Height and Distance

This type is the root of the chapter. There is one vertical object, one observer on the ground and one angle. The figure is a single right triangle; the question gives one side and asks for another.

Look at the figure below and keep these names in your head: the vertical side AB is the height h, the ground side CB is the distance d, and the slant side CA is the line of sight l. In a kite question CA is the string, in a ladder or ramp question it is the ladder or ramp, and in a wire question it is the wire.

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The whole method fits in one sentence: pick the ratio that connects the given side and the asked side.

• A tower is 40 m away and its top is seen at 60°; find the height. Height and distance, so tan 60° = h40 ⇒ h = 40√3 m.
• A 30 m rope tied to the top of a pole makes 45° with the ground; find the height. Height and slant, so sin 45° = h30 ⇒ h = 30√2 = 15√2 m.
• A 9 m ramp rises at 30°; find the horizontal length it covers. Distance and slant, so cos 30° = d9 ⇒ d = 9√32 m.

Sometimes the angle itself is asked. A 12 m pole is seen from 12√3 m away: tan θ = 1212√3 = 1√3, so θ = 30°.

The broken tree. A tree breaks at some point, the broken part stays attached, and its top touches the ground. The standing part is the height of the triangle, the broken part is the hypotenuse, and the distance from the root to where the top touches is the base. The broken part was standing on top before it fell, so original height of the tree = standing part + broken part.

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Radhika sees a tree broken by a storm. Its top touches the ground 6 m from the root and makes 45° with the ground. The standing part is 6 tan 45° = 6 m and the broken part is 6√2 m, so the tree was (6 + 6√2) m tall.

Now the observer goes up, on a tower, a lighthouse, a cliff, a bridge or an aeroplane, and looks down at something on the ground. The angle is measured below the horizontal line through his eye.

There is only one new fact, and the foundation has already proved it. The horizontal line at the top is parallel to the ground, so the angle of depression at the top = the angle of elevation at the ground point. In the figure, 30° below the horizontal at A is the same as 30° at C. Copy the angle down and solve the triangle exactly as in Type 1. Since the height is usually given, the distance comes from

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d = h cot θ (or tan θ = hd)

Two objects on the same side. From the top of a tower of height h, two cars are seen on the same side at depressions θ₁ and θ₂ (θ₁ < θ₂). The car with the smaller angle is farther away. Their distances are h cot θ₁ and h cot θ₂, so

distance between them = h (cot θ₁ − cot θ₂)

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Two objects on opposite sides. If the cars are on the two sides of the tower, the distances add:

distance between them = h (cot θ₁ + cot θ₂)

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The bridge question is the same figure read backwards. From a point on a bridge the two banks are seen at θ₁ and θ₂, and the width of the river is given. Then width = h (cot θ₁ + cot θ₂), so h = widthcot θ₁ + cot θ₂.

Take Radhika on a 40 m high flyover. She sees two buses on the road, one on each side, both at 45°. Each bus is 40 cot 45° = 40 m away, so the buses are 80 m apart. If both were on the same side at 45°, they would be at the same point and the gap would be 0. This small case shows clearly why we add for opposite sides and subtract for the same side.

A shadow question is a Type 1 triangle in which the sun is the observer. The vertical object is the height, its shadow on the ground is the distance, and the sun's ray passing over the top of the object to the end of the shadow is the line of sight. The angle of elevation of the sun (also called the sun's altitude) is the angle at the tip of the shadow.

tan θ = heightshadow

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So a 10 m pole with a 10 m shadow means tan θ = 1, and the sun is at 45°. As the sun rises higher, the angle grows and the shadow gets shorter; at 60° the shadow is only 1√3 of the height.

Two objects at the same time. The sun is so far away that its rays are parallel everywhere. At the same moment, therefore, every vertical object makes the same angle θ with its shadow. The triangles are similar, and

Formula (Shadows at the same moment):
>> H₁S₁ = H₂S₂ (height ÷ shadow is the same for all objects)

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You do not even need the angle. A 2 m stick with a 3 m shadow tells you that at that moment every height is 23 of its shadow, so a tower with a 45 m shadow is 30 m tall. Keep both lengths in the same unit (cm with cm, m with m).

Line of sight over an object. When a man's eye, the top of a tree and the top of a building lie on one straight line, the small triangle (eye to tree) and the big triangle (eye to building) are similar. If the eye is on the ground, heights are simply proportional to distances from the eye. If the eye is at height e, subtract e from every height first and then use the proportion.

Mirror on the ground. Light from the top of a tower hits a mirror lying on the ground and reaches the man's eye. It leaves the mirror at the same angle at which it arrives, so the two triangles on either side of the mirror are similar:

height of towerdistance of tower from mirror = height of man's eyedistance of man from mirror

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Radhika, whose eye is 1.5 m above the ground, stands 2 m from a mirror on the ground and sees the top of a pole in it. The pole is 8 m from the mirror, so its height is 1.5 × 82 = 6 m.

