ICSE Class 10 Maths Heights and Distances Solutions
ICSE Class 10 Maths Heights and Distances: direct summary
ICSE Class 10 Maths Chapter 22 Heights and Distances uses trigonometric ratios to find heights, widths and distances that cannot be measured directly. The main skill is to convert a word problem into a right-angled triangle, identify the angle of elevation or depression, choose the correct ratio, and solve the unknown length with proper units.
This replacement page is written for the Selina Concise Mathematics Class 10 ICSE treatment of Heights and Distances. It explains the method first, then gives school-style worked solutions for the published exercise questions available from the existing page, followed by teacher-made practice examples for the same chapter type.
Formula and method reference for Concise Mathematics Selina Solutions Class 10 ICSE Chapter 22 Heights and Distances
In this chapter, almost every question reduces to one of the following right-triangle relations. Use the ratio that connects the two sides given in the problem.
| Situation | Use this ratio | Meaning in Heights and Distances |
|---|---|---|
| Height and horizontal distance are involved | \tan \theta=\dfrac{\text{perpendicular}}{\text{base}} | \tan \theta=\dfrac{\text{height}}{\text{distance}} |
| Height and slant length are involved | \sin \theta=\dfrac{\text{perpendicular}}{\text{hypotenuse}} | \sin \theta=\dfrac{\text{height}}{\text{line of sight}} |
| Horizontal distance and slant length are involved | \cos \theta=\dfrac{\text{base}}{\text{hypotenuse}} | \cos \theta=\dfrac{\text{distance}}{\text{line of sight}} |
| Angle of depression from a horizontal line | Use the alternate interior angle with the ground line | The angle of depression from the top equals the angle of elevation from the point on level ground. |
Useful exact values used in this chapter are \tan 30^\circ=\dfrac{1}{\sqrt{3}}, \tan 45^\circ=1, \tan 60^\circ=\sqrt{3}, \sin 45^\circ=\dfrac{1}{\sqrt{2}}, \cos 60^\circ=\dfrac{1}{2}, and \cos 30^\circ=\dfrac{\sqrt{3}}{2}.
Concept snapshot for angles of elevation and depression
Think of the observer’s eye as a hinge and the horizontal line of sight as a straight ruler. If the eye turns upward to see the top of a tower, the angle made with the ruler is the angle of elevation. If the eye turns downward from a cliff to see a buoy, the angle made below the ruler is the angle of depression.
The distance on the ground is normally the base of the right triangle. The height of the tower, pole or cliff is normally the perpendicular. Once you label these two sides, the tangent ratio usually becomes the shortest route.
How to draw the triangle in Heights and Distances
For a word problem, do not start with the formula. Start with the diagram.
- Draw the vertical object as a straight vertical line.
- Draw the ground or water level as a horizontal line.
- Mark the observer’s line of sight as the slanting line.
- Put a right-angle mark between the vertical and horizontal lines.
- Write the given length and angle before choosing the ratio.
An important edge case appears when the observer has a height, such as a boy of height 1.6\,\text{m}. The triangle begins at eye level, not ground level. After solving the height above eye level, add the observer’s height to get the full height of the object.
Exercise 22(A): worked solutions
The following solutions use the question headings and recoverable statements from the existing published page. Each solution is rewritten with full MathJax working and the final answer is stated separately.
Question 1(a): Find x when \tan 60^\circ=\dfrac{15}{x}
Step 1: Use the tangent ratio because the perpendicular and base are involved.
\tan 60^\circ=\frac{15}{x}
Step 2: Substitute \tan 60^\circ=\sqrt{3}.
\sqrt{3}=\frac{15}{x}
Step 3: Cross-multiply and simplify.
x=\frac{15}{\sqrt{3}}=\frac{15}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=5\sqrt{3}
Final answer: x=5\sqrt{3}\,\text{cm}. The correct option is 5\sqrt{3}\,\text{cm}.
Question 1(b): Find x when the hypotenuse is 40\,\text{m} and the angle is 60^\circ
Step 1: Use the cosine ratio because the base and hypotenuse are involved.
\cos 60^\circ=\frac{x}{40}
Step 2: Substitute \cos 60^\circ=\dfrac{1}{2}.
\frac{1}{2}=\frac{x}{40}
Step 3: Solve for x.
x=\frac{40}{2}=20
Final answer: x=20\,\text{m}. The correct option is 20\,\text{m}.
Question 1(c): Given AB=BD, BC=20\,\text{cm} and \angle D=45^\circ, find AC
Step 1: In \triangle BCD, use the sine ratio.
