Chapter 22 — Applications of Trigonometry

Long before there were tape measures for tall buildings or instruments for aircraft, people needed to know heights and distances that no one could pace out: the height of a temple tower, the width of a river without crossing it, the distance to a ship at sea, the height of a cloud or a mountain peak. The method that solved all these problems is the same one used by surveyors today. Stand at a known distance from the object, measure the angle at which you must look up (or down) to see the point in question, and the height follows from a trigonometric ratio.

Why does this work? The observer's eye, the foot of the object and the top of the object form a triangle. If the ground is level and the object is vertical, the triangle has a right angle at the foot. In a right triangle, once one side and one acute angle are known, every other side is determined: tan θ = opposite/adjacent gives the vertical side from the horizontal side, and sin θ or cos θ gives the slanted line of sight. The previous chapter showed that these ratios depend only on the angle, so a value looked up in the table applies to a real tower just as it applies to a diagram in a notebook.

Three lines. The line of sight is the straight line from the observer's eye to the object being viewed. The horizontal line is the line through the observer's eye parallel to the level ground. The vertical line is the object itself, or the perpendicular dropped from the object to the ground. These three lines form the right triangle of every problem.

An example to fix the idea. A student stands 20 m from the foot of a flagpole and finds that the line of sight to its top makes an angle of 45° with the horizontal. Then tan 45° = height/20, so the height is 20 × 1 = 20 m. If the angle had been 60°, the height would be 20 tan 60° = 20√3 ≈ 34.64 m. If the angle had been 30°, the height would be 20/√3 ≈ 11.55 m. A larger angle, for the same distance, means a taller object.

A note on the observer's height. If the angle is measured from the eye of a person 1.5 m tall, the triangle gives the height of the object above the eye, and 1.5 m must be added to obtain the height above the ground. Problems say 'from a point on the ground' when the observer's height is to be ignored, and give the observer's height when it is to be included.

Where it is used. Surveyors use this method to map land and to set out roads and railways. Navigators use it to find the distance of a lighthouse from its known height and its angle of elevation. Astronomers used it to find the distance of the moon. Architects and engineers use it to check the height of structures and the slope of ramps and roofs.

Two words describe the angle between the line of sight and the horizontal, depending on whether the observer looks up or down.

Angle of elevation. When the object is above the horizontal line through the observer's eye, the observer must raise the line of sight to see it. The angle between the line of sight and the horizontal is the angle of elevation of the object. Looking at the top of a tower, a kite, a bird or an aeroplane involves an angle of elevation.

Angle of depression. When the object is below the horizontal line through the observer's eye, the observer lowers the line of sight, and the angle between the line of sight and the horizontal is the angle of depression of the object. Looking from the top of a tower at a car on the road, from a cliff at a boat, or from an aeroplane at a village involves an angle of depression.

The key fact about depression. The horizontal line through the eye of the observer on the tower is parallel to the ground. The line of sight is a transversal cutting these two parallel lines. Therefore the angle of depression at the top equals the angle of elevation of the top as seen from the object on the ground, since they are alternate angles. In the figure, if the angle of depression of a boat from the top of a cliff is 30°, then the angle of elevation of the cliff top from the boat is also 30°. This lets every depression problem be solved in the right triangle at the base, where the angle sits at the ground point.

Measuring the angle. A simple instrument called a clinometer can be made from a protractor, a straw and a weighted string: sight the object along the straw and read where the string crosses the scale. Surveyors use a theodolite, a telescope mounted on graduated circles, which measures angles to a fraction of a degree. In the examination the angles are always standard ones, 30°, 45° or 60°, so that the answers involve only 1, √3 and 1/√3.

Worked example 1. From the top of a lighthouse 100 m high, the angle of depression of a boat is 30°. How far is the boat from the foot of the lighthouse? The angle of elevation of the top from the boat is 30°. tan 30° = 100/d, so d = 100/(1/√3) = 100√3 ≈ 173.2 m.

Worked example 2. A kite is flying at a height of 60 m and its string makes an angle of 60° with the ground. Find the length of the string, assuming it is straight. sin 60° = 60/L, so L = 60/(√3/2) = 120/√3 = 40√3 ≈ 69.28 m. The angle here is the angle of elevation of the kite from the person holding the string.

Worked example 3. A person on a bridge 12 m above a river sees a boat at an angle of depression of 45°. The horizontal distance to the boat is 12/tan 45° = 12 m.

