Geometry · G.SRT.C.8

Angles of elevation and depression

You can see the top of a tower but cannot walk to its foot. How can angles, measured from where you are allowed to stand, tell you how tall it is?

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Angles of elevation and depression

When you reach for it

Some lengths cannot be measured with a tape: the height of a cliff, a tower on the far bank of a river, the distance from a lighthouse out to a boat. Angles can be measured from where you stand, with a clinometer or a surveyor’s instrument, and solving a right triangle turns one angle and one length into all the rest. This lesson is about where those right triangles are in the world, and what to do when one triangle is not enough.

Every problem here is built from the same three lines: a horizontal line through the eye of the person looking, a vertical line through the thing looked at, and the line of sight joining the two. A horizontal line and a vertical line meet at a right angle, so the three always make a right triangle, and the angle that gets measured is always taken from the horizontal.

Reach for one sighting when one of the two legs is known: the distance along the ground to the foot of a building, or the height of the cliff you are standing on. Reach for the same triangle, read as rise over run, when a road or a roof is described by its grade. And reach for two sightings when the foot of the object cannot be reached, so that neither leg is known. One angle then gives only the shape of the triangle, not its size, and a second angle measured from a second place, a known distance away, is what fixes the height.

The procedure

Angles of elevation and depression

When you look up at something, the angle of elevation is the angle between the horizontal line through your eye and your line of sight to it. When you look down at something, the angle of depression is the angle between the horizontal line through your eye and your line of sight, this time measured downward. Both are measured from the horizontal, never from the vertical, and both lie strictly between 0^\circ and 90^\circ.

Figure 1Looking up from A and looking down from B give the same angle, because the two horizontals are parallel.

Elevation from one end is depression from the other. Suppose from A the angle of elevation of B is 32^\circ. Then from B the angle of depression of A is also 32^\circ. The horizontal through A and the horizontal through B are both perpendicular to the same vertical, so they are parallel. The line of sight AB crosses both of them, and the two marked angles lie between the parallels on opposite sides of it: they are alternate interior angles, so they are equal.

That fact is the first move in almost every depression problem. The angle of depression at B sits outside the right triangle, between the line of sight and a horizontal that is not one of the triangle’s sides. The equal angle of elevation at A sits inside it, as one of its acute angles. So move the angle to the lower end, and the problem becomes an ordinary right triangle with the angle where opposite and adjacent are easy to name.

One sighting

Put the angle \theta at the lower end. The horizontal distance d is adjacent to it, the difference in height v is opposite it, and the line of sight s is the hypotenuse:

\tan\theta = \frac{v}{d}, \quad \sin\theta = \frac{v}{s}, \quad \cos\theta = \frac{d}{s}

Most problems give d and want v, or give v and want d, so the tangent is the ratio used most often. The sine and cosine come in when the straight-line distance s is known or wanted.

Heights are measured from the eye. The vertical leg runs from the level of the eye to the object, not from the ground. If the eye is 1.6 meters above level ground, the height found from the triangle is the height above the eye, and the object’s height above the ground is 1.6 meters more. When a problem says the angle is measured at ground level, there is nothing to add.

Grade and slope

The steepness of a road, a roof or a railway is its rise divided by its run, which is the Slope of the line it follows. Put the angle \theta that the road makes with the horizontal at the bottom of a right triangle: the run is adjacent to \theta and the rise is opposite it, so

\text{slope} = \frac{\text{rise}}{\text{run}} = \tan\theta

Figure 2The slope is rise over run, and that is the tangent of the angle the slope makes with the horizontal.

The grade of a road is the same slope written as a percentage. A road with a grade of 8\% rises 8 meters for every 100 meters it runs, so its slope is 0.08 and \tan\theta = 0.08. Two things follow at once. The run is the horizontal distance, not the distance along the road, which is the hypotenuse and a little longer. And the grade is not a share of a right angle: a grade of 100\% means rise equals run, which is \tan 45^\circ = 1, so it is a 45^\circ slope, not a vertical wall.

Going from an angle to a grade is one tangent. Going from a grade back to an angle asks which acute angle has a given tangent, which is the inverse tangent: \theta = \tan^{-1}(0.08) \approx 4.57^\circ.

