Heights and distances | ICSE Class 10 Maths Notes

This note covers lines of sight, angles of elevation and depression, right-angled triangles, trigonometric ratios and tables, heights above eye level, ladders and ropes, shadows, buildings, and distances across a river.

What do heights and distances problems represent?

Trigonometry studies relationships between the sides and angles of a triangle. It lets us calculate a height or distance that is difficult to measure directly, using other measurements in a suitable triangle.

A right-angled triangle contains a right angle, measuring 90°. The symbol ° means degrees, the unit used here for angles. Horizontal means level; vertical means upright, perpendicular to the horizontal. Perpendicular directions meet at a right angle. A vertical object on horizontal ground therefore supplies the right angle.

How is a real situation simplified?

A tower, its distance from an observer, and the observer's view of its top can form a triangle in a vertical plane. A two-dimensional problem represents the relevant lengths and directions in one plane, rather than combining views in different planes.

The foot of a tower is its bottom point on the ground. A horizontal line through an observer's eyes is at eye level. If the eyes are above the ground, the triangle's vertical side measures the height above those eyes, rather than the entire tower.

The drawing shows a minar beside a simplified triangle. A marks the observer's eye, E the ground point below it, C the minar's top, D its foot and B the point on the minar level with A. AC is the sloping line of sight; AB is horizontal.

See Fig. 9.1 in your NCERT textbook

For two labelled points, such as A and B, AB denotes the segment joining them or its length, according to the context. The height CD splits into CB above eye level and BD below it. Here BD equals AE, the observer's eye height.

Which assumptions belong in the diagram?

Mark objects as vertical and ground as level when the problem gives that arrangement. A taut rope means a rope stretched without slack, so it can represent a straight side. Keep the distinction between a horizontal distance and a sloping length throughout the calculation.

Note: Work with configurations involving one or two right-angled triangles. Identify the common height or distance before combining two triangles.

How are angles of elevation and depression measured?

Definition: A line of sight joins an observer's eye to the point being viewed. An angle of elevation lies between this line and the horizontal when the viewed point is above eye level.

An angle of depression is the angle between the horizontal through the observer's eye and the line of sight to a point below that horizontal. The observer looks down, but the reference direction remains horizontal.

Where should the angle be marked?

Start at the eye position. Draw a horizontal reference line and then the line to the object. Mark the angle between them. In an elevation problem, that angle may lie inside the main triangle; in a depression problem, it may initially lie outside it.

Result: Corresponding elevation and depression angles are equal

For the same line of sight, the angle of depression from the higher point equals the angle of elevation from the lower point. The horizontal lines through the two points are parallel, meaning they remain the same distance apart in their plane. The sight line acts as a transversal, a line crossing both parallel lines.

The relevant alternate interior angles, lying between the parallel lines on opposite sides of that transversal, are equal. This explains why a depression angle can be transferred to the lower vertex of the right triangle.

Be precise about which point is viewed. Looking at the top of a shorter building gives a triangle based on the difference between the buildings' heights. Looking at its foot gives a triangle based on the full height of the taller building.

These are different sight lines, so their angles need not match. Equality applies to the elevation and depression angles at the ends of the same sight line, not to every angle in a diagram.

Which trigonometric ratio should you choose?

Let θ, read as theta, denote the chosen acute angle in a right-angled triangle. An acute angle is greater than 0° and less than 90°. The hypotenuse is the side opposite the right angle and is the triangle's longest side.

The opposite side faces θ. The adjacent side is the side beside θ other than the hypotenuse. Opposite and adjacent depend on the chosen acute angle; the hypotenuse remains opposite the right angle.

The symbol ÷ means division. A reciprocal is one divided by the original non-zero quantity. The expression sin θ means the sine of angle θ, rather than multiplication of a quantity called sin by θ.

Result: A fixed angle has fixed trigonometric ratios

The trigonometric ratios do not change when the triangle's side lengths change while its angles remain the same. The triangles are similar, meaning their corresponding angles are equal and their corresponding sides are proportional, meaning the ratios of matching side lengths are equal.