\sin 45^\circ=\frac{BC}{BD}
Step 2: Substitute BC=20\,\text{cm} and \sin 45^\circ=\dfrac{1}{\sqrt{2}}.
\frac{1}{\sqrt{2}}=\frac{20}{BD}
Step 3: Solve for BD.
BD=20\sqrt{2}\,\text{cm}
Step 4: Since AB=BD, AB=20\sqrt{2}\,\text{cm}. Hence
AC=AB+BC=20\sqrt{2}+20
AC=20(1.414)+20=28.28+20=48.28\,\text{cm}
Final answer: AC=48.28\,\text{cm}. The correct option is 48.28\,\text{cm}.
Question 1(d): Find x to the nearest metre when x=70\cos 30^\circ
Step 1: Use the cosine ratio.
\cos 30^\circ=\frac{x}{70}
Step 2: Substitute \cos 30^\circ=\dfrac{\sqrt{3}}{2}.
\frac{\sqrt{3}}{2}=\frac{x}{70}
Step 3: Solve and round only at the end.
x=70\times\frac{\sqrt{3}}{2}=35\sqrt{3}=60.62\ldots
Final answer: x\approx61\,\text{m}. The correct option is 61\,\text{m}.
Question 1(e): Find DC when AB=10\,\text{m}, \angle C=45^\circ and \angle D=30^\circ
Step 1: In \triangle ABC, use \tan 45^\circ=\dfrac{AB}{BC}.
1=\frac{10}{BC}
BC=10\,\text{m}
Step 2: In \triangle ABD, use \tan 30^\circ=\dfrac{AB}{BD}.
\frac{1}{\sqrt{3}}=\frac{10}{BD}
BD=10\sqrt{3}\,\text{m}
Step 3: Subtract the shorter base from the longer base.
DC=BD-BC=10\sqrt{3}-10=10(\sqrt{3}-1)\,\text{m}
Final answer: DC=10(\sqrt{3}-1)\,\text{m}. The correct option is 10(\sqrt{3}-1)\,\text{m}.
Question 2: The height of a tree is \sqrt{3} times the length of its shadow. Find the angle of elevation of the sun.
Step 1: Let the length of the shadow be x\,\text{m}. Then the height of the tree is \sqrt{3}x\,\text{m}.
Step 2: Let the angle of elevation of the sun be \theta. Use tangent.
\tan\theta=\frac{\text{height}}{\text{shadow}}=\frac{\sqrt{3}x}{x}=\sqrt{3}
Step 3: Compare with the standard value \tan 60^\circ=\sqrt{3}.
\theta=60^\circ
Final answer: The angle of elevation of the sun is 60^\circ.
Question 3: A tower is 160\,\text{m} away and the angle of elevation is 60^\circ. Find the height.
Step 1: Let the height of the tower be h\,\text{m}.
Step 2: Use the tangent ratio.
\tan 60^\circ=\frac{h}{160}
Step 3: Substitute \tan 60^\circ=\sqrt{3}.
\sqrt{3}=\frac{h}{160}
h=160\sqrt{3}
Step 4: If a decimal value is required, use \sqrt{3}\approx1.732.
h=160\times1.732=277.12
Final answer: The height of the tower is 160\sqrt{3}\,\text{m}, i.e. approximately 277.12\,\text{m}.
Question 4: A ladder is 2.4\,\text{m} from a wall and makes 68^\circ with the ground. Find the height reached.
Step 1: Let the height reached on the wall be h\,\text{m}.
Step 2: Since the ground distance is 2.4\,\text{m}, use tangent.
\tan 68^\circ=\frac{h}{2.4}
Step 3: Use \tan 68^\circ\approx2.475.
2.475=\frac{h}{2.4}
h=2.4\times2.475=5.94
Final answer: The ladder reaches a height of 5.94\,\text{m}.
Question 5: Two persons stand on opposite sides of a 50\,\text{m} tower. Their angles of elevation are 30^\circ and 38^\circ. Find the distance between them.
Step 1: Let the distances from the foot of the tower be x and y metres.
Step 2: For the person seeing the top at 30^\circ,
\tan 30^\circ=\frac{50}{x}
\frac{1}{\sqrt{3}}=\frac{50}{x}
x=50\sqrt{3}=86.60\,\text{m}
Step 3: For the person seeing the top at 38^\circ, use \tan 38^\circ\approx0.7813.
\tan 38^\circ=\frac{50}{y}
0.7813=\frac{50}{y}
y=\frac{50}{0.7813}=64.00\,\text{m}
Step 4: Since the persons are on opposite sides, add the distances.
x+y=86.60+64.00=150.60\,\text{m}
Final answer: The distance between the two persons is 150.6\,\text{m}.
