Chapter 23 — Applications of Trigonometry Optional

The optional exercise does not introduce a new kind of figure; every problem still reduces to right triangles with the ground horizontal, the object vertical and the angle at the observer. What it tests is the ability to handle three complications at once.

Complication 1: two unknowns. When the distance to the object is not given, one observation gives one equation in two unknowns, the height and the distance. A second observation gives a second equation, and the two are solved together, usually by writing the distance from each equation in terms of the height and equating, or by dividing one equation by the other.

Complication 2: symbolic angles. Several problems ask for a proof rather than a number: 'show that the height of the tower is d tan α tan β/(tan α + tan β)', 'prove that the height of the house is h(1 + tan α cot β)'. The method is identical to the numerical case, but the values 1/√3 and √3 are replaced by tan α and tan β, and the algebra must be carried through in symbols. Such a result, once proved, can be checked by putting in 30° and 60° and comparing with a known numerical answer.

Complication 3: a shifted observer. The observer may be a girl 1.2 m tall, a man on a deck 10 m above the water, a person at a window h metres up, or a point 60 m above a lake. The triangle then gives heights above the observer's eye, and the eye's height must be added, or the figure must include a rectangle whose top is the observer's horizontal line.

The plan for every problem. (1) Draw and label the figure with the ground, the verticals, the observer's horizontal and the lines of sight. (2) Mark each angle at the right vertex; transfer depressions to the ground by alternate angles. (3) Name the unknowns, usually h and d. (4) Write one tan equation per triangle; use sin or cos only when a line of sight length is given or asked. (5) Eliminate d. (6) Solve for h, or for the quantity asked: a difference of distances for a moving object, a difference of heights for an object on an object. (7) Add the observer's height if needed. (8) State the answer with units and check its size.

Checks that catch most errors. The nearer observation point has the larger angle. At 45° the height equals the distance; at 30° it is less; at 60° it is more. A height obtained as h(√3 − 1) is about 0.73h and one obtained as h(√3 + 1) is about 2.73h; if the figure suggests otherwise, the signs have been swapped. A speed should be a few metres per second for a car or boat and a few hundred for a jet.

The topics that follow take the exercise's problems in families, prove the general formulas, and work each standard case in full.

The two most common two-triangle figures have neat general formulas, and the exercise asks for them to be proved.

Formula 1: a tower between two points. A tower PQ of height h stands between two points A and B on a straight road, with AB = d. The angles of elevation of the top P from A and B are α and β. Prove that h = d tan α tan β/(tan α + tan β).

Proof. In triangle PQA, tan α = h/AQ, so AQ = h/tan α. In triangle PQB, tan β = h/BQ, so BQ = h/tan β. Since Q lies between A and B, AQ + BQ = d: h/tan α + h/tan β = d, so h(tan β + tan α)/(tan α tan β) = d, giving h = d tan α tan β/(tan α + tan β). Equivalently h = d/(cot α + cot β).

Check. α = 30°, β = 60°, d = 100 m: h = 100 × (1/√3)(√3)/(1/√3 + √3) = 100/(4/√3) = 25√3 ≈ 43.3 m, the answer found numerically in the main chapter.

Formula 2: two points on the same side. A tower AB of height h; from a point C the angle of elevation of the top is α, and from a point D on the line CB produced, with CD = d, it is β (β < α). Prove that h = d tan α tan β/(tan α − tan β).

Proof. BC = h/tan α and BD = h/tan β. BD − BC = d: h/tan β − h/tan α = d, so h(tan α − tan β)/(tan α tan β) = d, and h = d tan α tan β/(tan α − tan β) = d/(cot β − cot α).

Check. α = 60°, β = 30°, d = 20 m: h = 20 × √3 × (1/√3)/(√3 − 1/√3) = 20/(2/√3) = 10√3 ≈ 17.32 m, as before.

Formula 3: an object on an object. A flagstaff of height b stands on a tower of height a; from a point at distance d the elevations of the tower's top and the flagstaff's top are α and β. Then a = d tan α, a + b = d tan β, so b = d(tan β − tan α) and, if d is unknown but a is known, d = a cot α and b = a(tan β cot α − 1).

Worked example. A pole of height 5 m stands on a tower; the elevations of the bottom and top of the pole from a point on the ground are 30° and 60°. Tower height a: d = a√3 and a + 5 = d√3 = 3a, so a = 2.5 m. The formula b = a(tan β cot α − 1) = a(√3 × √3 − 1) = 2a confirms a = b/2 = 2.5 m.