This is why an angle value can be used for towers of different heights. The angle supplies a ratio between lengths, rather than an actual length. An additional length measurement determines the scale of the triangle.

How can the available data guide the choice?

1. Identify the given acute angle and the right angle.
2. Name the opposite, adjacent and hypotenuse relative to that acute angle.
3. Find the ratio containing the known length and the required length.
4. Substitute the angle value, rearrange the equation and attach the length unit.

Use tangent for height and horizontal distance. Use sine when height and a ladder or taut rope are involved. Use cosine when the horizontal distance and sloping length are involved. Avoid introducing an unnecessary unknown side.

How do exact values and trigonometric tables enter the calculation?

A trigonometric table supplies numerical values of ratios for angles. First choose the ratio from the geometry, then read its value for the given angle. The table does not decide whether a length is opposite, adjacent or the hypotenuse.

The values below include √, the positive square-root symbol. Thus √3 is the positive number whose square is 3. Fractions such as 1/√3 mean one divided by that square root. The entries retain their exact form.

How should you use a table supplied for other angles?

Locate the required ratio and angle using the table's headings. Where an angle includes minutes, a minute is one sixtieth of a degree; use the relevant minute column. Follow the table's own instructions if an adjustment between listed entries is needed.

Write the chosen ratio as an equation before inserting the table value. Keep enough digits during intermediate calculations, then round the final length to the requested precision. A decimal approximation is a nearby numerical value, rather than an exact replacement.

Why retain exact values when possible?

Using √3 directly keeps the algebra exact. If a question supplies a decimal value for √3, use that value for the requested approximation. State “approximately” when reporting a rounded length, so the answer does not imply exact equality.

The endpoint entries 0° and 90° complete the reference table. A non-degenerate right triangle, meaning a triangle with positive side lengths and area, has two acute angles. Do not use an undefined table entry as though it were zero.

How do you find a height from one angle of elevation?

When the observation point is on the same horizontal level as the foot of a vertical object, the triangle's opposite side is the whole height. If the observation point is above ground, it is the height above the observation point.

Result: Total height includes the observer's eye height

Let H be the object's total height, e the observer's eye height above level ground, and d the horizontal distance to the object. Let θ be the angle of elevation of its top. Use the same unit for all three lengths.

The triangle's vertical side is H − e, where − means subtraction. Therefore tan θ = (H − e)/d, and rearrangement gives H = e + d tan θ. The height found from tangent must be measured from the correct horizontal reference.

Worked example 1. A vertical tower stands on level ground. Its top has an angle of elevation of 60° from a ground point 15 m from its foot. Find its height. Here m means metres.

Answer: Let h be the tower's height in metres. The height is opposite 60° and the 15 m ground distance is adjacent. Therefore tan 60° = h/15. Substituting tan 60° = √3 gives h = 15√3. The tower is 15√3 m high.

The distance in this example is measured along the horizontal ground. Treating it as the hypotenuse would change the triangle and lead to an incorrect use of sine. The sloping view of the tower's top is a separate length.

Worked example 2. An observer whose eyes are 1.5 m above level ground stands 28.5 m horizontally from a vertical chimney. The angle of elevation of its top is 45°. Find its height.

Answer: Let y be the chimney's height above eye level, in metres. Then tan 45° = y/28.5. Since tan 45° = 1, y = 28.5. Add the eye height: the chimney's total height is 28.5 + 1.5 = 30 m.

The second calculation first finds an upper segment of the chimney. Adding the observer's eye height completes the vertical length from the ground to the top. The horizontal separation stays 28.5 m at both ground level and eye level.

How do ladders and ropes form right-angled triangles?

A ladder leaning against a vertical pole forms a triangle with the pole and horizontal ground. The ladder is sloping, so it represents the hypotenuse. The point where it touches the pole determines the vertical height reached.

If the required working point is below the pole's top, subtract that upper gap before choosing a ratio. Using the entire pole height would answer a different question. The angle with the ground is between the ladder and the horizontal.

A is the pole's top, D its foot, B the point reached by the ladder and C the ladder's foot. BC slopes from the ground to the pole. The ground angle at C is marked 60°.