A ladder leaning on a wall is the cleanest right triangle in the chapter. The wall is vertical and the ground is horizontal, so the right angle is at the foot of the wall. The ladder itself is always the hypotenuse. The height at which it touches the wall is the perpendicular, and the distance of its foot from the wall is the base. The angle θ of the ladder with the ground is at the foot of the ladder.

• Ladder l, angle θ → height = l sin θ, foot distance = l cos θ.
• Height h and angle θ → ladder = hsin θ.

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When a ratio is given instead of an angle. A line like "sin x = 2425" or "tan θ = 815" is really telling you the shape of the triangle. Write the triangle with those numbers, complete the third side by Pythagoras, and scale. These triplets keep coming:

For example, Radhika's ladder is 10 m long and makes tan θ = 43 with the ground. So height : base : ladder = 4 : 3 : 5. 5 units = 10, so 1 unit = 2. The ladder reaches 8 m up the wall, and its foot is 6 m from the wall.

Sliding ladder. The length of the ladder does not change when it slips. Find the first position by Pythagoras, change the height (or the foot) as told, and use Pythagoras again with the same hypotenuse. Sliding of the foot = new foot distance − old foot distance.

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Ladder turned to the other side. If the foot stays at one point in the street and the ladder is turned to touch windows on both sides, then street width = base on one side + base on the other side. The same ladder is the hypotenuse both times.

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Direction questions ("goes 6 km east and then 8 km north") use the same Pythagoras: the two movements are the legs, and the straight distance from the start is the hypotenuse.

Now the question gives two angles. The height is common, and the two angles make two right triangles that share it. So the method is: write both distances in terms of h, then use the given difference or sum.

Moving towards the object (same side). From D the top A is seen at α. After walking a distance d towards the tower to C, the angle becomes β, and β > α because you have come closer. Then BD = h cot α and BC = h cot β, so

Formula (Two angles, same side):
>> d = h (cot α − cot β) ⇒ h = dcot α − cot β

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The same figure comes in other clothes too: two observers on the same side; a shadow that is 30 m longer when the sun's altitude is 30° than when it is 45° (the sun is the observer and the two shadow tips are D and C); or angles seen from points a and b metres away.

The 30°–60° case. With 30° at D and 60° at C, ∠DAC = 60° − 30° = 30° = ∠ADC. So △ACD is isosceles and AC = CD. In units: BC = 1, BD = 3, gap = 2, height = √3. Hence height = √32 × gap, and the farther distance is 3 times the nearer one.

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Complementary angles. Suppose two points on the same side are at distances a and b from the foot, and the angles from them are complementary (θ and 90° − θ). Then tan θ = ha and tan (90° − θ) = cot θ = hb. Multiplying, 1 = h²ab, so

Formula (Complementary angles):
>> h = √(ab)

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Opposite sides. If the two points are on opposite sides of the tower, the distances add: total = h (cot α + cot β).

Two angles from the same point (object on top). Think of a statue on a pedestal, a flagstaff on a building or a poster on a wall. From one point P at distance d, the top of the lower part is seen at α and the top of the upper part at β. Then lower height = d tan α and total height = d tan β, so the upper object = d (tan β − tan α).

Radhika stands 30 m from her school building and sees its top at 30°. A flagstaff on the building has its top at 45°. The building is 30 tan 30° = 10√3 m and the top of the flagstaff is 30 tan 45° = 30 m above the ground, so the flagstaff is (30 − 10√3) m. If the building were raised until the angle became 60°, the new height would be 30√3 m, exactly 3 times 10√3, because tan 60° ÷ tan 30° = 3.

Now there are two vertical objects on the same ground: a tower and a building, a hill and a pole, or two poles. The horizontal distance d between them is the same at every level, and this is the key to the whole type.

From the top of the taller one. A man at the top A of a tower of height H looks at a shorter building CD. He sees its top C at depression α and its foot D at depression β (β > α). Draw a horizontal dashed line from C to the tower. Now there are two right triangles:

• big triangle (A, B, D): tan β = Hd ⇒ d = H cot β
• small triangle above the level of C: tan α = H − hd ⇒ H − h = d tan α

So find d from the bottom angle first, then the difference of heights from the top angle, and subtract: h = H − d tan α.

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From the top of the shorter one. A man at the top C of a pole of height h sees the top A of a taller tower at elevation α and its foot B at depression β. Then d = h cot β (lower triangle) and H − h = d tan α (upper triangle), so H = h + d tan α.

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If the distance d is given directly and only the elevation from the smaller pole is given, the same line H = h + d tan α finishes it.

A point between the two. From a point E on the ground between the two objects, their tops are seen at θ₁ and θ₂. Each object makes its own Type 1 triangle: H₁ = d₁ tan θ₁ and H₂ = d₂ tan θ₂. If E is exactly midway (d₁ = d₂), then H₁ : H₂ = tan θ₁ : tan θ₂.

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Radhika is on the roof of her 20 m high building. The foot of the temple opposite is seen at a depression of 45° and the top of the temple at an elevation of 30°. Distance = 20 cot 45° = 20 m. The part of the temple above her roof = 20 tan 30° = 20√3 m. So the temple is 20 + 20√3 ≈ 20 + 11.55 = 31.55 m tall.