Students should always ask: where is the observer, is the object above or below, and therefore is the angle one of elevation or depression?

Every problem in this chapter is solved by first turning the words into a figure. A correct figure makes the solution almost automatic; an incorrect figure makes it impossible. These are the rules.

Rule 1: draw the ground as a horizontal line. The observer stands on it, and the foot of the object stands on it. Unless the problem says otherwise, the ground is level.

Rule 2: draw the object as a vertical line. Towers, buildings, trees, poles, cliffs and the height of a kite or aeroplane are vertical segments, perpendicular to the ground. Mark the right angle at the foot.

Rule 3: draw the line of sight. Join the observer's eye (or the ground point, if the observer's height is ignored) to the top of the object. This is the hypotenuse of the right triangle.

Rule 4: put the angle at the observer. The angle of elevation is at the ground point between the ground and the line of sight. The angle of depression is at the top, between the horizontal line drawn through the top and the line of sight downward; then transfer it, by alternate angles, to the ground point.

Rule 5: label everything. Write the known distance on the horizontal side, the known or unknown height on the vertical side, and the angle in degrees. Give letters to the vertices: A for the top of the tower, B for its foot, C for the observer, is a common choice.

Rule 6: two observations give two triangles sharing the vertical side. Draw both lines of sight from the two ground points, mark both angles, and mark the distance between the two ground points. The nearer point has the larger angle.

Rule 7: an object on an object gives two triangles sharing the horizontal side. A flagstaff on a building gives two lines of sight from one ground point, to the bottom and to the top of the flagstaff, with two angles at the same vertex.

Worked example. 'A tower stands vertically on the ground. From a point on the ground 15 m away from the foot of the tower, the angle of elevation of the top is 45°. Find the height.' Draw the ground line, the tower AB with B on the ground, the point C with BC = 15 m, the line CA and the angle ACB = 45°. In the right triangle ABC, tan 45° = AB/BC, so AB = 15 tan 45° = 15 m.

Worked example. 'From the top of a tower 50 m high, the angle of depression of a car is 30°.' Draw the tower AB, the horizontal AX through the top A, the car at C on the ground, the line of sight AC, and the angle XAC = 30°. Since AX is parallel to BC, ∠ACB = 30° as well. In triangle ABC, tan 30° = 50/BC, BC = 50√3 ≈ 86.6 m.

A figure drawn neatly with a ruler, even if not to scale, earns marks on its own and prevents the two most common errors: putting the angle at the wrong vertex, and using the line of sight where the horizontal distance is meant.

Once the right triangle is drawn, the solution needs one equation, and the equation comes from the ratio that connects the side that is known with the side that is wanted, for the angle that is given. The rule of choice:

• Known horizontal distance, wanted height (or the reverse): use tan θ = height/distance.
• Known line of sight (ladder, string, rope, slope), wanted height: use sin θ = height/line of sight.
• Known line of sight, wanted horizontal distance: use cos θ = distance/line of sight.
• Known height, wanted line of sight: use sin θ; known distance, wanted line of sight: use cos θ.

The hypotenuse is involved whenever the problem mentions a ladder, a string, a rope, a wire, a slide, a slope or the distance 'from the observer to the top'. It is not involved when the problem gives the distance 'from the foot' or 'along the ground'. This distinction is the single most important reading skill of the chapter.

Worked example 1. A ladder 10 m long rests against a wall making 30° with the ground. Height reached: sin 30° = h/10, h = 5 m. Distance of the foot from the wall: cos 30° = d/10, d = 5√3 ≈ 8.66 m.

Worked example 2. A tower is 30 m high; from a point on the ground the angle of elevation of its top is 60°. Distance of the point from the foot: tan 60° = 30/d, d = 30/√3 = 10√3 ≈ 17.32 m. Distance of the point from the top: sin 60° = 30/L, L = 30/(√3/2) = 20√3 ≈ 34.64 m.

Worked example 3. A rope is tied from the top of a 20 m pole to a peg on the ground and makes 45° with the ground. Length of the rope: sin 45° = 20/L, L = 20√2 ≈ 28.28 m.

Worked example 4. A slide in a park is 8 m long and inclined at 30° to the ground. Its top is at height 8 sin 30° = 4 m and its foot is 8 cos 30° = 4√3 ≈ 6.93 m from the point below its top.

