Chapter 9 — Some Applications of Trigonometry

Suppose a student stands on the ground and looks at the top of a tall building. The straight line from her eye to the top of the building is called the line of sight. The horizontal line through her eye is the horizontal level. The angle that the line of sight makes with the horizontal level, when the object viewed is above the horizontal, is the angle of elevation of the object. To look at the top of the building she raises, or elevates, her head through this angle.

Now suppose the student is on the balcony of a tall building and looks down at a ball lying in the garden. The line of sight now slopes downward. The angle that this line of sight makes with the horizontal level through her eye is the angle of depression of the ball. To see it she lowers, or depresses, her head through this angle.

Both angles are measured from the horizontal, never from the vertical. A common examination error is to mark the angle of depression between the line of sight and the vertical wall of the building; that angle is the complement of the true angle of depression. In the figure, always draw the horizontal line through the observer's eye first, then the line of sight, and mark the angle between them.

A crucial fact connects the two angles. If a person at the top of a tower observes a point on the ground with angle of depression θ, then a person at that point on the ground observes the top of the tower with angle of elevation θ, the same angle. The reason is that the horizontal line at the top and the ground are parallel, and the line of sight is a transversal, so the two angles are alternate interior angles. Thus every angle of depression can be transferred into the right triangle as the angle of elevation at the ground vertex, where it is useful.

Surveyors measure these angles with a theodolite, an instrument consisting of a telescope mounted on a graduated circle, which reads the angle between the line of sight and the horizontal to a fraction of a degree. In the eighteenth and nineteenth centuries the Great Trigonometrical Survey of India used such instruments and the methods of this chapter to measure the whole subcontinent and to compute the height of the highest peak of the Himalaya.

Unless the question states otherwise, the observer's eye is taken to be at ground level, so that the observer is a point. If the height of the observer is given, the horizontal level is at that height and the observer's height must be added to the computed height at the end.

Every problem in this chapter is solved by the same four steps, and the examination rewards each step separately.

Step 1: Draw the figure. Represent the vertical object (tower, pole, building, cliff, tree) as a vertical segment, the ground as a horizontal segment, and the line of sight as the hypotenuse. Mark the right angle where the vertical meets the ground. Mark the given angle in its correct place: an angle of elevation at the ground observer, an angle of depression at the top, and then transfer it to the ground vertex as an equal angle of elevation. Label the known length and denote the unknown by a letter such as h or x. A clear figure is worth a mark on its own.

Step 2: Choose the ratio. In the right triangle you now have one known angle, one known side and one wanted side. If the two sides are the opposite and the adjacent, use tangent. If one of them is the hypotenuse, use sine (with the opposite) or cosine (with the adjacent). Most problems involve the vertical height and the horizontal distance, so tangent is by far the most used ratio.

Step 3: Solve. Substitute the standard value of the ratio (tan 30° = 1/√3, tan 45° = 1, tan 60° = √3, and so on) and solve the resulting equation. Rationalise surds in the denominator.

Step 4: Answer. State the answer with its unit. If the observer has a height, add it. If an approximate value is asked, use √3 ≈ 1.732 and √2 ≈ 1.414.

Worked example. A tower stands vertically on the ground. From a point on the ground 15 m from its foot, the angle of elevation of the top is 60°. Find the height of the tower.
Let AB be the tower with foot B, and C the point on the ground with BC = 15 m and ∠ACB = 60°. In right triangle ABC, AB is opposite the angle and BC is adjacent, so tan 60° = AB/BC. Hence √3 = AB/15 and AB = 15√3 m ≈ 25.98 m.

Worked example. An electrician has to repair a fault on a pole 5 m high, and must reach a point 1.3 m below the top. What length of ladder, inclined at 60° to the horizontal, is needed, and how far from the pole should its foot be?
The point to reach is 5 − 1.3 = 3.7 m above the ground. If the ladder is L and its foot is d from the pole: sin 60° = 3.7/L gives L = 3.7 × 2/√3 = 7.4/√3 ≈ 4.28 m, and tan 60° = 3.7/d gives d = 3.7/√3 ≈ 2.14 m.