Two sightings from the same side

Here is the case that one triangle cannot handle. A tower stands on the far bank of a river, so its foot F cannot be reached. From a point P on level ground the angle of elevation of the top T is \alpha. You walk a measured distance d straight toward the tower, to Q, and the angle of elevation is now \beta. The height h = TF is wanted, and so is the distance x = QF from Q to the foot.

Figure 3Two sightings in line with the foot of the tower. The triangles TPF and TQF share the side TF = h.

Why one sighting is not enough. From P alone, \displaystyle \tan\alpha = \frac{h}{x + d} gives only the ratio of the height to a distance nobody knows. A tall tower far away and a short one close by can look exactly the same. Every right triangle with the angle \alpha has the same shape, so one angle fixes the shape and not the size.

The two triangles. Triangles TQF and TPF both have their right angle at F, and they share the side TF = h. Each gives one equation, from the tangent at its own observer:

h = x\tan\beta \qquad\text{and}\qquad h = (x + d)\tan\alpha

That is a system of two equations in the two unknowns h and x. Both right sides equal h, so they equal each other:

\begin{aligned} x\tan\beta &= x\tan\alpha + d\tan\alpha \\ x\,(\tan\beta - \tan\alpha) &= d\tan\alpha \\ x &= \frac{d\tan\alpha}{\tan\beta - \tan\alpha} \end{aligned}

and then h = x\tan\beta gives

h = \frac{d\tan\alpha\tan\beta}{\tan\beta - \tan\alpha}

The division is allowed because \tan\beta - \tan\alpha is not 0. From Q the same height stands over a shorter horizontal distance, so \displaystyle \frac{h}{x} is bigger than \displaystyle \frac{h}{x + d}, and \tan\beta > \tan\alpha: the nearer point always sees the larger angle.

Another way to read it. Divide the top and bottom of the formula by \tan\alpha\tan\beta:

h = \frac{d}{\dfrac{1}{\tan\alpha} - \dfrac{1}{\tan\beta}}

which says \displaystyle d = \frac{h}{\tan\alpha} - \frac{h}{\tan\beta}: the walk is the difference between the two horizontal distances to the foot. The ratio \displaystyle \frac{1}{\tan\theta} has a name of its own, the cotangent, and some books write the formula with it.

The formula is worth knowing, but it is the two equations that are worth remembering. They change with the layout, and the formula does not. If the object stands between the two observers, the two ground distances add up to the distance between them instead of differing by it (Example 5). If both angles are measured from one window, one looking up and one looking down, the horizontal through the window splits the view into two right triangles that share a horizontal side instead of a vertical one (Example 7). In every case, draw the horizontal and the vertical, find the right triangles, write one equation per triangle, and solve the system.

Rounding

The angles are measured to the nearest degree or so, which already limits the accuracy. Keep the calculator’s full values through the working and round once, at the end, so the rounding does not add a second error to the first. All lengths below are given to two decimal places.

Worked examples

Example 1

One sighting, with the height of the eye

A surveyor stands on level ground 40 m from the foot of a building and measures the angle of elevation of its top as 52^\circ. Her eye is 1.6 m above the ground. How tall is the building?

Solution

The horizontal through her eye meets the building 1.6 m above the ground. The right triangle has its angle of 52^\circ at her eye, the horizontal 40 m adjacent to it, and the part of the building above eye level, v, opposite it. Opposite and adjacent means the tangent:

\tan 52^\circ = \frac{v}{40} \quad\Longrightarrow\quad v = 40\tan 52^\circ \approx 51.20 \text{ m}

That is the height above her eye. The building’s height is 1.6 m more:

40\tan 52^\circ + 1.6 \approx 52.80 \text{ m}

Leaving out the eye height would understate the building by exactly 1.6 m, however far away she stood.

Example 2

An angle of depression

A lighthouse keeper’s eye is 32 m above the sea. She sees a boat at an angle of depression of 7^\circ. How far is the boat from the foot of the lighthouse, and how far is it from her eye in a straight line?