Worked example 5. An electric pole is 10 m high. A steel wire tied to its top is fixed to the ground at a point making 45° with the horizontal. Length of the wire: 10/sin 45° = 10√2 ≈ 14.14 m.

Handling surds. The standard values bring √2 and √3 into the answers. Keep them exact through the working (10√3, 20√2) and convert to decimals only in the last line, using √2 ≈ 1.414 and √3 ≈ 1.732 when the question asks for a decimal. Rationalise denominators: 30/√3 = 10√3.

The simplest problems give the distance from the foot of a vertical object and the angle of elevation of its top, and ask for the height; or give the height and the angle and ask for the distance. One right triangle and one use of tan θ suffice.

Worked example 1. The angle of elevation of the top of a tower from a point 30 m from its foot is 30°. Find the height. tan 30° = h/30, h = 30/√3 = 10√3 ≈ 17.32 m.

Worked example 2. A tree breaks due to a storm, and the broken part bends so that the top touches the ground making an angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Let the standing part be BC = h and the broken part AC = L, with the top touching at A, AB = 8 m. tan 30° = h/8, so h = 8/√3; cos 30° = 8/L, so L = 8/(√3/2) = 16/√3. Height of the tree = h + L = 8/√3 + 16/√3 = 24/√3 = 8√3 ≈ 13.86 m.

Worked example 3. A person 1.5 m tall stands 28.5 m from a chimney and finds the angle of elevation of the top to be 45°. Height of the chimney = 28.5 tan 45° + 1.5 = 28.5 + 1.5 = 30 m.

Worked example 4. The shadow of a tower is 40 m long when the sun's altitude (angle of elevation) is 30°. Height = 40 tan 30° = 40/√3 ≈ 23.09 m. When the altitude is 45° the shadow equals the height; when it is 60° the shadow is h/√3.

Worked example 5. A contractor plans two slides in a park, both with the top at a height of 3 m: one inclined at 60° for older children and one inclined at 30° for younger children. Lengths: 3/sin 60° = 6/√3 = 2√3 ≈ 3.46 m for the steeper slide and 3/sin 30° = 6 m for the gentler one. The gentler slope needs the longer slide.

Worked example 6. The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. Distance between them: tan 60° = 50/d, d = 50/√3. Building height: tan 30° = h/d, h = (50/√3)(1/√3) = 50/3 ≈ 16.67 m.

Worked example 7. A bridge across a river makes an angle of 45° with the bank. If the bridge is 150 m long, the width of the river is 150 sin 45° = 150/√2 = 75√2 ≈ 106.07 m.

Each of these is a two- or four-mark question; the four-mark ones are those with an extra step, such as the broken tree, the observer's height, or two towers looking at each other.

When the observer is on top of a tower, a cliff, a lighthouse or a building and looks down, the angle given is an angle of depression. The method is to draw the horizontal line through the observer's eye, mark the angle of depression between it and the line of sight, and then transfer the angle to the ground point by alternate angles. After that the problem is exactly a tan θ problem in the triangle at the base.

Worked example 1. From the top of a 75 m high lighthouse, the angles of depression of two ships on the same side are 30° and 45°. Find the distance between the ships. Let the lighthouse be AB with A at the top. For the nearer ship C, ∠ACB = 45°, so BC = 75/tan 45° = 75 m. For the farther ship D, ∠ADB = 30°, so BD = 75/tan 30° = 75√3 ≈ 129.9 m. Distance CD = 75√3 − 75 = 75(√3 − 1) ≈ 54.9 m.

Worked example 2. From the top of a building 60 m high, the angles of depression of the top and the bottom of a tower are 30° and 60°. Find the height of the tower. Let the building be AB (A top) and the tower CD (C top), with BD the ground distance. From the bottom D: tan 60° = 60/BD, BD = 60/√3 = 20√3. From the top C, draw the horizontal CE to meet AB at E; then CE = BD = 20√3 and tan 30° = AE/CE, AE = 20√3/√3 = 20 m. The tower's height CD = EB = AB − AE = 60 − 20 = 40 m.

Worked example 3. A man on the deck of a ship, 10 m above the water, observes the angle of elevation of the top of a cliff as 45° and the angle of depression of its base as 30°. Find the height of the cliff and its distance from the ship. Let the eye be at E, 10 m above the water, and the cliff be CB with base B. Depression of B is 30°: tan 30° = 10/d, d = 10√3 ≈ 17.32 m. Elevation of the top C: tan 45° = (h − 10)/d, so h − 10 = 10√3, h = 10 + 10√3 ≈ 27.32 m.