See Fig. 9.5 in your NCERT textbook

How do you calculate both ladder length and ground distance?

Worked example 3. An electrician must reach a point 1.3 m below the top of a vertical 5 m pole. A ladder makes 60° with level ground. Find its length and the distance of its foot from the pole. Use √3 = 1.73 for approximation.

Answer: The required vertical height is 5 − 1.3 = 3.7 m. Let L be the ladder length and d its horizontal distance from the pole, both in metres. Since sin 60° = 3.7/L = √3/2, L = 7.4/√3, approximately 4.28 m.

For the ground distance, cot 60° = d/3.7 = 1/√3. Thus d = 3.7/√3, approximately 2.14 m. These calculations refer to the same triangle and the same point reached on the pole.

What changes when the sloping length is already known?

Worked example 4. A 20 m rope is stretched tightly from the top of a vertical pole to level ground, making 30° with the ground. Find the pole's height.

Answer: Let h be the pole's height in metres. The rope is the hypotenuse, so sin 30° = h/20. As sin 30° = 1/2, h = 20 × 1/2 = 10 m. The symbol × means multiplication.

The height is shorter than the rope because the hypotenuse is the longest side. This provides a useful check after substitution. The no-slack condition matters: the calculation uses the rope as a straight segment rather than as a curved length.

How do two triangles reveal a flagstaff's height?

A flagstaff on a building creates two vertical heights from the same ground level: the building alone and the building together with the flagstaff. From one observation point, both right triangles share the same horizontal distance.

Draw the building and flagstaff on one vertical line. Join the observation point separately to the building's top and the flagstaff's top. The larger elevation angle corresponds to the upper endpoint in this arrangement.

Which triangle should be solved first?

Begin with the triangle containing a known vertical height. It supplies the horizontal distance shared by the second triangle. The second angle then gives the combined height. Subtracting the building height leaves the length of the flagstaff.

Worked example 5. From a ground point on level ground, the top of a vertical 10 m building has an elevation of 30°. A vertical flagstaff on its roof has its top at an elevation of 45° from the same point. Find the horizontal distance and flagstaff length. Use √3 = 1.732.

Answer: Let d be the horizontal distance and x the flagstaff length, both in metres. For the building triangle, tan 30° = 10/d. Therefore d = 10√3 = 17.32 m using the given approximation.

For the larger triangle, tan 45° = (10 + x)/d = 1. Hence 10 + x = 10√3, giving x = 10(√3 − 1), approximately 7.32 m.

Why must the two heights be distinguished?

The second tangent equation uses 10 + x because its opposite side reaches from the ground to the top of the flagstaff. Writing x alone would place the base of that triangle at roof level while leaving the observer on the ground.

The unknown d has the same meaning in both equations. Using a different horizontal distance for the second line of sight would contradict the fixed observation point. The two sloping sight lines are different lengths, but they end above the same building foot.

A final answer should identify each requested length by name. The distance from the observer is 17.32 m approximately; the additional height above the roof is 7.32 m approximately. Neither should be confused with the known 10 m building height.

How do changing shadows connect two distances?

The Sun's altitude is its angular elevation above the horizontal. For a vertical tower on level ground, the ray from the tower's top to the shadow's tip forms a right triangle with the tower and its shadow.

At different solar altitudes, the tower's height remains the same while the shadow length changes. In the configuration below, the shadow at 30° is longer than the shadow at 60°. The given 40 m is the difference between the two lengths.

How should the two shadow lengths be labelled?

Let h be the tower's height and x its shorter shadow length, both in metres. The longer shadow is then x + 40. Writing the difference this way retains the relationship provided by the problem.

Worked example 6. A vertical tower on level ground casts a shadow 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the tower's height.

Answer: For the shorter shadow, tan 60° = h/x, so h = x√3. For the longer shadow, tan 30° = h/(x + 40), so x + 40 = h√3.

Substitute h = x√3 into the second equation: x + 40 = 3x. Hence 2x = 40 and x = 20. Therefore h = 20√3 m.