The examination expects the figure, the statement in triangle ABC, tan 60° = AB/BC, the substitution and the answer, each on its own line.

The simplest class of problems gives one distance and one angle in a single right triangle. The unknown may be the height, the horizontal distance, the length of the line of sight, or the angle itself. Let us see all four types.

Unknown height. The angle of elevation of the top of a tower from a point 30 m from its foot is 30°. Height h satisfies tan 30° = h/30, so h = 30/√3 = 10√3 m ≈ 17.32 m.

Unknown distance. The angle of elevation of the top of a 50 m tower from a point on the ground is 45°. Distance d satisfies tan 45° = 50/d, so d = 50 m. At 45° the height and the distance are always equal.

Unknown line of sight. A kite is flying at a height of 60 m and the string, assumed straight, makes 60° with the ground. The length L of the string satisfies sin 60° = 60/L, so L = 60 × 2/√3 = 120/√3 = 40√3 m ≈ 69.28 m.

Unknown angle. A circus artist climbs a 20 m rope tied to the top of a vertical pole of height 10 m. The rope is the hypotenuse of length 20 m and the pole the opposite side of 10 m, so sin θ = 10/20 = 1/2, so θ = 30° is the angle the rope makes with the ground. Recognising the angle requires the value to be one of the standard ratios.

A broken tree. A tree breaks due to a storm and the broken part bends so that its top touches the ground making an angle of 30° with the ground; the distance between the foot of the tree and the point where the top touches the ground is 8 m. Find the height of the tree. Let the standing part be AB = h₁ and the broken part AC = h₂, with BC = 8 m and ∠ACB = 30°. tan 30° = h₁/8 gives h₁ = 8/√3; cos 30° = 8/h₂ gives h₂ = 16/√3. Total height = 24/√3 = 8√3 m ≈ 13.86 m. Two ratios were needed because the tree's height is the sum of a leg and the hypotenuse.

Slide for children. A slide for younger children is 1.5 m high and inclined at 30°; for older children it is 3 m high and inclined at 60°. The lengths are 1.5/sin 30° = 3 m and 3/sin 60° = 3 × 2/√3 = 2√3 m ≈ 3.46 m.

Every one of these uses one triangle and one ratio at a time. Master them before moving to two-triangle problems, where the same steps are simply done twice.

When the observer is at a height, the angle given is an angle of depression, and the first thing to do is to convert it into an angle of elevation at the ground point using the alternate-angle fact. After that the problem is exactly as before.

Worked example. From the top of a 75 m high lighthouse, the angle of depression of a ship is 30°. Find the distance of the ship from the foot of the lighthouse.
Let AB = 75 m be the lighthouse and C the ship. The angle of depression at A is 30°, so the angle of elevation ∠ACB = 30°. tan 30° = AB/BC gives 1/√3 = 75/BC, so BC = 75√3 m ≈ 129.9 m.

Worked example. A man on the top of a cliff 80 m high observes a boat at an angle of depression of 60°. Later the boat is seen at 30°. How far did the boat move? (This two-position situation is developed fully in a later topic.) The distances are 80/tan 60° = 80/√3 and 80/tan 30° = 80√3, so the boat moved 80√3 − 80/√3 = (240 − 80)/√3 = 160/√3 ≈ 92.4 m.

Worked example. From a point on a bridge across a river, the angles of depression of the banks on opposite sides are 30° and 45°. If the bridge is 3 m above the banks, find the width of the river.
Let P be the point on the bridge, 3 m above the level of the banks, with foot Q directly below. Let A and B be the points on the two banks. In ΔPQA, tan 30° = 3/QA gives QA = 3√3 m. In ΔPQB, tan 45° = 3/QB gives QB = 3 m. The width AB = QA + QB = 3√3 + 3 = 3(√3 + 1) m ≈ 8.196 m. Here the two triangles lie on opposite sides of the vertical PQ, so the distances add.

Worked example. Two cars on the same side of a tower are seen at angles of depression 30° and 45° from its top, which is 100 m high. Distance of the nearer car = 100/tan 45° = 100 m; distance of the farther car = 100/tan 30° = 100√3 m; distance between the cars = 100(√3 − 1) ≈ 73.2 m. Here both triangles lie on the same side, so the distances subtract.