Solution

The angle of depression is outside the right triangle, between her horizontal and the line of sight. By the alternate angles fact, the angle of elevation from the boat up to her eye is also 7^\circ, and that one is inside the triangle, at the boat. From the boat, the 32 m height is opposite the angle and the horizontal distance d is adjacent to it:

\tan 7^\circ = \frac{32}{d} \quad\Longrightarrow\quad d = \frac{32}{\tan 7^\circ} \approx 260.62 \text{ m}

The line of sight s is the hypotenuse, and the 32 m is opposite the angle, so the sine, again from the given height rather than the rounded distance:

\sin 7^\circ = \frac{32}{s} \quad\Longrightarrow\quad s = \frac{32}{\sin 7^\circ} \approx 262.58 \text{ m}

A small angle of depression means a long, flat triangle, and then the hypotenuse is only a little longer than the horizontal leg, as here.

Example 3

From a grade to an angle

A sign at the top of a hill warns of a 12\% grade. (a) What angle does the road make with the horizontal? (b) A car drives 800 m down the road. How much height does it lose?

Solution

(a) A grade of 12\% is a slope of 0.12, and the slope is the tangent of the angle with the horizontal:

\tan\theta = 0.12 \quad\Longrightarrow\quad \theta = \tan^{-1}(0.12) \approx 6.84^\circ

(b) The 800 m is measured along the road, so it is the hypotenuse, not the run. The height lost is the side opposite \theta, so the sine:

\text{height} = 800\sin\theta \approx 95.32 \text{ m}

with \theta the unrounded angle from part (a). Taking 12\% of 800 would give 96 m, which is too much: the grade is rise over the horizontal run, and the horizontal run here is 800\cos\theta \approx 794.30 m, a little less than 800. As a check, \displaystyle \frac{95.32}{794.30} is 0.12 to two places.

Example 4

Two sightings from the same side

A tower stands on the far bank of a river. From a point P on level ground the angle of elevation of its top is 28^\circ. A surveyor walks 50 m straight toward the tower, to Q, where the angle of elevation is 41^\circ. Both angles are measured at ground level. Find the height of the tower and the distance from Q to its foot.

Solution

Neither the height h nor the distance x from Q to the foot is known, so neither right triangle can be solved on its own. Write one equation for each, as in the figure of the procedure:

h = x\tan 41^\circ \qquad\text{and}\qquad h = (x + 50)\tan 28^\circ

Set the right sides equal, gather the x terms, and divide by \tan 41^\circ - \tan 28^\circ, which is positive:

\begin{aligned} x\tan 41^\circ &= x\tan 28^\circ + 50\tan 28^\circ \\ x &= \frac{50\tan 28^\circ}{\tan 41^\circ - \tan 28^\circ} \\ &\approx 78.75 \text{ m} \end{aligned}

Then, using the unrounded x,

h = x\tan 41^\circ \approx 68.46 \text{ m}

Check with the other triangle. From P the horizontal distance is about 78.75 + 50 = 128.75 m, and \displaystyle \frac{68.46}{128.75} \approx 0.532, which matches \tan 28^\circ \approx 0.532.

Example 5

The object between the observers

Two observers P and Q stand 300 m apart on level ground. A hot-air balloon hangs between them, directly above a point F on the line PQ. From P its angle of elevation is 35^\circ, and from Q it is 52^\circ. Both are measured at ground level. How high is the balloon?

Figure 4Example 5. The balloon is between the observers, so the two ground distances add up to 300 m.

Solution

Call the height h and the distance PF x. The balloon is between the observers, so FQ = 300 - x: this time the two ground distances add up to the distance between the observers. Each observer gives one equation:

h = x\tan 35^\circ \qquad\text{and}\qquad h = (300 - x)\tan 52^\circ

Set them equal and gather the x terms on one side:

\begin{aligned} x\tan 35^\circ + x\tan 52^\circ &= 300\tan 52^\circ \\ x &= \frac{300\tan 52^\circ}{\tan 35^\circ + \tan 52^\circ} \\ &\approx 193.92 \text{ m} \end{aligned}

Then h = x\tan 35^\circ \approx 135.78 m.

The balloon is nearer Q, about 300 - 193.92 = 106.08 m away horizontally, which fits: Q sees it at the steeper angle. The denominator is now a sum, so it can never be 0, and two observers on opposite sides can measure any two angles at all.

Example 6

Two boats seen from a cliff

From the top of a cliff 85 m above the sea, the angles of depression of two boats are 34^\circ and 19^\circ. The boats and the foot of the cliff lie on one straight line, with both boats on the same side. How far apart are the boats?