So far the 1 : √3 : 2 ratio has solved one triangle at a time. In this last type the figure has two or more triangles, and something is moving: a car, a boat, an aeroplane, a balloon. The approach is to put the whole figure on one common scale of units, find the distance moved in units, and only then convert to metres. Speed comes at the very end, from speed = distance movedtime.

Step 1: choose the common side. In most figures the height is common. Give it the value that makes both triangles easy. For 30° and 60° take height = √3 (the distances are then 3 and 1). For 30° and 45° take height = 1 (distances √3 and 1). For 45° and 60° take height = √3 (distances √3 and 1).

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Step 2: read the moving distance in units. An aeroplane at a fixed height moving from 60° to 30° covers 3 − 1 = 2 units while the height is √3 units. A car on the ground moving away from a cliff, with the depression changing from 45° to 30°, covers (√3 − 1) units for a height of 1 unit.

Step 3: convert and divide by time. If the height is 1500 m in the first case, √3 units = 1500, so 2 units = 3000√3 = 1000√3 m. If this took 20 s, the speed is 50√3 m/s.

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When the object moves vertically (a balloon rising straight up, a climber going up a cliff), the observer is fixed, so the distance is common. At each moment height = d tan θ, and the rise = d (tan θ₂ − tan θ₁).

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Extra information. Some tough-looking questions give movement, speeds and new angles, but ask only for something the first position already fixes (for example the starting distance between two boats). Read the last line first and use only what it needs.

Radhika is on a 50 m high cliff and sees a boat sailing straight towards it. In one minute the depression changes from 30° to 60°. Height √3 units = 50, so 1 unit = 50√3. Distance moved = 3 − 1 = 2 units = 100√3 ≈ 57.7 m in 60 s, so the boat moves at about 0.96 m/s.

If an unusual angle turns up, these help: tan 15° = 2 − √3, tan 75° = 2 + √3, and for 15°–75°–90° the sides are (√3 − 1) : (√3 + 1) : 2√2.

How much longer? A car or a boat moves at a steady speed straight towards the foot of a tower. From the top the angle of depression changes from α to β in t seconds. The question then asks how much more time it needs to reach the foot. Because the speed is constant, time is proportional to distance, so you only need the two distances in units:

time left = t × distance from the second position to the footdistance covered in the t seconds

With 30° and 60° (height √3 units): the distance covered is 3 − 1 = 2 units and 1 unit is left, so the time left is t2. With 45° and 60°: the car covers 1 − 1√3 and has 1√3 left, so the time left is t(√3 + 1)2.

The same idea gives the speed when the height is known: speed = distance covered in metrestime.

Radhika watches a boat from a lighthouse. The angle of depression goes from 30° to 60° in 12 minutes, so the boat needs 6 more minutes to reach the foot of the lighthouse.

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So far every figure was a straight vertical tower with a horizontal ground. Three other pictures come up often, and each of them is still a right triangle: a tree broken by the wind, a road or ramp that climbs along a slope, and a heap or a tent whose slanting side makes an angle with the ground. Find the right triangle first, and the rest is the usual 1 : √3 : 2 and 1 : 1 : √2 work.

A broken tree or pole. A tree of height H breaks at a point x above the ground. The upper part does not fall off; it bends over and its tip touches the ground at a distance d from the foot, making an angle θ with the ground. The picture is one right triangle:

• the standing part x is the perpendicular (vertical side),
• the broken part y is the hypotenuse,
• the distance d on the ground is the base.

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Whatever is given, the total height is H = x + y, and the broken part and the standing part were one piece, so we add them to get H, not subtract. With y = xsin θ we get

H = x + xsin θ = x(1 + cosec θ)

For θ = 30° we have sin 30° = 12, so y = 2x, and H = 3x. The tree broke at one-third of its height. For θ = 45°, y = x√2 and H = x(1 + √2); for θ = 60°, y = 2x√3.

A slope. A ramp, a road up a hill or a ladder on an inclined plane is the hypotenuse of a right triangle. If you move a distance s along a slope that makes an angle θ with the horizontal, then

• you rise by s sin θ,
• you cover s cos θ on the ground.

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A slope of 30° gives a rise of exactly half the distance travelled along it. Sometimes the slope is given as a ratio: "the road rises 5 m for every 13 m of road" means sin θ = 513, and the 5-12-13 triplet gives the horizontal part at once.

A cone: a heap of sand or a tent. The cross-section of a conical heap is an isosceles triangle. The slanting side makes an angle θ with the ground, so with height h, radius r and slant height l:

[[fig:2]]

From r the area of the ground covered is πr², the volume is 13πr²h and the curved surface (the canvas of a tent) is πrl, as in Mensuration (3D).

Radhika's flagpole is 36 m tall. A storm breaks it so that the top touches the ground at 30°. It broke at 363 = 12 m above the ground, the broken part is 24 m long, and the tip is 12√3 ≈ 20.8 m from the foot.