Worked example 4. From an aeroplane flying at a height of 1200 m, the angles of depression of two boats on the same side are 60° and 30°. Distance between the boats: 1200 cot 30° − 1200 cot 60° = 1200√3 − 1200/√3 = 1200(√3 − 1/√3) = 1200 × 2/√3 = 800√3 ≈ 1385.6 m.

Worked example 5. The angle of depression of a car from the top of a 50 m tower is 60°. How far is the car from the tower? 50/tan 60° = 50/√3 ≈ 28.87 m. The distance from the top of the tower to the car (the line of sight) is 50/sin 60° = 100/√3 ≈ 57.74 m.

Common error. Marking the angle of depression at the foot of the tower between the tower and the line of sight. The angle of depression is measured from the horizontal, not from the vertical; the angle between the tower and the line of sight is 90° minus the depression.

When the distance from the object is not given, one observation is not enough: the height and the distance are two unknowns. A second observation from a different point on the same line gives a second equation, and the two are solved together. Both triangles share the vertical side, which is the height h; their horizontal sides differ by the known distance between the two observation points.

The setup. Tower AB of height h. Observation from C, at distance x from B, gives angle α (the larger angle, since C is nearer). Observation from D, farther away by the known distance d, gives angle β. Then tan α = h/x and tan β = h/(x + d). From the first, x = h cot α; substitute in the second: h = (h cot α + d) tan β, which gives h(1 − cot α tan β) = d tan β, or more neatly h = d/(cot β − cot α).

Worked example 1. The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 20 m towards the tower the angle becomes 60°. Find the height of the tower. Let the nearer point be at distance x. tan 60° = h/x gives h = x√3. tan 30° = h/(x + 20) gives h = (x + 20)/√3. Equating: x√3 = (x + 20)/√3, 3x = x + 20, x = 10 m, h = 10√3 ≈ 17.32 m. By the formula, h = 20/(cot 30° − cot 60°) = 20/(√3 − 1/√3) = 20/(2/√3) = 10√3.

Worked example 2. The angles of elevation of the top of a tower from two points at distances 4 m and 9 m from its foot, on the same side, are complementary. Find the height. Let the angles be θ and 90° − θ. tan θ = h/4 and tan(90° − θ) = cot θ = h/9. Multiply: tan θ cot θ = h²/36 = 1, so h² = 36 and h = 6 m. In general h = √(ab) for complementary angles at distances a and b.

Worked example 3. A 1.5 m tall boy is some distance from a 30 m building. The angle of elevation of the top from his eyes is 30°; he walks towards the building and the angle becomes 60°. How far did he walk? The height above his eyes is 30 − 1.5 = 28.5 m. First distance = 28.5 cot 30° = 28.5√3; second = 28.5 cot 60° = 28.5/√3. Distance walked = 28.5(√3 − 1/√3) = 28.5 × 2/√3 = 57/√3 = 19√3 ≈ 32.9 m.

Worked example 4. From a point P the angle of elevation of the top of a tower is 45°; from a point Q, 50 m farther from the tower on the same line, it is 30°. Height h = 50/(cot 30° − cot 45°) = 50/(√3 − 1) = 50(√3 + 1)/2 = 25(√3 + 1) ≈ 68.3 m.

Worked example 5. The angle of elevation of a cloud from a point 60 m above a lake is 30°, and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud above the lake. Let the cloud be at height H above the lake and the horizontal distance be d. The cloud is H − 60 above the eye. The reflection is at depth H below the surface, so it is H + 60 below the eye. tan 30° = (H − 60)/d and tan 60° = (H + 60)/d. Dividing: tan 60°/tan 30° = 3 = (H + 60)/(H − 60), so 3H − 180 = H + 60, H = 120 m.

In all these problems write the two tan equations first, then eliminate the distance. Keep surds exact until the end.

When two observers stand on opposite sides of a tower, or a tower stands between two points on a road, the two right triangles share the vertical side and their horizontal sides add up to the distance between the observers.

The setup. Tower AB of height h, points C and D on opposite sides with angles α at C and β at D. Then BC = h cot α and BD = h cot β, so the distance CD = h(cot α + cot β). If CD is known, h = CD/(cot α + cot β).