Solution

Move each angle of depression down to its boat, where it becomes an angle of elevation of the clifftop, inside a right triangle with the 85 m cliff opposite it. The horizontal distance from each boat to the foot of the cliff is adjacent to that angle:

\begin{aligned} \frac{85}{\tan 34^\circ} &\approx 126.02 \text{ m} \\ \frac{85}{\tan 19^\circ} &\approx 246.86 \text{ m} \end{aligned}

The smaller angle belongs to the farther boat. Both distances are measured from the same foot, along the same line, so the boats are

\frac{85}{\tan 19^\circ} - \frac{85}{\tan 34^\circ} \approx 120.84 \text{ m}

apart. This is the same-side picture of Example 4 turned upside down: there the walk was known and the height wanted, here the height is known and the walk is wanted.

Example 7

Looking up and down from one window

From a window 18 m above level ground, the angle of elevation of the top of a taller building across the street is 40^\circ, and the angle of depression of its foot is 32^\circ. How wide is the street, and how tall is the other building?

Figure 5Example 7. The horizontal through the window cuts the view into two right triangles, one above it and one below.

Solution

The horizontal from the window W meets the far building at H, 18 m above the ground, and it splits the view into two right triangles, both with their right angle at H. They share the horizontal side WH, which is as long as the street is wide; call it d.

Below the horizontal. Triangle WHG has the angle of depression 32^\circ at W, between WH and the line of sight to the foot G. It is inside this triangle, since WH is one of its sides. The side HG is 18 m, opposite the angle, and d is adjacent:

\tan 32^\circ = \frac{18}{d} \quad\Longrightarrow\quad d = \frac{18}{\tan 32^\circ} \approx 28.81 \text{ m}

Above the horizontal. Triangle WHT has the 40^\circ angle at W, the same d adjacent to it, and the part of the building above the window, HT, opposite it: HT = d\tan 40^\circ \approx 24.17 m.

The building is the part above the window plus the part below it:

GT = 18 + d\tan 40^\circ \approx 42.17 \text{ m}

The angle of depression did not need moving this time: the horizontal through the window is a side of the triangle it belongs to.

Example 8

How fast is the plane?

A plane flies level at a height of 1200 m on a straight course that will take it directly over an observer on the ground. At one moment its angle of elevation is 30^\circ, and 15 seconds later it is 60^\circ. How fast is the plane flying?

Solution

At each moment the plane, the point on the ground below it and the observer make a right triangle, with the height 1200 m opposite the angle of elevation and the horizontal distance adjacent to it. These angles have exact tangents, \displaystyle \tan 30^\circ = \frac{1}{\sqrt{3}} and \displaystyle \tan 60^\circ = \sqrt{3}, so the distances are exact:

\begin{aligned} \frac{1200}{\tan 30^\circ} &= 1200\sqrt{3} \\ \frac{1200}{\tan 60^\circ} &= \frac{1200}{\sqrt{3}} = 400\sqrt{3} \end{aligned}

The plane is heading straight for the point above the observer, so the distance it flew in the 15 seconds is the difference:

1200\sqrt{3} - 400\sqrt{3} = 800\sqrt{3} \approx 1385.64 \text{ m}

Speed is distance divided by time:

\frac{800\sqrt{3}}{15} = \frac{160\sqrt{3}}{3} \approx 92.38 \text{ m per second}

Multiplying by 3.6 turns meters per second into kilometers per hour, giving \displaystyle 192\sqrt{3} \approx 332.55 km per hour. The two sightings are the same as in Example 4, with the roles swapped: the height is known, and the unknown is how far the thing moved between them.

Boundaries

The ground must be level, or the levels known. The right angle at the foot of the object comes from a horizontal meeting a vertical. If the ground between the observers slopes, the triangle of Example 4 is no longer right-angled at the foot, and the problem needs the laws of sines and cosines instead.

The two sightings must be in line with the foot. The method of Example 4 assumes P, Q and F lie on one straight line, so that the two horizontal distances differ by exactly the walk. If the second point is off to one side, the three points form a triangle on the ground, and locating the object becomes Applications of oblique triangles.