How can substitution check the result?

The shorter shadow is 20 m and the longer shadow is 20 + 40 = 60 m. Their difference is 40 m, as required. The ratio of height to the shorter shadow is √3, which matches tan 60°.

For the longer shadow, dividing 20√3 by 60 gives 1/√3, matching tan 30°. Checking both equations confirms that the answer fits both observations. Checking just one would not test the relationship between the shadow lengths.

The same reasoning about a common height applies when an observer changes position along level ground. First decide from the stated arrangement whether distances must be added or subtracted. Do not infer that relationship merely from the presence of two angles.

How do depression angles determine building heights and river widths?

Depression problems use the same ratios as elevation problems once the angles are placed in suitable triangles. The main decision is which vertical segment and which horizontal distance belong to each sight line.

What happens when the top and foot of another building are viewed?

AB is the shorter building and PC the taller one. A and C are their feet; B and P are their tops. BD is horizontal to point D on PC, and PQ is horizontal through P. Sight lines PB and PA carry depression angles of 30° and 45° respectively.

See Fig. 9.9 in your NCERT textbook

Worked example 7. Two vertical buildings stand on the same horizontal ground. From the taller building's top, the depression angles to the top and foot of the 8 m shorter building are 30° and 45°, respectively. Find the taller height and horizontal separation.

Answer: Let H be the taller height and d the separation, both in metres. The sight line to the foot gives tan 45° = H/d = 1, so H = d. The sight line to the top gives tan 30° = (H − 8)/d = 1/√3.

Substituting d = H gives H√3 − 8√3 = H. Thus H = 8√3/(√3 − 1) = 4(3 + √3) m. The separation d is also 4(3 + √3) m.

When must two horizontal distances be added?

A cross-section here is a vertical slice through the observation point and the two bank points. It places the river-width problem in one plane.

P is the observation point on the bridge, A and B are opposite bank points, and D lies between A and B below P. PD is marked 3 m, with depression angles of 30° towards A and 45° towards B.

See Fig. 9.10 in your NCERT textbook

Worked example 8. A bridge observation point is 3 m above the common horizontal level of two opposite river banks. Its vertical projection lies between the banks in the cross-section considered. Depression angles to the banks are 30° and 45°. Find the width across that section.

Answer: Let a be the horizontal distance towards the 30° bank and b that towards the 45° bank, both in metres. Then tan 30° = 3/a, giving a = 3√3. Also tan 45° = 3/b, giving b = 3.

The vertical projection lies between the banks, so the width is a + b = 3(√3 + 1) m. Each triangle uses the same 3 m vertical height.

In the building problem the triangles share a horizontal separation. In the river problem the two horizontal segments together make the required width. Reading that distinction from the arrangement prevents a correct ratio from being followed by an incorrect final addition or subtraction.

Glossary

• Trigonometry — The study of relationships between the angles and side lengths of triangles.
• Line of sight — The line joining an observer's eye to the particular point being viewed.
• Angle of elevation — The angle between the horizontal and an upward line of sight from the observer.
• Angle of depression — The angle between the horizontal and a downward line of sight from the observer.
• Hypotenuse — The longest side of a right-angled triangle, lying opposite its right angle.
• Opposite side — The side facing the chosen acute angle in a right-angled triangle.
• Adjacent side — The side beside the chosen acute angle, excluding the triangle's hypotenuse.
• Tangent — The ratio of the opposite side to the adjacent side for a chosen acute angle.
• Sine — The ratio of the opposite side to the hypotenuse for a chosen acute angle.
• Cosine — The ratio of the adjacent side to the hypotenuse for a chosen acute angle.
• Sun's altitude — The angle of elevation of the Sun above the horizontal reference direction.
• Trigonometric table — A reference listing numerical values of trigonometric ratios for specified angles.