These formulas need not be memorised; they are two lines each. But knowing their shape lets the student check a numerical answer at once.

Result. The angles of elevation of the top of a tower from two points at distances a and b from its base, on the same side and in the same straight line, are complementary. Prove that the height of the tower is √(ab).

Proof. Let the height be h and the angle at the point distant a be θ; then the angle at the point distant b is 90° − θ. In the first triangle tan θ = h/a. In the second, tan(90° − θ) = h/b, and tan(90° − θ) = cot θ, so cot θ = h/b. Multiplying the two equations: tan θ × cot θ = h²/(ab). The left side is 1, so h² = ab and h = √(ab), the geometric mean of a and b.

Worked example 1. Distances 4 m and 9 m: h = √36 = 6 m. Check the angles: tan θ = 6/4 = 1.5 and cot θ = 6/9 = 2/3 = 1/1.5, consistent.

Worked example 2. Distances 16 m and 25 m: h = √400 = 20 m.

Worked example 3. A tower is 12 m high; from a point 9 m from its base the elevation is θ. Where is the point from which the elevation is 90° − θ? Since 12² = 9 × b, b = 144/9 = 16 m.

Variant: angles θ and 2θ. A problem in which the angle of elevation is θ from one point and 2θ from a nearer point needs the formula tan 2θ = 2 tan θ/(1 − tan²θ), which belongs to the next course; so the board keeps to complementary angles or to the standard angles 30°, 45° and 60°, where the doubled angle 30° → 60° can be handled directly from the table.

Variant: the angle subtended by a segment. A vertical pole AB stands on the ground. From a point P on the ground the angle of elevation of the top is 60°; the pole is 10 m high. The angle subtended by the pole at P is the angle of elevation itself, 60°, since the foot is on the ground. If instead the pole stands on a 10 m building, the angle subtended at P by the pole is the difference of the elevations of its top and bottom: with d = 10 m, the bottom is at 45° and the top at tan⁻¹(20/10) = tan⁻¹ 2, so the pole subtends tan⁻¹ 2 − 45° ≈ 18.4°. Examination problems choose the numbers so that both elevations are standard.

Why the geometric mean appears. The two right triangles with angles θ and 90° − θ at the base are similar to each other with the roles of the legs interchanged, so h/a = b/h, which is the proportion defining the geometric mean. The same proportion arises in the altitude-to-hypotenuse theorem for a right triangle, where the altitude is the geometric mean of the two segments of the hypotenuse.

When the problem gives the height of the observer, the angles are measured from the observer's eyes, and the right triangle stands on the horizontal line through the eyes, not on the ground. Draw that horizontal line, subtract the eye height from every object height, solve, and add it back only if the question asks for a height above the ground.

Worked example 1. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from her eyes is 60° at one instant and 30° after some time. Find the distance travelled by the balloon.

Height of the balloon above her eyes = 88.2 − 1.2 = 87 m. First horizontal distance = 87 cot 60° = 87/√3 = 29√3 m. Second = 87 cot 30° = 87√3 m. Distance travelled = 87√3 − 29√3 = 58√3 ≈ 100.46 m.

Worked example 2. A 1.5 m tall boy stands at some distance from a 30 m building. The angle of elevation of the top from his eyes increases from 30° to 60° as he walks towards it. Find the distance walked. Height above his eyes = 28.5 m. Distances: 28.5√3 and 28.5/√3. Walked = 28.5(√3 − 1/√3) = 28.5 × 2/√3 = 19√3 ≈ 32.9 m.

Worked example 3. A man 1.8 m tall stands 10 m from a lamp post and finds the angle of elevation of the lamp to be 45°. Height of the lamp above the ground = 10 tan 45° + 1.8 = 11.8 m.

Worked example 4. An observer 1.5 m tall is 28.5 m from a chimney; the angle of elevation of its top is 45°. Chimney height = 28.5 + 1.5 = 30 m. The number 28.5 is chosen so that the answer is round; notice that forgetting the 1.5 gives 28.5 m and loses a mark.

Worked example 5. From the eyes of a person 1.6 m tall the angle of elevation of the top of a tree is 60°, and the person stands 12 m from the tree. Tree height = 12√3 + 1.6 ≈ 20.78 + 1.6 = 22.38 m.