The two angles must differ. On the same side, equal angles would make the denominator \tan\beta - \tan\alpha zero. That happens only for something so far away that walking toward it changes nothing measurable, such as the sun, whose rays reach every nearby point at the same angle. When the two angles are close, a small error in either one makes a large error in h, because the answer divides by their small difference.

Angles lie strictly between 0^\circ and 90^\circ. At 0^\circ the line of sight is horizontal and at 90^\circ it is vertical, and either way there is no triangle.

Measure from the eye. The vertical leg starts at the level of the eye. When the eye is above the ground, add its height, as in Example 1.

Grade is not a share of a right angle. A 20\% grade is about 11.31^\circ, not 18^\circ, because the tangent does not grow in step with the angle. A 100\% grade is 45^\circ, and no grade at all is vertical, since \tan 90^\circ is not defined. A downhill grade is sometimes written with a minus sign; the angle it makes with the horizontal is then an angle of depression, and the same tangent applies to its size.

The earth is treated as flat. Horizontals at two places are taken to be parallel, which is very nearly true over a few kilometers and stops being true over long distances, where the curvature of the earth matters.

Where this goes next

Example 4 can also be solved with one non-right triangle instead of two right ones. In triangle PQT the angle at T is \beta - \alpha, and the law of sines gives QT directly, then h = QT\sin\beta. Both routes give the same number, and putting them side by side shows that

\begin{aligned} &\frac{d\tan\alpha\tan\beta}{\tan\beta - \tan\alpha} \\ &\qquad = \frac{d\sin\alpha\sin\beta}{\sin(\beta - \alpha)} \end{aligned}

Working in one non-right triangle is the route that carries on when the sightings are not in line with the foot, which is the start of Applications of oblique triangles: locating a point from the two ends of a measured baseline. Turn the same triangles on their side, onto a map, and they become bearings.

Before trigonometry, heights were found with shadows, comparing the shadow of a tall object with the shadow of a stick of known height. That works because the sun’s angle of elevation is the same for both, so the two triangles are similar, and it is the subject of Measuring what you cannot reach.

Example 8 measured an average speed between two sightings. Asking how fast the angle of elevation changes at a single instant, as the plane approaches, is a question about rates of change, and it is answered by related rates.

Exercises

Give lengths to two decimal places, and angles to two decimal places of a degree.

One sighting

From a point on level ground 25 m from the foot of a tree, the angle of elevation of its top, measured at ground level, is 38^\circ. How tall is the tree?
A person whose eye is 1.5 m above level ground stands 60 m from a flagpole and sees its top at an angle of elevation of 21^\circ. How tall is the flagpole?
From the top of a building 45 m tall, the angle of depression of a parked car is 24^\circ. How far is the car from the foot of the building?
A drone hovers 120 m above a level field. From a point on the field, at ground level, its angle of elevation is 55^\circ. Find the horizontal distance to the point below the drone, and the straight-line distance to the drone.
From a lookout, a campsite in the valley is seen at an angle of depression of 15^\circ. The campsite is 900 m away horizontally. How far below the lookout is it?
From the top of a cliff 30 m above the sea, a buoy is seen at an angle of depression of 12^\circ. At what angle of elevation is the clifftop seen from the buoy, and how far is the buoy from the foot of the cliff?

Grade and slope

A road rises 6 m for every 100 m of horizontal distance. What is its grade, and what angle does it make with the horizontal?
A ski slope makes an angle of 18^\circ with the horizontal. What is its grade?
A railway climbs at a grade of 2.5\%. How much does it climb over 4 km of horizontal distance, and what angle does it make with the horizontal?
A street has a grade of 20\%. Find the angle it makes with the horizontal, and say whether it is more or less than 18^\circ, which is 20\% of a right angle.
Find the grade of a slope that makes 45^\circ with the horizontal, and of one that makes 60^\circ.

