using that shadow in a proportion with a post on level ground gives too large a height.
Explain why a scale drawing and the tangent give the same height in an angle-and-distance problem, and why the tangent’s answer is the more exact one.

Answers

16 m. Height over shadow is \displaystyle \frac{2}{2.5} = 0.8, and 20 \times 0.8 = 16.
54 ft. Height over shadow is \displaystyle \frac{6}{4} = 1.5, and 36 \times 1.5 = 54.
3.6 m. Write 150 cm as 1.5 m first. Height over shadow is \displaystyle \frac{1.5}{2} = 0.75, and 4.8 \times 0.75 = 3.6.
16.5 m. \displaystyle \frac{h}{22} = \frac{3}{4}, so \displaystyle h = \frac{3 \times 22}{4} = 16.5.
12 m. Shadow over height is \displaystyle \frac{0.9}{1.2} = 0.75, and 16 \times 0.75 = 12.
Yes, by 1 m. Shadow over height is 1.75, so the tree’s shadow is 12 \times 1.75 = 21 m long, 1 m more than the 20 m to the pond.
6 m. The top is above the center, 4 m in from the edge, so the triangle’s base is 4 + 5 = 9 m. Height over shadow is \displaystyle \frac{1}{1.5}, so \displaystyle h = \frac{9}{1.5} = 6.
12 m. \displaystyle \frac{1.6}{2} = \frac{h}{15}, so h = 15 \times 0.8 = 12.
40 m. Call the distance d: \displaystyle \frac{1.5}{3} = \frac{20}{d}, so \displaystyle d = \frac{20 \times 3}{1.5} = 40.
About 21 m. (a) The base is 5 cm, and the upright side measures about 4.2 cm, which is 4.2 \times 5 = 21 m. (b) 25\tan 40^\circ \approx 20.98 m. Measured at ground level, so nothing is added.
About 33.60 m. 60\tan 28^\circ \approx 31.90 m above her eye, plus 1.7 m.
About 30.21 m. \displaystyle \tan 50^\circ = \frac{36}{d}, so \displaystyle d = \frac{36}{\tan 50^\circ}.
About 33.69^\circ. \displaystyle \tan\theta = \frac{2}{3}, so \displaystyle \theta = \tan^{-1}\left(\frac{2}{3}\right).
\displaystyle 5\sqrt{3} \approx 8.66 m. \displaystyle \tan 60^\circ = \frac{15}{s} and \displaystyle \tan 60^\circ = \sqrt{3}, so \displaystyle s = \frac{15}{\sqrt{3}} = 5\sqrt{3}.
About 53.13^\circ, and the tree is 12 m tall. \displaystyle \tan\theta = \frac{1.6}{1.2} = \frac{4}{3}, so \displaystyle \theta = \tan^{-1}\left(\frac{4}{3}\right), and the tree is \displaystyle 9 \times \frac{4}{3} = 12 m.
Both objects are upright on level ground, so each triangle has a right angle at the foot. The Sun’s rays are parallel, so each ray meets the level ground at the same angle, the Sun’s angle of elevation (corresponding angles, when the shadows lie along one line). Two pairs of angles are equal, so the third pair is too, since each triangle’s angles add up to 180^\circ. Equal angles make the triangles similar, so their sides are in proportion.
The Sun’s angle of elevation changes during the day, and with it the ratio of height to shadow. The post’s triangle and the tree’s triangle then have different angles at the shadow tip, so they are not similar and the proportion between them is false.
At 45^\circ. A shadow equals its object’s height exactly when height over shadow is 1, and that ratio is the tangent of the Sun’s angle. \tan 45^\circ = 1: the triangle is a right triangle with two 45^\circ angles, so its two shorter sides are equal.
On level ground the shadow tip is where the ray meets the ground. When the ground drops away below that level, the ray has further to fall before it meets the ground, so it lands further out, and the shadow is longer than it would be on level ground. The post’s ratio times a shadow that is too long gives a height that is too large.
The drawn triangle and the real one both have a right angle and the measured angle \theta, so they are similar, and the drawing’s upright side over its base is the real ratio. Every right triangle with the angle \theta has that same ratio, and \tan\theta is its value. A drawing can be measured to about a millimeter, while a calculator gives \tan\theta to many decimal places, so the tangent loses nothing to the drawing.