Worked example 6: a person's shadow under a lamp. A person 1.7 m tall stands x metres from a lamp post 21.7 m high. The tip of the shadow is where the line from the lamp through the person's head meets the ground. By similar triangles, the shadow length s satisfies s/1.7 = (x + s)/21.7, so 21.7s = 1.7x + 1.7s, 20s = 1.7x, s = 0.085x. At x = 40 m the shadow is 3.4 m long. This uses similar triangles rather than a standard angle, and it shows that the observer's height enters as a ratio.

The rule. Draw the eye-level line; the triangle sits on it; the rectangle below it, of height equal to the observer's height, contributes nothing to the angles but must be added to a final height above the ground.

Not every problem in the exercise is about a horizontal distance and a height. When a ladder, a rope, a kite string, a wire, a slide or a sloping road is involved, the slant length is the hypotenuse of the right triangle, and sin θ or cos θ is the ratio to use: sin θ = height/slant length and cos θ = horizontal distance/slant length. The optional exercise adds a second observation or a movement, so that two triangles share the slant length or the height.

Worked example 1: a ladder that slips. A ladder 10 m long leans against a wall making 60° with the ground. Its foot slips outward until the ladder makes 45° with the ground. How far does the foot move, and how far does the top slide down? At 60°: foot at 10 cos 60° = 5 m from the wall, top at 10 sin 60° = 5√3 ≈ 8.66 m. At 45°: foot at 10 cos 45° = 5√2 ≈ 7.07 m, top at 5√2 ≈ 7.07 m. The foot moves 7.07 − 5 = 2.07 m outward and the top slides 8.66 − 7.07 = 1.59 m down. The ladder's length is the shared hypotenuse of the two triangles.

Worked example 2: a ladder in a lane. A ladder rests against one wall of a lane at an angle of 60° with the ground, reaching a height of 6√3 m. Turned about its foot to rest against the opposite wall, it makes 45° with the ground. Find the width of the lane. Length of ladder = 6√3/sin 60° = 6√3 × 2/√3 = 12 m. Foot to first wall = 12 cos 60° = 6 m; foot to second wall = 12 cos 45° = 6√2 ≈ 8.49 m. Width = 6 + 6√2 ≈ 14.49 m.

Worked example 3: a kite. A kite is flying at a height of 60 m and its string, assumed straight, makes 60° with the ground. Length of string = 60/sin 60° = 40√3 ≈ 69.28 m. If more string is let out until the kite is 90 m high at the same angle, the string is 60√3 ≈ 103.9 m, an extra 34.64 m.

Worked example 4: a rope to a pole. A rope from the top of a vertical pole to a peg on the ground is 20 m long and makes 30° with the ground. Height of the pole = 20 sin 30° = 10 m; the peg is 20 cos 30° = 10√3 ≈ 17.32 m from the foot. The rope is tightened so that it makes 45° with the ground, the peg being moved: new peg distance = 10 cot 45° = 10 m and new rope length = 10/sin 45° = 10√2 ≈ 14.14 m.

Worked example 5: an inclined road. A road rises at 30° to the horizontal. To gain 100 m of height a vehicle must travel 100/sin 30° = 200 m along the road and cover 100 cot 30° = 100√3 ≈ 173.2 m horizontally. On a road at 45° the same rise needs 100√2 ≈ 141.4 m along the road.

Worked example 6: a slide. A slide is to be built so that its top is 3 m high and it makes 30° with the ground. Its length is 3/sin 30° = 6 m and its foot is 3√3 ≈ 5.2 m from the point below its top. A steeper slide at 60° of the same height is only 2√3 ≈ 3.46 m long.

The signal for sin and cos is the phrase 'the ladder is 10 m long', 'the string is 60 m long', 'the rope makes 30° with the ground': a slant length is given or wanted. The signal for tan is 'from a point 30 m from the foot': a horizontal distance is given or wanted.

A moving object at a known constant height (a plane) or seen from a known height (a boat from a lighthouse, a car from a tower) is located twice by its angle. The two horizontal distances differ by the distance moved; dividing by the time gives the speed.