Same side, subtract; opposite sides, add. Deciding which applies is the whole art of a two-triangle depression problem, and the figure decides it for you.

The most frequently examined configuration has two observation points on the same side of a vertical object, with two known angles and one known distance between the points, and both the height and one distance unknown. Two right triangles share the vertical side, giving two equations in two unknowns.

Worked example. 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 of elevation becomes 60°. Find the height of the tower and the distance of the second point from the tower.
Let AB = h be the tower, D the first point and C the second, with DC = 20 m and CB = x. From ΔABC: tan 60° = h/x, so h = x√3. From ΔABD: tan 30° = h/(x + 20), so h = (x + 20)/√3. Equating, x√3 = (x + 20)/√3, so 3x = x + 20, x = 10 m. Then h = 10√3 m ≈ 17.32 m.

Observe the pattern: the nearer point gives the larger angle. The tangent ratios give h = x tan β and h = (x + d) tan α, and elimination of h yields x = d tan α/(tan β − tan α). With α = 30°, β = 60°, d = 20: x = 20 × (1/√3)/(√3 − 1/√3) = 20/(3 − 1) = 10, as found.

Worked example. The shadow of a tower standing on level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower.
Let the height be h and the shorter shadow x. tan 60° = h/x gives h = x√3; tan 30° = h/(x + 40) gives h = (x + 40)/√3. So 3x = x + 40, x = 20 m and h = 20√3 m ≈ 34.64 m.

Worked example. A 1.5 m tall boy is some distance from a 30 m tall building. The angle of elevation from his eyes to the top rises from 30° to 60° as he walks towards the building. How far did he walk?
The vertical side is 30 − 1.5 = 28.5 m (the height above his eye level). Distances: 28.5/tan 30° = 28.5√3 and 28.5/tan 60° = 28.5/√3. Distance walked = 28.5√3 − 28.5/√3 = 28.5(3 − 1)/√3 = 57/√3 = 19√3 m ≈ 32.9 m.

Worked example. The angles of elevation of the top of a tower from two points at distances 4 m and 9 m from its base, in the same line, are complementary. Prove the height is 6 m. Let the angles be θ and 90° − θ. tan θ = h/4 and tan(90° − θ) = cot θ = h/9. Multiplying, tan θ × cot θ = h²/36 = 1, so h² = 36 and h = 6 m.

Always write both tangent equations, eliminate h, solve for the distance, and then find h. Check that the answer for h is the same from both triangles.

When the two observation points, or the two observed objects, lie on opposite sides of the vertical, the two right triangles stand back to back on the same vertical side, and the horizontal distances add up to the total distance between the points.

Worked example. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Let AB = 7 m be the building and CD the tower, with the horizontal from A meeting CD at E, so that CE = AB = 7 m and AE = BC = x. In ΔAEC (angle of depression 45° transferred to C as an elevation, or directly ∠EAC = 45°): tan 45° = CE/AE = 7/x, so x = 7 m. In ΔAED: tan 60° = DE/AE = DE/7, so DE = 7√3 m. Height of the tower = CD = CE + ED = 7 + 7√3 = 7(1 + √3) m ≈ 19.12 m. Here both angles are at the same observer, one looking up and one looking down, so the two triangles share the horizontal AE, and the heights add.

Worked example. Two poles of equal heights are standing opposite each other on either side 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 distances of the point from the poles.
Let the height be h and the distances x and 80 − x. tan 60° = h/x gives h = x√3; tan 30° = h/(80 − x) gives h = (80 − x)/√3. So 3x = 80 − x, x = 20 m. The point is 20 m from one pole and 60 m from the other, and h = 20√3 m ≈ 34.64 m.

Worked example. As observed from the top of a 75 m high lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the ships. Distances from the foot: 75/tan 45° = 75 m and 75/tan 30° = 75√3 m; separation 75(√3 − 1) ≈ 54.9 m. This is a same-side problem and is placed here for contrast: the phrase behind the other tells you to subtract.

Worked example. A statue 1.6 m tall stands on the 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 x. tan 45° = h/x gives x = h; tan 60° = (h + 1.6)/x = (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) m ≈ 2.19 m.