Common errors and misconceptions

• Misconception: Depression is measured from a vertical line. Correct: Measure it from the horizontal through the observer's eye to the downward sight line.
• Misconception: Tangent uses the sloping sight-line length. Correct: Tangent relates opposite and adjacent sides; the hypotenuse belongs in sine or cosine ratios.
• Misconception: A height calculated from eye level is the whole object's height. Correct: Add eye height when the object and observer stand on the same level ground.
• Misconception: The top-of-flagstaff equation uses the flagstaff alone. Correct: From ground level, its opposite side includes both building and flagstaff heights.
• Misconception: A difference between shadows is one entire shadow. Correct: Label the shorter shadow first and add the stated difference to obtain the longer one.
• Misconception: Any two horizontal distances should be added. Correct: Add segments that together form the required length; subtract when the required length is their difference.
• Misconception: “Not defined” in a table means zero. Correct: An undefined ratio has no numerical value there and cannot be substituted as zero.

Exam-style questions with model answers

Q1. Define the angle of elevation and the angle of depression, specifying the reference line for each. [2 marks]

1. An angle of elevation is measured from the horizontal through the observer's eye to a sight line above that horizontal.
2. An angle of depression is measured from the horizontal through the observer's eye to a sight line below that horizontal.

Q2. A vertical tower on level ground is viewed from a ground point 15 m horizontally from its foot. The elevation of its top is 60°. Find its height, using tan 60° = √3. [3 marks]

1. Let h be the tower's height in metres. The height is opposite the 60° angle, while the given horizontal distance of 15 m is adjacent to it.
2. Using tangent, tan 60° = h/15. Substitute the supplied ratio value to obtain √3 = h/15.
3. Multiply both sides by 15. Hence h = 15√3, so the tower's height is 15√3 m.

Q3. An observer's eyes are 1.5 m above level ground. A vertical chimney is 28.5 m horizontally away, and its top has an elevation of 45°. Find the chimney's height, using tan 45° = 1. [3 marks]

1. Let y be the chimney's height above the observer's horizontal eye level, in metres. This segment is the opposite side of the right triangle.
2. The adjacent side is 28.5 m. Therefore tan 45° = y/28.5 = 1, giving y = 28.5 m.
3. The chimney extends another 1.5 m below eye level to the ground. Its total height is therefore 28.5 + 1.5 = 30 m.

Q4. A ladder must reach 1.3 m below the top of a vertical 5 m pole. It makes 60° with level ground. Find its length and the horizontal distance of its foot from the pole. Use sin 60° = √3/2, cot 60° = 1/√3 and √3 = 1.73. [4 marks]

1. The height to be reached is 5 − 1.3 = 3.7 m. Let L be the ladder length and d the horizontal distance, both in metres.
2. The ladder is the hypotenuse. Thus sin 60° = 3.7/L = √3/2, giving L = 7.4/√3.
3. Using the given approximation, L = 7.4/1.73, approximately 4.28 m.
4. Cotangent gives d/3.7 = 1/√3. Hence d = 3.7/1.73, approximately 2.14 m from the pole.

Q5. A vertical 10 m building stands on level ground with a vertical flagstaff on its roof. From the same ground point, the elevations of the building's top and flagstaff's top are 30° and 45°, respectively. Find the horizontal distance to the building and the flagstaff's length. Use tan 30° = 1/√3, tan 45° = 1 and √3 = 1.732. [5 marks]

1. Let d be the horizontal distance from the observation point to the building and x the flagstaff length, both in metres. Both triangles have the same adjacent side d.
2. The building triangle gives tan 30° = 10/d = 1/√3. Rearranging gives d = 10√3, approximately 17.32 m.
3. The larger triangle reaches the flagstaff's top, so its vertical height is 10 + x. Thus tan 45° = (10 + x)/d.
4. Since tan 45° = 1, the combined height equals d. Therefore 10 + x = 10√3 and x = 10√3 − 10.
5. Using √3 = 1.732 gives x approximately 7.32 m. The horizontal distance is approximately 17.32 m and the flagstaff length approximately 7.32 m.