Worked example 1: the jet plane. The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds the angle of elevation changes to 30°. If the plane is flying at a constant height of 1500√3 m, find its speed. First horizontal distance = 1500√3 cot 60° = 1500√3/√3 = 1500 m. Second = 1500√3 cot 30° = 1500√3 × √3 = 4500 m. Distance flown = 3000 m in 15 s. Speed = 200 m/s = 200 × 18/5 = 720 km/h.

Worked example 2: the boat. From the top of a lighthouse 150 m high, a boat is seen at an angle of depression of 30°; two minutes later the angle is 60°. Distances: 150√3 and 150/√3 = 50√3; moved 100√3 ≈ 173.2 m in 2 minutes; speed ≈ 86.6 m/min ≈ 5.2 km/h.

Worked example 3: the car. From the top of a 50 m tower a car's angle of depression changes from 30° to 60° in 6 s. Distances 50√3 and 50/√3; moved 100/√3 in 6 s; remaining 50/√3 takes 3 s. Speed = (100/√3)/6 ≈ 9.62 m/s ≈ 34.6 km/h.

Worked example 4: symbolic. A plane at height H is seen at elevations α and then β (β < α) after t seconds. Its speed is H(cot β − cot α)/t. With H = 1500√3, α = 60°, β = 30°, t = 15: 1500√3(√3 − 1/√3)/15 = 1500√3 × (2/√3)/15 = 3000/15 = 200 m/s.

Worked example 5: a balloon rising. A balloon rises vertically from a point 40 m from an observer. Its angle of elevation changes from 30° to 60° in 20 seconds. Height gained = 40(tan 60° − tan 30°) = 40 × 2/√3 = 80/√3 ≈ 46.19 m; rate = 2.31 m/s.

Worked example 6: time to reach. A boat approaching a 100 m cliff is at depression 30°; how far must it travel before the depression is 45°? Distances 100√3 and 100; travel 100(√3 − 1) ≈ 73.2 m. At 2 m/s this takes 36.6 s.

Units. Convert m/s to km/h by multiplying by 18/5; km/h to m/s by 5/18. Keep the time in the unit given and convert the answer if the question asks for km/h.

A classic problem: from a point at height e above the surface of a still lake, the angle of elevation of a cloud is α and the angle of depression of its reflection in the lake is β. Find the height of the cloud.

Setup. Let the cloud be at height H above the lake and let the horizontal distance from the observer to the cloud's vertical line be d. The reflection of the cloud is at depth H below the surface. The observer's eye is e above the surface. So the cloud is H − e above the eye level, and the reflection is H + e below the eye level.

Equations. tan α = (H − e)/d and tan β = (H + e)/d.

Solution. Dividing, tan β/tan α = (H + e)/(H − e). Cross-multiplying, (H − e) tan β = (H + e) tan α, so H(tan β − tan α) = e(tan β + tan α), and H = e(tan α + tan β)/(tan β − tan α). Also d = (H − e)/tan α.

Worked example 1. e = 60 m, α = 30°, β = 60°. H = 60(1/√3 + √3)/(√3 − 1/√3) = 60 × (4/√3)/(2/√3) = 60 × 2 = 120 m. The cloud is 120 m above the lake; d = (120 − 60)/tan 30° = 60√3 ≈ 103.9 m.

Worked example 2. e = 2500 m (a hill above a lake), α = 30°, β = 45°. H = 2500(1/√3 + 1)/(1 − 1/√3) = 2500(1 + √3)/(√3 − 1) = 2500(1 + √3)²/2 = 1250(4 + 2√3) = 2500(2 + √3) ≈ 9330 m.

Worked example 3. e = 20 m, α = 30°, β = 60°: H = 40 m by the same computation as Worked example 1 scaled down (the formula is linear in e).

Why the depression is larger. The reflection is farther below the eye than the cloud is above it, by 2e, at the same horizontal distance, so its angle is larger: β > α always, and the formula's denominator tan β − tan α is positive.

A note on what has a reflection. A boat on the surface has no separate reflection to observe, since it sits on the mirror itself; the lake problem is specific to objects above the water, such as clouds, birds, balloons, kites and the tops of towers on the far bank.

Worked example 4. From a point 12 m above a lake, the angle of elevation of the top of a tower on the far bank is 30° and the angle of depression of its reflection is 60°. Height of the tower above the lake = 12(1/√3 + √3)/(√3 − 1/√3) = 12 × 2 = 24 m; horizontal distance = (24 − 12)/tan 30° = 12√3 ≈ 20.78 m.