Before writing equations, decide from the figure whether the horizontal distances add (opposite sides), subtract (same side), or are equal (same observer looking up and down), and whether the vertical heights add (object on object).

Many questions give the height of the observer or of the observation platform. The eye is then not at ground level, the horizontal level passes through the eye, and the right triangle's vertical side is the height of the object above the eye, not above the ground. The observer's height is added back at the end for a height above ground, and subtracted at the start when the height of an object is used.

Worked example. A person 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. Find the height of the chimney.
Let the eye be at E, 1.5 m above the ground, and the horizontal from E meet the chimney at D. Then ED = 28.5 m and the top C satisfies tan 45° = CD/ED, so CD = 28.5 m. Height of the chimney = CD + 1.5 = 30 m.

Worked example. 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 at one instant is 60°; after some time it reduces to 30°. Find the distance travelled by the balloon.
The balloon is 88.2 − 1.2 = 87 m above her eye level throughout. Horizontal distances: 87/tan 60° = 87/√3 = 29√3 and 87/tan 30° = 87√3. Distance travelled = 87√3 − 29√3 = 58√3 m ≈ 100.46 m.

Worked example. From a point on the ground the angle of elevation of the top of a 20 m building is 30°. From the top of the building a flagstaff stands; the angle of elevation of the top of the flagstaff from the same point is 45°. Find the height of the flagstaff. Distance x = 20/tan 30° = 20√3; total height = x tan 45° = 20√3; flagstaff = 20√3 − 20 = 20(√3 − 1) ≈ 14.64 m. This is an object on an object, handled by subtracting the lower height.

Worked example. An observer 1.5 m tall is 20.5 m from a tower 22 m high. Find the angle of elevation of the top of the tower from his eye. Height above eye = 22 − 1.5 = 20.5 m, distance 20.5 m, so tan θ = 20.5/20.5 = 1 and θ = 45°.

Worked example. From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 30°. Following the same-observer pattern: horizontal x = 10/tan 30° = 10√3; extra height = x tan 60° = 10√3 × √3 = 30; tower = 10 + 30 = 40 m.

The rule: draw the horizontal through the eye, work in the triangle above it, and account for the eye height separately.

A moving object observed twice gives two positions, two angles and a distance travelled; combined with a time, this yields a speed, and with a speed it yields a time. The geometry is the two-triangle same-side configuration; the extra step is speed = distance/time.

Worked example. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with uniform speed. Six seconds later the angle of depression is 60°. Find the time taken by the car to reach the foot of the tower from this point.
Let the height be h. Distances from the foot: first position d₁ = h/tan 30° = h√3; second position d₂ = h/tan 60° = h/√3. Distance covered in 6 s = h√3 − h/√3 = h(3 − 1)/√3 = 2h/√3. Speed = (2h/√3)/6 = h/(3√3) per second. Remaining distance d₂ = h/√3 takes time (h/√3) ÷ (h/(3√3)) = 3 s. Note that h cancels; the answer does not depend on the tower's height.

Worked example. An aeroplane flying horizontally at a height of 1500√3 m is observed at an angle of elevation of 60°; after 15 seconds the angle of elevation is 30°. Find the speed of the aeroplane.
Horizontal distances: 1500√3/tan 60° = 1500 m and 1500√3/tan 30° = 4500 m. Distance flown = 3000 m in 15 s, so speed = 200 m/s = 720 km/h.

Worked example. A boat is observed from the top of a 150 m cliff at an angle of depression of 30°, and after 2 minutes at 45°, moving towards the cliff. Speed? Distances 150√3 and 150 m; distance moved 150(√3 − 1) ≈ 109.8 m in 120 s; speed ≈ 0.915 m/s ≈ 3.3 km/h.

Worked example. From a lighthouse 100 m high, a boat is seen at 60° and later at 30°. Distance moved = 100√3 − 100/√3 = 200/√3 ≈ 115.5 m. If the boat's speed is known to be 6 km/h = 100 m/min, the time between observations is 1.155 min ≈ 69 s.

Convert units consistently: m/s to km/h by multiplying by 18/5, km/h to m/s by multiplying by 5/18. Keep the working in exact surds until the last step, and note how often the unknown height cancels.