Question 6: A boy 1.6\,\text{m} tall is 20\,\text{m} from a tower. Find the tower height when the angle of elevation is 45^\circ and 60^\circ.
Step 1: The angle is measured from the boy’s eye level. Let the height above eye level be h\,\text{m}.
Step 2: For 45^\circ,
\tan 45^\circ=\frac{h}{20}
1=\frac{h}{20}
h=20\,\text{m}
Step 3: Add the boy’s height.
\text{Tower height}=20+1.6=21.6\,\text{m}
Step 4: For 60^\circ,
\tan 60^\circ=\frac{h}{20}
\sqrt{3}=\frac{h}{20}
h=20\sqrt{3}=34.64\,\text{m}
Step 5: Add the boy’s height again.
\text{Tower height}=34.64+1.6=36.24\,\text{m}
Final answer: The tower heights are 21.6\,\text{m} for 45^\circ and 36.24\,\text{m} for 60^\circ.
Question 7: A broken tree makes 45^\circ with the ground and the top touches the ground 15\,\text{m} from the root. Find the original height.
Step 1: Let AB be the standing part and AC the broken part. The ground distance is BC=15\,\text{m}.
Step 2: Use \tan 45^\circ=\dfrac{AB}{BC}.
1=\frac{AB}{15}
AB=15\,\text{m}
Step 3: Use Pythagoras to find the broken part AC.
AC^2=AB^2+BC^2=15^2+15^2=450
AC=\sqrt{450}=15\sqrt{2}\,\text{m}
Step 4: Original height equals standing part plus broken part.
AB+AC=15+15\sqrt{2}=36.21\,\text{m}
Final answer: The height of the tree before it broke was 15+15\sqrt{2}\,\text{m}, approximately 36.21\,\text{m}.
Question 8: An unfinished tower subtends 30^\circ from 80\,\text{m}. How much higher must it be raised so that the angle becomes 60^\circ?
Step 1: Let the present height be h_1. Use \tan 30^\circ=\dfrac{h_1}{80}.
\frac{1}{\sqrt{3}}=\frac{h_1}{80}
h_1=\frac{80}{\sqrt{3}}\,\text{m}
Step 2: Let the new height be h_2. Use \tan 60^\circ=\dfrac{h_2}{80}.
\sqrt{3}=\frac{h_2}{80}
h_2=80\sqrt{3}\,\text{m}
Step 3: Required extra height is h_2-h_1.
h_2-h_1=80\sqrt{3}-\frac{80}{\sqrt{3}}
=138.56-46.19=92.37\,\text{m}
Final answer: The tower must be raised by approximately 92.37\,\text{m}.
Question 9: When the sun’s altitude is 30^\circ, the shadow of a tower is 45\,\text{m}. Find the height and the shadows for 45^\circ and 60^\circ.
Step 1: Let the tower height be h.
\tan 30^\circ=\frac{h}{45}
\frac{1}{\sqrt{3}}=\frac{h}{45}
h=\frac{45}{\sqrt{3}}=15\sqrt{3}=25.98\,\text{m}
Step 2: When the sun’s altitude is 45^\circ, let the shadow be s_1.
\tan 45^\circ=\frac{h}{s_1}
1=\frac{15\sqrt{3}}{s_1}
s_1=15\sqrt{3}=25.98\,\text{m}
Step 3: When the sun’s altitude is 60^\circ, let the shadow be s_2.
\tan 60^\circ=\frac{h}{s_2}
\sqrt{3}=\frac{15\sqrt{3}}{s_2}
s_2=15\,\text{m}
Final answer: Height of tower =15\sqrt{3}\,\text{m} or 25.98\,\text{m}; shadow at 45^\circ=25.98\,\text{m}; shadow at 60^\circ=15\,\text{m}.
Question 10: A 30\,\text{m} ladder rests against two poles on opposite sides of a road. The angles are 32^\circ24' with the pole and 32^\circ24' with the road. Find the road width.
Step 1: Convert the information into two horizontal projections of the same ladder.
Step 2: In the first position, the ladder makes 32^\circ24' with the vertical pole. The horizontal part is opposite this angle.
\text{First horizontal distance}=30\sin 32^\circ24'
=30(0.536)=16.08\,\text{m}
Step 3: In the second position, the ladder makes 32^\circ24' with the road. The horizontal part is adjacent to this angle.
\text{Second horizontal distance}=30\cos 32^\circ24'
=30(0.844)=25.32\,\text{m}
Step 4: Add both parts to get the width of the road.
16.08+25.32=41.40\,\text{m}
Final answer: The width of the road is 41.4\,\text{m}.