The figure must show the surface line, the observer's eye-level line above it, the cloud above and the reflection below at equal distances from the surface; then the two triangles share the horizontal side d.

When the observer stands at a known height and sees both the top and the foot of another object, the depression of the foot gives the horizontal distance and the elevation of the top gives the extra height. The total height of the object is the observer's height plus the extra.

Result (the window problem). A window of a house is at height h above the ground. From the window the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are α and β. Prove that the height of the opposite house is h(1 + tan α cot β).

Proof. Let the width of the street be d. From the depression of the foot: tan β = h/d, so d = h cot β. From the elevation of the top: the top is d tan α above the window, so the opposite house has height h + d tan α = h + h cot β tan α = h(1 + tan α cot β).

Worked example 1. h = 10 m, α = 60°, β = 30°: height = 10(1 + √3 × √3) = 10 × 4 = 40 m; street width = 10 cot 30° = 10√3 ≈ 17.32 m.

Worked example 2: the deck of a ship. A man on the deck of a ship, 10 m above the water, sees the top of a cliff at elevation 45° and its base at depression 30°. Cliff height = 10(1 + tan 45° cot 30°) = 10(1 + √3) ≈ 27.32 m; distance = 10√3 ≈ 17.32 m.

Worked example 3: from a building to a tower. From the top of a 7 m building, the elevation of the top of a cable tower is 60° and the depression of its foot is 45°. Tower = 7(1 + √3 × 1) = 7(1 + √3) ≈ 19.12 m.

Worked example 4: from a taller to a shorter building. From the top of a 60 m building the depressions of the top and the bottom of a tower are 30° and 60°. Distance = 60 cot 60° = 20√3 m; the tower's top is 20√3 tan 30° = 20 m below the building's top, so the tower is 40 m high. Here both angles are depressions, and the tower's height is the building's height minus the drop to the tower's top.

Worked example 5: from one pole top to another. Two poles of heights 20 m and 14 m stand 8 m apart. From the top of the taller pole the angle of depression of the top of the shorter one is found from tan θ = (20 − 14)/8 = 6/8 = 3/4, so θ ≈ 37°, not a standard angle. The examination version gives the angle and asks for a distance or a height: with a depression of 30° and a horizontal separation of 6√3 m, the difference in heights is 6√3 tan 30° = 6 m.

Worked example 6: from a hill. From the top of a hill 300 m high the angles of depression of two boats on the same side, in line with the hill, are 30° and 45°. Distance between the boats = 300(cot 30° − cot 45°) = 300(√3 − 1) ≈ 219.6 m.

In all raised-observer problems, draw the observer's horizontal line first; the depression is below it and the elevation above it, and the two triangles share the horizontal distance d.

Some problems place two objects on the same vertical line at different heights: one aeroplane flying above another, a flag on a tower, a bird above a boat. From a ground point the two objects are seen at two angles, and the two triangles share the horizontal distance.

Worked example 1. An aeroplane flying at a height of 4000 m passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from a point on the ground are 60° and 45°. Find the vertical distance between them. Horizontal distance d = 4000 cot 60° = 4000/√3. Height of the lower plane = d tan 45° = 4000/√3 ≈ 2309.4 m. Vertical distance = 4000 − 4000/√3 = 4000(1 − 1/√3) ≈ 1690.6 m.

Worked example 2. A flagstaff on a tower: from a point 50 m from the tower's foot the elevations of the tower's top and the flag's top are 45° and 60°. Tower = 50 m, flag top at 50√3, flagstaff = 50(√3 − 1) ≈ 36.6 m.

Worked example 3. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point on the ground is 45°. It flies off horizontally and after 1 second the elevation from the same point is 30°. Distance flown = 20(cot 30° − cot 45°) = 20(√3 − 1) ≈ 14.64 m; speed ≈ 14.64 m/s.

Worked example 4: symbolic. Two objects at heights H (upper) and h (lower) on the same vertical line are seen from a point at elevations α and β. Then H cot α = h cot β = d, so h = H tan β cot α and the gap is H(1 − tan β cot α). With H = 4000, α = 60°, β = 45°: gap = 4000(1 − 1/√3), as in Worked example 1.