Worked example 1. Two points A and B are on opposite sides of a tower on a straight road, 100 m apart. The angles of elevation of the top of the tower from A and B are 30° and 60°. Find the height of the tower and its distances from A and B. h = 100/(cot 30° + cot 60°) = 100/(√3 + 1/√3) = 100/(4/√3) = 25√3 ≈ 43.3 m. Distance from A = h cot 30° = 25√3 × √3 = 75 m; from B = h cot 60° = 25 m. Check: 75 + 25 = 100.

Worked example 2. A 100 m high tower stands between two points on the ground such that the angles of elevation of its top from them are 45° and 30°. Distance between the points = 100(cot 45° + cot 30°) = 100(1 + √3) ≈ 273.2 m.

Worked example 3. Two towers of heights 20 m and 30 m stand on opposite sides of a road, and from a point on the road between them the angles of elevation of their tops are both 45°. The distances from the point are 20 cot 45° = 20 m and 30 cot 45° = 30 m, so the road is 50 m wide at that place.

Worked example 4. Two poles of equal height stand on opposite sides of a road 80 m wide. From a point between them on the road the angles of elevation of their tops are 60° and 30°. Find the height of the poles and the position of the point. Let the point be x from the first pole. h = x tan 60° = x√3 and h = (80 − x) tan 30° = (80 − x)/√3. So 3x = 80 − x, x = 20 m, h = 20√3 ≈ 34.64 m. The point is 20 m from one pole and 60 m from the other.

Worked example 5. From the top of a 7 m building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. The horizontal distance d satisfies tan 45° = 7/d, so d = 7 m. The part of the tower above the building's top is d tan 60° = 7√3. Height of tower = 7 + 7√3 = 7(1 + √3) ≈ 19.12 m. (Here the two triangles share the horizontal side rather than the vertical, but the method is the same: two tan equations and one common length.)

Worked example 6. The angles of elevation of the top of a hill from two points on opposite sides, on level ground and in line with the hill, are 60° and 30°. If the points are 400 m apart, the hill is 400/(√3 + 1/√3) = 100√3 ≈ 173.2 m high.

Draw the tower in the middle and the two points on either side; label both angles and the total distance, and the equations write themselves.

A flagstaff on a building, a statue on a pedestal, a water tank on a tower, a tower on a hill: in each case one vertical object stands on another, and the observer at a single ground point sees the bottom and the top of the upper object at two different angles. The two right triangles share the horizontal side, and the difference of their vertical sides is the height of the upper object.

The setup. Lower object of height a, upper object of height b, observer at distance d. Elevation of the junction: tan α = a/d. Elevation of the top: tan β = (a + b)/d. Hence b = d(tan β − tan α), and if a is known, d = a cot α.

Worked example 1. A flagstaff stands on a building 20 m high. From a point on the ground the angles of elevation of the bottom and the top of the flagstaff are 45° and 60°. Find the height of the flagstaff. d = 20 cot 45° = 20 m. Total height = 20 tan 60° = 20√3. Flagstaff = 20√3 − 20 = 20(√3 − 1) ≈ 14.64 m.

Worked example 2. A statue 1.6 m tall stands on top of a pedestal. From a point on the ground the angle of elevation of the top of the statue is 60° and of the top of the pedestal is 45°. Find the height of the pedestal. Let the pedestal be h and the distance d. tan 45° = h/d gives d = h. tan 60° = (h + 1.6)/d = (h + 1.6)/h, so h√3 = h + 1.6, h(√3 − 1) = 1.6, h = 1.6/(√3 − 1) = 1.6(√3 + 1)/2 = 0.8(√3 + 1) ≈ 2.19 m.

Worked example 3. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 5 m. From a point on the plane the angles of elevation of the bottom and top of the flagstaff are 30° and 60°. Find the height of the tower. d = h cot 30° = h√3; also (h + 5) = d tan 60° = h√3 × √3 = 3h. So 2h = 5, h = 2.5 m.

Worked example 4. A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite, the angle of elevation of the top is 60°. From a point 20 m farther from the tower on the same line the angle is 30°. Find the height of the tower and the width of the canal. This is the two-observation problem: h = 20/(cot 30° − cot 60°) = 10√3 ≈ 17.32 m, and the width = h cot 60° = 10 m.