A flagstaff on a tower, a statue on a pedestal, a transmission tower on a building, a water tank on a house: in each of these the vertical side is made of two parts, and the observer sees two angles of elevation from the same point, one to the top of the lower object and one to the top of the upper object. The two right triangles share the horizontal side, and the heights are related by subtraction.

General setting. Let the lower object have height h₁, the upper object height h₂, and the observation point be at distance x. Then tan α = h₁/x and tan β = (h₁ + h₂)/x. Dividing, (h₁ + h₂)/h₁ = tan β/tan α, so h₂ = h₁(tan β/tan α − 1) if h₁ is known, or h₁ = h₂ tan α/(tan β − tan α) if h₂ is known.

Worked example. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top is 30°. Find the height of the tower and the width of the canal. This is the same-side pattern: h = x√3 and h = (x + 20)/√3 give x = 10 m (the width) and h = 10√3 m.

Worked example. From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from P. Distance x = 10/tan 30° = 10√3 ≈ 17.32 m. Total height = x tan 45° = 10√3 m. Flagstaff = 10√3 − 10 = 10(√3 − 1) ≈ 7.32 m.

Worked example. The angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively from a point on the ground. Find the height of the tower. x = 20/tan 45° = 20 m; total = 20 tan 60° = 20√3; tower = 20√3 − 20 = 20(√3 − 1) ≈ 14.64 m.

Worked example. A pedestal carries a statue 1.6 m tall. From a ground point the angles of elevation of the top of the statue and the top of the pedestal are 60° and 45°. Pedestal height h: x = h, and (h + 1.6)/h = √3, so h = 1.6/(√3 − 1) = 0.8(√3 + 1) ≈ 2.19 m.

The critical step is to write the total height as the sum of the two parts and to use the same x in both triangles.

Some questions involve two vertical objects and the angle between their tops, or the angle of elevation of one top from the other. The horizontal distance between them and the difference of their heights form the legs of a right triangle whose hypotenuse joins the two tops.

Worked example. 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.
Let the distance between them be x. From the foot of the building, tan 60° = 50/x, so x = 50/√3. From the foot of the tower, tan 30° = h/x, so h = x/√3 = 50/3 m ≈ 16.67 m.

Worked example. Two poles of heights 6 m and 11 m stand on a plane ground 12 m apart. Find the angle of elevation of the top of the taller pole from the top of the shorter one, given that tan θ = 5/12 corresponds to θ ≈ 22.6°, and find the distance between their tops. The horizontal from the top of the shorter pole meets the taller pole 6 m up, leaving a vertical of 11 − 6 = 5 m; with a horizontal of 12 m, tan θ = 5/12, and the distance between the tops is √(25 + 144) = 13 m.

Worked example. From the top of a tower 60 m high the angles of depression of the top and the bottom of a building are 30° and 60°. Find the height of the building.
Let the horizontal distance be x and the building height h. From the bottom: tan 60° = 60/x, x = 60/√3 = 20√3. From the top of the building, the vertical difference is 60 − h and tan 30° = (60 − h)/x, so 60 − h = 20√3/√3 = 20, h = 40 m.

Worked example. Two towers of equal height stand on either side of a road 100 m wide. From a point on the road between them, the angles of elevation of their tops are 30° and 60°. The distances are 100 − x and x with h = x√3 = (100 − x)/√3, so 3x = 100 − x, x = 25 m, h = 25√3 ≈ 43.3 m.

Worked example. The angles of elevation of the top of a tower from the top and the bottom of a 15 m building are 30° and 60°. Height of the tower H and distance x: from the bottom, H = x√3; from the top, H − 15 = x/√3. So x√3 − 15 = x/√3, 3x − 15√3 = x, x = 15√3/2 and H = 45/2 = 22.5 m.

In all of these the difference of heights is the vertical leg of the upper triangle, while the full height is the vertical leg of the lower triangle, and both share the same horizontal distance.