Worked example 5. A kite is flying at a height of 75 m with the string, assumed straight, making 60° with the ground; a second kite from the same point flies at 75 m with its string at 30°. String lengths: 75/sin 60° = 50√3 ≈ 86.6 m and 75/sin 30° = 150 m. Horizontal distances 25√3 and 75√3 m, so the kites are 50√3 ≈ 86.6 m apart horizontally if in the same direction.

Worked example 6. The angle of elevation of the top of a vertical tower from a point on the ground is 60°; from a point 10 m vertically above the first point (on a platform) the angle is 45°. Find the height of the tower. Let the horizontal distance be d and the height h. From the ground, h = d√3. From the platform, h − 10 = d. So d√3 − 10 = d, d(√3 − 1) = 10, d = 10/(√3 − 1) = 5(√3 + 1) ≈ 13.66 m, and h = d√3 = 5(3 + √3) ≈ 23.66 m. Here the two observers are on the same vertical line and the triangles share d with different vertical sides.

Draw the common vertical line, mark both heights on it, and draw both lines of sight from the single ground point; the equations follow.

An observer high above the ground, in an aeroplane or on a tall tower, may see two points on opposite sides. The two angles of depression give two horizontal distances on opposite sides, and their sum is the distance between the points.

Worked example 1. An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. Find the width of the river. Width = 200 cot 45° + 200 cot 60° = 200 + 200/√3 = 200(1 + 1/√3) ≈ 200 × 1.577 = 315.5 m.

Worked example 2. From an aeroplane vertically above a straight road, the angles of depression of two consecutive kilometre stones on opposite sides are 60° and 30°. Find the height of the plane. The stones are 1000 m apart: h cot 60° + h cot 30° = 1000, h(1/√3 + √3) = 1000, h × 4/√3 = 1000, h = 250√3 ≈ 433 m.

Worked example 3. A lighthouse 100 m high sees two ships on opposite sides at depressions 30° and 45°. Distance between the ships = 100(√3 + 1) ≈ 273.2 m.

Worked example 4: symbolic. From a height H the depressions of two points on opposite sides are α and β. Distance = H(cot α + cot β). If the distance D is known, H = D/(cot α + cot β) = D tan α tan β/(tan α + tan β), the same formula as for a tower between two points, which is the same figure turned upside down.

Worked example 5. From the top of a tower 60 m high, the angles of depression of two cars on opposite sides of the tower on a straight road are 60° and 30°. Find the distance between the cars. Distances from the foot: 60 cot 60° = 60/√3 and 60 cot 30° = 60√3. Being on opposite sides, the cars are 60(1/√3 + √3) = 60 × 4/√3 = 80√3 ≈ 138.6 m apart.

Worked example 6: same side versus opposite sides. A common trap: the problem says 'on the same side' or 'on opposite sides', and the distances are subtracted in the first case and added in the second. From a 100 m tower with depressions 30° and 60°: same side, the objects are 100(√3 − 1/√3) = 200/√3 ≈ 115.5 m apart; opposite sides, 100(√3 + 1/√3) = 400/√3 ≈ 230.9 m apart. Read the phrase, draw the figure accordingly, and the sign takes care of itself.

Worked example 7. A bridge over a river: from a point on the bridge 20 m above the water, two boats directly under the line of the bridge on opposite sides are seen at depressions 45° and 60°. Distance between the boats = 20(1 + 1/√3) ≈ 31.5 m.

Draw the observer at the top of a vertical line with the two points on the ground on either side; the two triangles share the vertical side and the ground distances add.

Problem 1. A 1.2 m tall girl spots a balloon at a height of 88.2 m; the elevation from her eyes changes from 60° to 30°. Distance travelled = 87(cot 30° − cot 60°) = 87 × 2/√3 = 58√3 ≈ 100.46 m.

Problem 2. The angles of elevation of a tower from two points at 4 m and 9 m on the same side are complementary; height = √(4 × 9) = 6 m.

Problem 3. The angle of elevation of a jet from a point on the ground is 60°; after 15 s it is 30°; height 1500√3 m. Speed = (4500 − 1500)/15 = 200 m/s = 720 km/h.

Problem 4. From a window at height h, the top and the foot of the opposite house are at elevation α and depression β. Height of the opposite house = h(1 + tan α cot β).

Problem 5. A man on the deck of a ship 10 m above water sees the top of a cliff at 45° and its base at 30°. Cliff = 10(1 + √3) ≈ 27.32 m; distance 10√3 ≈ 17.32 m.