Worked example 5. A tower stands on the summit of a hill. From a point on level ground 100 m from the point directly below the summit, the angles of elevation of the summit and of the top of the tower are 30° and 45°. Find the heights of the hill and the tower. Let the hill be a and the tower b. tan 30° = a/100 gives a = 100/√3 ≈ 57.7 m. tan 45° = (a + b)/100 gives a + b = 100 m, so b = 100 − 57.7 ≈ 42.3 m. The hill is about 57.7 m high and the tower about 42.3 m.

Worked example 6. A pole on top of a tower 50 m high subtends an angle of 15° at a point 50 m from the foot: elevation of the tower's top is 45°, so the pole's top is at 60°; pole = 50 tan 60° − 50 = 50(√3 − 1) ≈ 36.6 m.

Draw the lower object, then the upper object on top of it as a continuation of the same vertical line, and draw two lines of sight from the one observation point.

When an object moves along the ground or through the air, its angle of elevation or depression changes with time. Two observations at known times give two distances, and the difference divided by the time gives the speed. These problems are two-triangle problems with a rate calculation at the end.

Worked example 1. A straight highway leads to the foot of a tower 50 m high. From the top of the tower the angle of depression of a car is 30°; six seconds later it is 60°. Find the time the car takes to reach the foot after the second observation, assuming uniform speed. Distances from the foot: 50 cot 30° = 50√3 and 50 cot 60° = 50/√3. Distance covered in 6 s = 50√3 − 50/√3 = 50(3 − 1)/√3 = 100/√3. Remaining distance = 50/√3, which is half of that, so the car takes 3 more seconds.

Worked example 2. A man on the top of a lighthouse 150 m high observes a boat approaching; its angle of depression changes from 30° to 60° in 2 minutes. Find the speed of the boat. Distances: 150√3 and 150/√3 = 50√3. Distance covered = 100√3 ≈ 173.2 m in 2 minutes, so speed = 86.6 m per minute = 5.196 km per hour.

Worked example 3. The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds the angle is 30°. If the plane flies horizontally at a height of 1500√3 m, find its speed. Horizontal distances: 1500√3 cot 60° = 1500 m and 1500√3 cot 30° = 4500 m. Distance flown = 3000 m in 15 s, speed = 200 m/s = 720 km/h.

Worked example 4. A boat is 100 m from the foot of a cliff, and the angle of elevation of the cliff top from the boat is 30°. The boat then moves 50 m closer. Cliff height = 100 tan 30° = 100/√3 ≈ 57.7 m. At the new distance of 50 m, tan θ = (100/√3)/50 = 2/√3 ≈ 1.155, so θ ≈ 49°, which is not a standard angle. The examination avoids this by giving the new angle and asking for the distance moved; the example shows that not every distance produces a standard angle, and that a calculator or table would be needed otherwise.

Worked example 5. An aeroplane at an altitude of 200 m observes the angles of depression of two opposite points on the two banks of a river to be 45° and 60°. Width of the river = 200(cot 45° + cot 60°) = 200(1 + 1/√3) ≈ 315.5 m.

Worked example 6. A balloon is rising vertically. From a point on the ground 40 m from the point directly below it, its angle of elevation is 30°; some time later it is 60°. Height gained = 40(tan 60° − tan 30°) = 40(√3 − 1/√3) = 80/√3 ≈ 46.2 m.

Worked example 7. A jet plane flying at a constant height sees two milestones on a straight road ahead, 1 km apart, at angles of depression 30° and 45°. Find the height of the plane. The horizontal distances to the milestones are h cot 30° and h cot 45°, so h(cot 30° − cot 45°) = 1000, h(√3 − 1) = 1000, h = 1000/(√3 − 1) = 500(√3 + 1) ≈ 1366 m.

Compute the two ground distances, subtract, then divide by the time; convert m/s to km/h by multiplying by 18/5.

The angle of elevation of the sun is called its altitude. A vertical object casts a shadow on level ground, and the shadow, the object and the sun's ray to the tip of the shadow form a right triangle whose angle at the tip of the shadow is the altitude. So height = shadow length × tan(altitude), and the shadow shrinks as the sun rises.

Worked example 1. A pole 6 m high casts a shadow 2√3 m long. Find the sun's altitude. tan θ = 6/(2√3) = 3/√3 = √3, so θ = 60°.