Two sightings

From a point on level ground the angle of elevation of the top of a tower is 30^\circ. After walking 40 m straight toward the tower, it is 45^\circ. Both are measured at ground level. Find the height of the tower exactly, and to two decimal places.
The summit of a hill is seen across a level plain at an angle of elevation of 14^\circ. After walking 400 m across the plain straight toward the point below the summit, it is seen at 21^\circ. How high is the summit above the plain, and how far is the second point from the point below the summit? Both angles are measured at ground level.
Two surveyors stand 250 m apart on level ground, on opposite sides of a radio mast and in line with its foot. Their angles of elevation of the top of the mast are 48^\circ and 39^\circ, measured at ground level. How tall is the mast?
From the top of a tower 60 m tall, the angles of depression of two cars on a straight road leading directly away from the foot of the tower are 42^\circ and 27^\circ. How far apart are the cars?
From a window 12 m above level ground, the angle of elevation of the top of the building across the street is 50^\circ and the angle of depression of its foot is 35^\circ. How wide is the street, and how tall is the building?
A building 20 m tall and a taller tower stand on level ground. From the foot of the building the angle of elevation of the top of the tower is 60^\circ, and from the top of the building it is 45^\circ. Find the distance between them and the height of the tower, exactly and to two decimal places.
A surveyor whose eye is 1.7 m above level ground sees the top of a chimney at an angle of elevation of 25^\circ. She walks 30 m straight toward it, and the angle is now 40^\circ. How tall is the chimney?

Things that move

A balloon rises straight up from a point on level ground 200 m from an observer. Measured at ground level, its angle of elevation goes from 20^\circ to 35^\circ. How far has it risen?
From the top of a cliff 70 m above the sea, a ship sailing straight toward the foot of the cliff is seen at an angle of depression of 9^\circ, and 2 minutes later at 16^\circ. How far did the ship sail, and what was its speed in meters per minute?

Using letters

Two observers stand a distance d apart on level ground, on opposite sides of an object and in line with its foot. Their angles of elevation of its top are \alpha and \beta, measured at ground level. Show that its height is \displaystyle h = \frac{d\tan\alpha\tan\beta}{\tan\alpha + \tan\beta}.

Reasoning

Explain why the angle of elevation of B from A is always equal to the angle of depression of A from B.
A tower’s foot cannot be reached. Explain why one angle of elevation of its top, from one point, cannot give its height, and what a second sighting from a known distance adds.
Two sightings are taken on the same side of a tower, in line with its foot. Explain why the nearer point always sees the larger angle, and what it would mean if the two angles came out equal.
Explain why a road with a grade of 100\% is not vertical, and why no grade describes a vertical wall.