Problem 6. From a point 60 m above a lake the elevation of a cloud is 30° and the depression of its reflection is 60°. Cloud height = 60(tan 30° + tan 60°)/(tan 60° − tan 30°) = 120 m.

Problem 7. Two poles of equal height stand on either side of an 80 m road; from a point between them the elevations of their tops are 60° and 30°. Poles 20√3 ≈ 34.64 m high; the point is 20 m from one and 60 m from the other.

Problem 8. A plane at 4000 m passes vertically above another when their elevations from a ground point are 60° and 45°. Vertical gap = 4000(1 − 1/√3) ≈ 1690.6 m.

Problem 9. From the top of a 7 m building the elevation of a tower's top is 60° and the depression of its foot is 45°. Tower = 7(1 + √3) ≈ 19.12 m.

Problem 10. A tower stands between two points 100 m apart with elevations 30° and 60°. Height = 100 tan 30° tan 60°/(tan 30° + tan 60°) = 100/(4/√3) = 25√3 ≈ 43.3 m.

Problem 11. A plane at 200 m sees the two banks of a river at depressions 45° and 60°. Width = 200(1 + 1/√3) ≈ 315.5 m.

Problem 12. A boy 1.5 m tall walks towards a 30 m building while the elevation of its top changes from 30° to 60°. Distance walked = 28.5 × 2/√3 = 19√3 ≈ 32.9 m.

Problem 13. The shadow of a tower is 40 m longer at solar altitude 30° than at 60°. Height = 40/(cot 30° − cot 60°) = 40/(2/√3) = 20√3 ≈ 34.64 m.

Problem 14. From the top of a hill 300 m high two boats on the same side are at depressions 30° and 45°. Distance apart = 300(√3 − 1) ≈ 219.6 m.

Problem 15. A statue 1.6 m tall on a pedestal: elevations 60° (top of statue) and 45° (top of pedestal). Pedestal = 1.6/(√3 − 1) = 0.8(√3 + 1) ≈ 2.19 m.

Each of these is a four-mark question. The figure, the two ratio equations, the elimination and the final answer with unit are the four steps the marking scheme rewards.

Reading. Underline every number and every angle in the problem, and the words 'from the top', 'from the foot', 'same side', 'opposite sides', 'above the ground', 'from his eyes', 'horizontally', 'vertically'. Each of these fixes something in the figure.

Drawing. Ground horizontal; objects vertical; observer's horizontal at eye height; lines of sight; angles at the observer; transfer depressions by alternate angles. Label the unknowns h and d. If there are two observers or two positions, draw both lines of sight and both angles; if there are two objects on one vertical, draw both heights on it.

Equations. One tan equation per triangle, in the form (vertical) = (horizontal) × tan(angle) or (horizontal) = (vertical) × cot(angle). Prefer cot for horizontal distances; then two-triangle problems become 'difference (or sum) of two cot-distances equals a known length'. Use sin or cos only when a slant length is involved.

Algebra. Keep surds exact: √3, 1/√3, √3 − 1, √3 + 1. Rationalise at the end: 1/(√3 − 1) = (√3 + 1)/2. Convert to decimals only if asked, with √3 ≈ 1.732.

Common errors. Placing the angle of depression between the tower and the line of sight instead of between the horizontal and the line of sight. Using the height above the ground when the angle is measured from the eyes. Adding distances that should be subtracted (same side) or subtracting those that should be added (opposite sides). Assigning the larger angle to the farther point. Forgetting the reflection is as far below the surface as the cloud is above it. Dividing by 15 s and then reporting km/h without converting. Writing the answer without the unit, or with the wrong unit.

Checking. Substitute the answer back into both tan equations. Compare with the 30°–45°–60° rules: at 30° the height is about 0.58 of the distance, at 60° about 1.73 times. Ask whether the size is reasonable: a cloud at 120 m, a jet at 720 km/h, a tower at 43 m, a river at 315 m wide are all plausible; a cloud at 1.2 m or a jet at 72 km/h is not.

What lies ahead. In the Intermediate course the sine rule and the cosine rule handle triangles without a right angle, so that heights can be found from two angles measured at points not in line with the object; bearings replace 'left and right' with angles from north; and the same resolution into horizontal and vertical components governs projectiles and inclined planes in physics. Surveying, navigation, architecture and astronomy all begin with the figures of this exercise.