Worked example 2. The shadow of a tower is 40 m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower. Shadow at 30° = h cot 30° = h√3; shadow at 60° = h cot 60° = h/√3. Difference: h(√3 − 1/√3) = h × 2/√3 = 40, so h = 20√3 ≈ 34.64 m.

Worked example 3. A tower is 30 m high. Find the length of its shadow when the altitude is 45° and when it is 30°. At 45° the shadow equals the height, 30 m. At 30° it is 30 cot 30° = 30√3 ≈ 51.96 m.

Worked example 4. The ratio of the height of a tower to the length of its shadow is √3 : 1. The altitude is given by tan θ = √3, so θ = 60°. If the ratio is 1 : √3 the altitude is 30°, and if it is 1 : 1 the altitude is 45°.

Worked example 5. A vertical stick 12 cm long casts a shadow 8 cm long on the ground; at the same time a tower casts a shadow 40 m long. Find the height of the tower. The sun's altitude is the same for both, so the ratio height : shadow is the same: h/40 = 12/8, h = 60 m. This is the similar-triangles version of the shadow problem and does not even need the angle.

Worked example 6. A boy 1.6 m tall stands at a distance of 3.2 m from a lamp post and casts a shadow 4.8 m long. Find the height of the lamp post. The lamp, the tip of the shadow and the boy's head form similar triangles with the lamp post and the boy: H/(3.2 + 4.8) = 1.6/4.8, so H = 8 × 1.6/4.8 = 2.67 m. Here the light source is the lamp, not the sun, so the rays are not parallel, but the same right-triangle reasoning works with the lamp as the apex.

Worked example 7. At what altitude of the sun is the shadow of a vertical pole equal to its height? tan θ = h/h = 1, θ = 45°. When is the shadow twice the height? tan θ = 1/2, which is not a standard angle; θ ≈ 26.6°.

Shadow problems are one- and two-mark favourites because they test the tan ratio directly; the two-shadow problem (Worked example 2) is a four-mark question.

The vocabulary. Line of sight; horizontal; angle of elevation (object above); angle of depression (object below); the depression at the top equals the elevation at the ground point by alternate angles.

The method. Draw the ground horizontal and the object vertical with a right angle at the foot; draw the line of sight; mark the angle at the observer; label the known lengths; choose the ratio (tan for height and ground distance, sin or cos when the line of sight is involved); write the equation; solve; state the answer with units.

The standard configurations. One triangle: height or distance from one observation; ladder, kite string, rope, slide. Two triangles sharing the vertical side: two observations from the same side (h = d/(cot β − cot α)); observations from opposite sides (CD = h(cot α + cot β)); complementary angles (h = √(ab)); moving object (speed from the difference of the cot distances); two shadows. Two triangles sharing the horizontal side: flagstaff on a building, statue on a pedestal, tower from the top of a building with an elevation and a depression, cloud and its reflection.

How the board asks. One-mark: define angle of elevation; give the altitude of the sun when the shadow equals the height; give the distance for a stated height and angle. Two-mark: a single-triangle calculation, typically a ladder, a kite, a tower seen from a given distance, or a lighthouse and a boat. Four-mark: a two-triangle problem with a figure, most often the tower with two observations 30° and 60° some metres apart, the flagstaff on a building, the two ships from a lighthouse, the statue on a pedestal, the building and the tower looking at each other, or the moving car. The figure carries one mark, the two equations one mark each, and the solution the remaining mark.

Common errors. Placing the angle of depression at the foot of the tower. Using the line of sight as the horizontal distance. Choosing sin when tan is needed. Confusing which point has the larger angle (the nearer one). Forgetting to add the observer's height. Leaving √3 in the denominator when a decimal is asked. Misreading 'from the top' and 'from the foot'. Writing metres for a speed.

Checking. A height found from a 30° angle must be less than the horizontal distance; from 60°, greater; from 45°, equal. A ladder must be longer than the height it reaches. An angle of elevation increases as the observer approaches. Estimate with √3 ≈ 1.7 and see that the answer is reasonable for the situation: a tower is tens of metres, a cloud hundreds, an aeroplane thousands.

What lies ahead. Surveying uses these ideas with angles measured to seconds and with the sine rule and cosine rule for triangles that are not right-angled, both met in the Intermediate course. Navigation uses bearings, which are angles measured from north. Physics uses the same resolution of a line of sight into horizontal and vertical components for forces, velocities and projectiles.