Q6. A vertical tower on level ground casts a shadow 40 m longer at a solar altitude of 30° than at 60°. Solar altitude is the Sun's elevation above the horizontal. Find the tower's height. Use tan 30° = 1/√3 and tan 60° = √3. [5 marks]

1. Let h be the tower's height and x the shorter shadow length, both in metres. The shadow at 30° then has length x + 40.
2. For the 60° triangle, tan 60° = h/x = √3. Thus h = x√3, relating the common height to the shorter shadow.
3. For the 30° triangle, tan 30° = h/(x + 40) = 1/√3. Rearranging gives x + 40 = h√3.
4. Substitute h = x√3 to obtain x + 40 = 3x. Therefore 2x = 40 and the shorter shadow is 20 m.
5. The height is h = 20√3 m. The longer shadow is 60 m, which exceeds the shorter shadow by the required 40 m.

Q7. Two vertical buildings stand on the same level ground. From the taller building's top, depression angles to the top and foot of the 8 m shorter building are 30° and 45°, respectively. Find the taller height and horizontal separation. Use tan 30° = 1/√3 and tan 45° = 1. [5 marks]

1. Let H be the taller height and d the horizontal separation, both in metres. Transfer each depression angle to its equal elevation angle using parallel horizontals.
2. The sight line to the shorter building's foot gives tan 45° = H/d = 1. Hence d = H.
3. The sight line to its top uses the height difference H − 8. Therefore tan 30° = (H − 8)/d = 1/√3.
4. Substitute d = H: √3(H − 8) = H. Rearranging gives H = 8√3/(√3 − 1) = 4(3 + √3) m.
5. Since d = H, the buildings' horizontal separation is also 4(3 + √3) m. The two equal final lengths represent different parts of the configuration.

Q8. A bridge observation point is 3 m above the common horizontal level of opposite river banks. Its vertical projection lies between the banks in the cross-section being measured. Depression angles towards the two banks are 30° and 45°. Find the width, using tan 30° = 1/√3 and tan 45° = 1. [4 marks]

1. Let a and b be the horizontal distances from the vertical projection to the banks viewed at 30° and 45°, respectively, in metres.
2. Using the equal elevation angle, tan 30° = 3/a = 1/√3. Thus a = 3√3 m.
3. For the other bank, tan 45° = 3/b = 1, giving b = 3 m.
4. The projection lies between the banks, so the width is their sum: a + b = 3√3 + 3 = 3(√3 + 1) m.

Key takeaways

• Measure elevation and depression from the horizontal through the observer's eye, identifying the particular point being viewed.
• Choose the trigonometric ratio that connects the known length with the required length in the labelled triangle.
• A ladder or taut rope is the hypotenuse when it slopes between vertical support and horizontal ground.
• A calculation from eye level gives the height above the eyes; add eye height when the ground levels match.
• For a building with a flagstaff, the larger triangle includes both heights and shares the smaller triangle's horizontal distance.
• When two shadows differ by a stated length, express the longer as the shorter plus that difference.
• Depression and elevation angles along the same sight line are equal because the corresponding horizontal reference lines are parallel.
• Use the arrangement to decide whether horizontal distances add or subtract, and keep exact values until approximation is needed.

Test yourself

What joins the observer's eye to the point being viewed?

The line of sight joins those two points and determines the viewing direction.

Which two sides appear in the tangent ratio of an acute angle?

Tangent is the opposite side divided by the adjacent side, with both identified relative to that angle.

Why can a depression angle be used as an elevation angle at the other end of the sight line?

The horizontal reference lines are parallel, so the sight line creates equal alternate interior angles.

A taut 20 m rope makes 30° with level ground and reaches a vertical pole's top. Using sin 30° = 1/2, what is the height?

The height is opposite the angle, so it equals 20 × 1/2 = 10 m.

What must be added to a chimney's height above the observer's eyes to obtain its total height on the same level ground?

Add the observer's eye height above the ground to the calculated upper segment.

Why does a flagstaff-on-building problem use two vertical heights?

One sight line reaches the building's top; the other reaches the top of the combined building and flagstaff.

When does the sum of two horizontal distances give a river's width?

When the observation point's vertical projection lies between the two bank points in the cross-section being measured.

What does “not defined” mean in a trigonometric table?

The ratio has no numerical value at that angle; the entry must not be treated as zero.