Question 11: Two climbers are on a vertical cliff. From a point 40\,\text{m} from the foot, their elevations are 48^\circ and 57^\circ. Find the distance between them.
Step 1: Let the lower climber’s height from the foot be h_1 and the upper climber’s height be h_2.
Step 2: For the upper climber, use \tan 57^\circ\approx1.539.
h_2=40\tan 57^\circ=40(1.539)=61.56\,\text{m}
Step 3: For the lower climber, use \tan 48^\circ\approx1.110.
h_1=40\tan 48^\circ=40(1.110)=44.40\,\text{m}
Step 4: Subtract the heights.
h_2-h_1=61.56-44.40=17.16\,\text{m}
Final answer: The distance between the climbers is approximately 17.17\,\text{m} depending on the trigonometric table rounding used.
Question 12: A man stands 9\,\text{m} from a flag-pole. The angle of elevation of the top is 28^\circ and the angle of depression of the bottom is 13^\circ. Find the pole height.
Step 1: The pole extends above and below the man’s eye level. Find both parts and add them.
Step 2: Height above eye level:
\tan 28^\circ=\frac{h_1}{9}
h_1=9\tan 28^\circ=9(0.532)=4.788\,\text{m}
Step 3: Height below eye level:
\tan 13^\circ=\frac{h_2}{9}
h_2=9\tan 13^\circ=9(0.231)=2.079\,\text{m}
Step 4: Add the two parts.
h_1+h_2=4.788+2.079=6.867\,\text{m}
Final answer: The height of the flag-pole is approximately 6.867\,\text{m}.
Question 13: From the top of a 92\,\text{m} cliff, the angle of depression of a buoy is 20^\circ. Find the distance of the buoy from the foot of the cliff to the nearest metre.
Step 1: The angle of depression from the cliff top equals the angle of elevation from the buoy to the cliff top.
Step 2: Let the horizontal distance from the foot of the cliff to the buoy be d\,\text{m}.
\tan 20^\circ=\frac{92}{d}
Step 3: Use \tan 20^\circ\approx0.3640.
0.3640=\frac{92}{d}
d=\frac{92}{0.3640}=252.75\,\text{m}
Final answer: The distance of the buoy from the foot of the cliff is 253\,\text{m} to the nearest metre.
Teacher-made worked examples for practice
These examples are not copied from the textbook. They are included to strengthen the same methods used in Exercise 22(A), Exercise 22(B), and Test Yourself questions.
Worked Example 1: Height of a tower with two observation points on the same line
A tower stands on level ground. From a point 40\,\text{m} from its foot, the angle of elevation is 60^\circ. From a second point farther away on the same line, the angle is 30^\circ. Find the distance between the two observation points.
Step 1: Let the height of the tower be h and the farther distance be x.
Step 2: From the nearer point,
\tan 60^\circ=\frac{h}{40}
\sqrt{3}=\frac{h}{40}\Rightarrow h=40\sqrt{3}\,\text{m}
Step 3: From the farther point,
\tan 30^\circ=\frac{h}{x}
\frac{1}{\sqrt{3}}=\frac{40\sqrt{3}}{x}\Rightarrow x=120\,\text{m}
Step 4: Distance between the two points is 120-40=80\,\text{m}.
Final answer: The two observation points are 80\,\text{m} apart.
Worked Example 2: Angle of depression from a building
From the top of a 30\,\text{m} building, the angle of depression of a car is 45^\circ. Find the distance of the car from the foot of the building.
Step 1: The angle of depression is equal to the angle of elevation from the car, so use 45^\circ at the car.
Step 2: Let the horizontal distance be d\,\text{m}.
\tan 45^\circ=\frac{30}{d}
1=\frac{30}{d}\Rightarrow d=30
Final answer: The car is 30\,\text{m} from the foot of the building.
Worked Example 3: Height of a pole above eye level
A student of height 1.5\,\text{m} observes the top of a pole at an angle of elevation 30^\circ from a point 18\,\text{m} away. Find the height of the pole.
Step 1: Let the height of the pole above the student’s eye level be h\,\text{m}.
\tan 30^\circ=\frac{h}{18}
\frac{1}{\sqrt{3}}=\frac{h}{18}\Rightarrow h=\frac{18}{\sqrt{3}}=6\sqrt{3}\,\text{m}
Step 2: Add the student’s height.
\text{Pole height}=6\sqrt{3}+1.5
=10.392+1.5=11.892\,\text{m}
Final answer: The pole is approximately 11.89\,\text{m} high.