Not every situation involves 30°, 45° or 60°. When the angle is another value, the examination provides the ratio, for instance take tan 35° = 0.7 or sin 22° = 0.375, and the method is unchanged. Sometimes the ratio itself is given in place of the angle, such as the angle whose tangent is 5/12, and you work with the ratio directly, never needing the angle in degrees.

Worked example. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7 m. From a point on the plane, the angle of elevation of the bottom of the flagstaff is such that its tangent is 3/4, and of the top of the flagstaff such that its tangent is 1. Let the tower be h and the distance x. h/x = 3/4 and (h + 7)/x = 1, so x = h + 7 and h = 3(h + 7)/4, giving 4h = 3h + 21, h = 21 m.

Worked example. 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 H above the lake; its reflection is H below the surface. The observer is 60 m above the lake, so the cloud is H − 60 above his horizontal and the reflection is H + 60 below it. With horizontal distance x: tan 30° = (H − 60)/x and tan 60° = (H + 60)/x. Dividing, (H + 60)/(H − 60) = 3, so H + 60 = 3H − 180, 2H = 240, H = 120 m.

Worked example. A ladder rests against a wall at an angle α to the horizontal. Its foot is pulled away a distance a so that it slides down the wall a distance b and now makes an angle β with the horizontal. Show a/b = (cos β − cos α)/(sin α − sin β). Let the ladder be L. Initially the foot is L cos α from the wall and the top L sin α up; finally L cos β and L sin β. So a = L cos β − L cos α and b = L sin α − L sin β; dividing gives the result.

Worked example. A person observes the top of a hill at an angle of elevation θ with tan θ = 1/2 from a point 400 m from the foot. Height = 400 × 1/2 = 200 m.

Worked example. The shadow of a vertical pole of height 12 m is 9 m long. tan θ = 12/9 = 4/3; the Sun's altitude is the angle with tangent 4/3 (about 53°), and the line from the tip of the shadow to the top of the pole is √(144 + 81) = 15 m.

When two angles are complementary, use tan(90° − θ) = cot θ = 1/tan θ, and when a reflection is involved, remember that the image is as far below the surface as the object is above it.

The board marks these problems by steps, so the layout of the solution matters as much as the arithmetic. Here is the layout that earns full marks, illustrated on a full problem.

Problem. The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Solution layout.

• Figure. Draw PC the multi-storeyed building, AB the 8 m building, with the horizontal from P and the horizontal from A meeting PC at D. Mark the 30° angle of depression to A and the 45° to B, and transfer them as ∠PAD = 30° and ∠PBC = 45°. Label PD = h, DC = AB = 8 m, AD = BC = x.
• Statement of the triangles. In right ΔPAD, tan 30° = PD/AD = h/x, so x = h√3. … (1)
• In right ΔPBC, tan 45° = PC/BC = (h + 8)/x, so x = h + 8. … (2)
• Solve. From (1) and (2), h√3 = h + 8, h(√3 − 1) = 8, h = 8/(√3 − 1) = 8(√3 + 1)/2 = 4(√3 + 1) m.
• Answer. Height PC = h + 8 = 4√3 + 4 + 8 = 4(√3 + 3) m ≈ 18.93 m. Distance BC = x = h + 8 = 4(3 + √3) m ≈ 18.93 m.

Marks are typically distributed as: one for the correct figure, one for each correct trigonometric equation, one for solving, one for the final answers with units. Omitting the figure or writing tan 30° = x/h with the sides inverted loses marks that no later correctness recovers.

Checklist before you finish.

• Are both angles measured from the horizontal in the figure?
• Is each ratio written as opposite/adjacent (tan), opposite/hypotenuse (sin) or adjacent/hypotenuse (cos) for the correct angle?
• Have surds been rationalised and the observer's height added back?
• Do the two triangles give consistent values (substitute back)?
• Are the units stated?

A second full problem. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. Figure: building AB = 20 m, tower BC = h, point P at distance x. In ΔABP, tan 45° = 20/x, x = 20. In ΔACP, tan 60° = (20 + h)/x = (20 + h)/20, so 20 + h = 20√3, h = 20(√3 − 1) ≈ 14.64 m. Answer: the tower is 20(√3 − 1) m, about 14.64 m, high.

Practise writing every solution in this pattern; in the examination the pattern is worth as much as the answer.