Answers

19.53 m. The height is opposite the angle and the 25 m is adjacent: 25\tan 38^\circ.
24.53 m. Above the eye the pole rises 60\tan 21^\circ \approx 23.03 m; add the 1.5 m of the eye.
101.07 m. The angle of elevation from the car is also 24^\circ, with the 45 m opposite it: \displaystyle \frac{45}{\tan 24^\circ}.
Horizontal 84.02 m, straight line 146.49 m. The 120 m is opposite the angle: \displaystyle \frac{120}{\tan 55^\circ} and \displaystyle \frac{120}{\sin 55^\circ}.
241.15 m. The angle of elevation from the campsite is 15^\circ, the 900 m is adjacent and the drop is opposite: 900\tan 15^\circ.
12^\circ, and 141.14 m. Elevation from the buoy equals depression from the clifftop (alternate angles), and the distance is \displaystyle \frac{30}{\tan 12^\circ}.
Grade 6\%, angle 3.43^\circ. The slope is \displaystyle \frac{6}{100} = 0.06, and \tan^{-1}(0.06) \approx 3.43^\circ.
32.49\%. The slope is \tan 18^\circ \approx 0.3249.
100 m, at 1.43^\circ. The rise is 0.025 \times 4000 = 100 m, since the grade is rise over horizontal run, and \tan^{-1}(0.025) \approx 1.43^\circ.
11.31^\circ, less than 18^\circ. \tan^{-1}(0.2) \approx 11.31^\circ. An 18^\circ slope has grade 100\tan 18^\circ \approx 32.49\%, as in Exercise 8: grade and angle do not grow in proportion.
100\% and 173.21\%. \tan 45^\circ = 1 and \displaystyle \tan 60^\circ = \sqrt{3} \approx 1.7321.
\displaystyle 20(\sqrt{3} + 1) \approx 54.64 m. With x the distance from the second point to the foot, h = x\tan 45^\circ = x and \displaystyle h = (x + 40)\tan 30^\circ = \frac{x + 40}{\sqrt{3}}. So \displaystyle x\sqrt{3} = x + 40, \displaystyle x = \frac{40}{\sqrt{3} - 1} = 20(\sqrt{3} + 1), and h = x.
Height 284.56 m, distance 741.30 m. With x the distance from the second point, x\tan 21^\circ = (x + 400)\tan 14^\circ, so \displaystyle x = \frac{400\tan 14^\circ}{\tan 21^\circ - \tan 14^\circ} and h = x\tan 21^\circ.
117.08 m. With x the distance from the first surveyor to the foot, x\tan 48^\circ = (250 - x)\tan 39^\circ, so \displaystyle x = \frac{250\tan 39^\circ}{\tan 48^\circ + \tan 39^\circ} and h = x\tan 48^\circ.
51.12 m. The distances from the foot are \displaystyle \frac{60}{\tan 27^\circ} \approx 117.76 m and \displaystyle \frac{60}{\tan 42^\circ} \approx 66.64 m; subtract.
Street 17.14 m, building 32.42 m. Below the window, \displaystyle d = \frac{12}{\tan 35^\circ}. Above it, the building rises d\tan 50^\circ more, so its height is 12 + d\tan 50^\circ.
Distance \displaystyle 10(\sqrt{3} + 1) \approx 27.32 m, tower \displaystyle 30 + 10\sqrt{3} \approx 47.32 m. From the foot, \displaystyle H = d\tan 60^\circ = d\sqrt{3}. From the top, the horizontal is 20 m up, so H - 20 = d\tan 45^\circ = d. Then \displaystyle d\sqrt{3} - d = 20 gives \displaystyle d = \frac{20}{\sqrt{3} - 1}, and \displaystyle H = d\sqrt{3}.
33.19 m. Above the eye, \displaystyle h = \frac{30\tan 25^\circ\tan 40^\circ}{\tan 40^\circ - \tan 25^\circ} \approx 31.49 m; add the 1.7 m of the eye.
67.25 m. The heights are 200\tan 20^\circ and 200\tan 35^\circ, both opposite the angle with the same 200 m adjacent; subtract: 200(\tan 35^\circ - \tan 20^\circ).
197.84 m, at 98.92 m per minute. The distances to the foot are \displaystyle \frac{70}{\tan 9^\circ} \approx 441.96 m and \displaystyle \frac{70}{\tan 16^\circ} \approx 244.12 m; the ship sailed the difference in 2 minutes.
Let x be the ground distance from the first observer to the foot, so the second is d - x from it. Then h = x\tan\alpha and h = (d - x)\tan\beta. So \displaystyle x = \frac{h}{\tan\alpha} and \displaystyle d - x = \frac{h}{\tan\beta}; adding, \displaystyle d = h\left(\frac{1}{\tan\alpha} + \frac{1}{\tan\beta}\right) = h \cdot \frac{\tan\alpha + \tan\beta}{\tan\alpha\tan\beta}, which rearranges to the formula.
The horizontals through A and B are both perpendicular to the vertical direction, so they are parallel. The line of sight AB is a transversal of the two, and the angle of elevation at A and the angle of depression at B lie between the parallels on opposite sides of it. They are alternate interior angles, so they are equal.
One angle fixes only the ratio of the height to the horizontal distance, \tan\alpha. Every right triangle with that angle is similar to every other, so a small near tower and a large far one give the same angle, and without the distance to the foot nothing fixes the size. A second sighting from a point a known distance d nearer gives a second ratio for the same height. Two equations then hold the two unknowns, the height and the distance, and the known d ties them to a definite size.
From the nearer point the same height stands over a shorter horizontal distance, so the ratio \displaystyle \frac{h}{x} is larger than \displaystyle \frac{h}{x + d}. The tangent of an acute angle grows as the angle grows, so the nearer angle is larger. Equal angles would need \displaystyle \frac{h}{x} = \frac{h}{x + d} with d > 0, which no finite distance allows: the formula would divide by 0. It means the object is so far away that walking d changes the angle by less than can be measured.
A grade of 100\% means the rise equals the run, so \tan\theta = 1 and \theta = 45^\circ. Grade is rise over horizontal run, and a vertical wall has no horizontal run at all: rise over a run of 0 is not a number. Every finite grade g gives the angle \tan^{-1}(g), which is less than 90^\circ, however large g is.