Examiner’s mindset for ICSE solutions
In ICSE Class 10 Maths, Heights and Distances answers are checked for method, not only for the final number. A clear diagram, the correct trigonometric ratio, substitution of the given data, correct rounding, and the final unit all matter.
- Ratio selection: If the question gives height and horizontal distance, write \tan\theta=\dfrac{\text{height}}{\text{distance}}.
- Angle of depression: State that it equals the corresponding angle of elevation when horizontal lines are parallel.
- Units: Keep metres and centimetres consistent. Do not mix them in one calculation.
- Rounding: Do not round \sqrt{3} or trigonometric values too early. Round at the last line when the question asks for a nearest value.
Common mistakes students make in Heights and Distances
- Mistake: Using \sin\theta whenever there is an angle. Correction: Choose the ratio from the sides involved. For height and ground distance, use \tan\theta.
- Mistake: Treating angle of depression as a new formula. Correction: Draw the horizontal at the top; the angle of depression equals the angle of elevation from the ground object.
- Mistake: Forgetting the observer’s height. Correction: If the angle is measured from eye level, add the observer’s height after finding the height above eye level.
- Mistake: Adding distances when the objects are on the same side of a tower. Correction: Add only when they are on opposite sides; subtract when they are on the same straight line on one side.
- Mistake: Rounding 160\sqrt{3} too soon. Correction: Keep exact surd form until the final decimal conversion.
Quick answer index
| Exercise | Question | Final answer |
|---|---|---|
| 22(A) | 1(a) | 5\sqrt{3}\,\text{cm} |
| 22(A) | 1(b) | 20\,\text{m} |
| 22(A) | 1(c) | 48.28\,\text{cm} |
| 22(A) | 1(d) | 61\,\text{m} |
| 22(A) | 1(e) | 10(\sqrt{3}-1)\,\text{m} |
| 22(A) | 2 | 60^\circ |
| 22(A) | 3 | 160\sqrt{3}\,\text{m}\approx277.12\,\text{m} |
| 22(A) | 4 | 5.94\,\text{m} |
| 22(A) | 5 | 150.6\,\text{m} |
| 22(A) | 6(i) | 21.6\,\text{m} |
| 22(A) | 6(ii) | 36.24\,\text{m} |
| 22(A) | 7 | 15+15\sqrt{2}\,\text{m}\approx36.21\,\text{m} |
| 22(A) | 8 | 92.37\,\text{m} |
| 22(A) | 9(i) | 15\sqrt{3}\,\text{m}\approx25.98\,\text{m} |
| 22(A) | 9(ii)(a) | 15\sqrt{3}\,\text{m}\approx25.98\,\text{m} |
| 22(A) | 9(ii)(b) | 15\,\text{m} |
| 22(A) | 10 | 41.4\,\text{m} |
| 22(A) | 11 | 17.17\,\text{m} approximately |
| 22(A) | 12 | 6.867\,\text{m} approximately |
| 22(A) | 13 | 253\,\text{m} to the nearest metre |
Related study links for ICSE Class 10 Maths
For chapter-level revision, use the ICSE Class 10 solutions page and the ICSE solutions index. Students revising the same textbook can also use Selina Concise Maths Class 10 solutions for neighbouring chapters.
For syllabus alignment, refer to the official CISCE website and the prescribed Selina Concise Mathematics Class 10 textbook used by the school.
Frequently Asked Questions
Which trigonometric ratio is used most often in ICSE Class 10 Maths heights and distances?
The tangent ratio is used most often because most questions give or ask for a vertical height and a horizontal distance. Use \tan \theta=\frac{\text{height}}{\text{horizontal distance}} when the triangle is right-angled and the angle is measured from the horizontal.
What is the difference between angle of elevation and angle of depression?
An angle of elevation is measured upward from the horizontal line of sight. An angle of depression is measured downward from the horizontal line of sight. In many Heights and Distances questions, the angle of depression from the top equals the angle of elevation from the object because the two horizontal lines are parallel.
How should I start a Selina Heights and Distances problem?
First draw a right-angled triangle, mark the vertical height and the horizontal distance, then choose \tan, \sin, or \cos according to the two sides involved. In most ICSE Class 10 Maths word problems, the correct diagram decides the rest of the solution.
Should answers be left in surd form or decimal form?
Leave the answer in exact surd form such as 160\sqrt{3}\,\text{m} unless the question asks for a decimal value or an answer to the nearest metre. If a decimal is needed, round only at the final step.
Why is the observer’s height added in some tower problems?
The angle of elevation from a person’s eye measures the height above the eye level, not always the full height from the ground. If the observer is 1.6\,\text{m} tall, add 1.6\,\text{m} after finding the height above eye level.