Class 10 Maths · Original EZT practice
Some Applications of Trigonometry (Class 10 Chapter 9): summary and practice
Some Applications of Trigonometry turns angles into measurements: line of sight, angle of elevation and depression, and the right triangle that hides inside every heights-and-distances word problem. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Angles of elevation and depression
The line of sight joins the observer's eye to the object. Above the horizontal it makes an angle of elevation; below it, an angle of depression.
2. Building the right triangle
Height goes on the vertical side, ground distance on the horizontal, and the line of sight becomes the hypotenuse. tan links height and distance directly, which is why it appears most often.
3. Standard values
The chapter runs on 30, 45 and 60 degrees: tan 30 = 1/sqrt(3), tan 45 = 1, tan 60 = sqrt(3). At 45 degrees the height equals the distance.
4. Two-position problems
Two observations from two points on one line give two triangles sharing the same height - write tan for each and solve the pair together.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Find a height from a distance and angle
Choose the method: height = distance x tan(angle of elevation).
EZT practice question
From a point 30 m from the foot of a tower, the angle of elevation of the top is 60 degrees. Find the height.
Show the worked solution
- tan 60 = height / 30.
- height = 30 x tan 60 = 30 x sqrt(3).
- sqrt(3) is about 1.732.
Answer: 30 sqrt(3) m, about 51.96 m (roughly 52 m).
Check: A 60-degree angle is steep, so the tower should be TALLER than its distance: 52 > 30, as expected.
Watch for this: tan is opposite over adjacent - height over distance. Swapping the ratio gives 17.3 m, which fails the steepness check.
Now try it yourself
Angle of elevation 45 degrees from 20 m away - height?
Check your answer
20 m.
tan 45 = 1, so height equals distance.
Find a distance from a height and angle
Choose the method: distance = height / tan(angle of elevation).
EZT practice question
The angle of elevation of the top of a 12 m building is 30 degrees. How far is the observer from the foot?
Show the worked solution
- tan 30 = 12 / distance.
- distance = 12 / tan 30 = 12 x sqrt(3).
- 12 x 1.732.
Answer: 12 sqrt(3) m, about 20.78 m.
Check: A shallow 30-degree angle means the observer stands FARTHER than the height: 20.78 > 12, as expected.
Watch for this: Dividing by tan 30 multiplies by sqrt(3) - do not multiply by tan 30 instead.
Now try it yourself
A 10 m pole at 45 degrees - distance?
Check your answer
10 m.
tan 45 = 1.
Use sin with the hypotenuse
Choose the method: When the slant length is given, sin(angle) = height / string.
EZT practice question
A kite flies with 75 m of string out, the string making 60 degrees with the ground. How high is the kite?
Show the worked solution
- The string is the hypotenuse.
- sin 60 = height / 75.
- height = 75 x sin 60 = 75 x 0.866 (sqrt(3)/2).
Answer: 37.5 sqrt(3) m, about 64.95 m.
Check: The height must be less than the string length: 64.95 < 75, as required.
Watch for this: sin, not tan - the string length is the HYPOTENUSE, and tan never involves the hypotenuse.
Now try it yourself
75 m of string at 30 degrees - height?
Check your answer
37.5 m.
sin 30 = 1/2.
Shadow problems
Choose the method: The shadow is the horizontal leg; the object is the vertical leg.
EZT practice question
Find the shadow of a 10 m pole when the sun's elevation is 45 degrees.
Show the worked solution
- tan 45 = pole / shadow.
- shadow = 10 / tan 45.
- tan 45 = 1.
Answer: 10 m.
Check: At 45 degrees object and shadow are equal - the classic sun-at-45 rule.
Watch for this: The sun's elevation is measured from the GROUND up, not from the vertical.
Now try it yourself
Shadow of a 10 m pole when the sun is at 30 degrees?
Check your answer
10 sqrt(3) m, about 17.3 m.
shadow = 10 / tan 30.
Tell elevation from depression
Choose the method: Both are measured from the horizontal - elevation looks up, depression looks down.
EZT practice question
From the top of a 40 m tower, the angle of depression of a car is 60 degrees. How far is the car from the foot?
Show the worked solution
- By alternate angles, the angle of elevation from the car to the top is also 60 degrees.
- tan 60 = 40 / distance.
- distance = 40 / sqrt(3).
Answer: 40/sqrt(3) m = 40 sqrt(3)/3 m, about 23.09 m.
Check: Steep 60 degrees from a 40 m tower means the car is CLOSER than the height: 23.09 < 40, as expected.
Watch for this: Angle of depression is measured below the horizontal at the TOP - it equals the elevation from the car by alternate angles.
Now try it yourself
Same tower, depression 45 degrees - distance?
Check your answer
40 m.
tan 45 = 1.
Two angles of elevation
Choose the method: Write tan for each position and combine.
EZT practice question
From one point the elevation of a tower's top is 45 degrees; from 20 m farther away it is 30 degrees. Find the height h.
Show the worked solution
- Near point: h = d (tan 45 = 1).
- Far point: h = (d + 20) tan 30 = (d + 20)/sqrt(3).
- d = (d + 20)/sqrt(3), so d sqrt(3) - d = 20, d(1.732 - 1) = 20.
Answer: d = 20/0.732 is about 27.3 m, and h = d is about 27.3 m.
Check: h = 27.3: from 27.3 m away tan is 27.3/27.3 = 1 (45 degrees); from 47.3 m away 27.3/47.3 = 0.577 = tan 30. Both match.
Watch for this: The two distances differ by exactly 20 m - write d and d + 20, not two unrelated letters.
Now try it yourself
If the far distance is 50 m and the angle there is 30 degrees, find h.
Check your answer
h = 50/sqrt(3), about 28.87 m.
tan 30 = h/50.
Try a mixed set
Choose a method or explanation before opening each answer.
1. tan of the elevation?
Show answer and reason
height / distance.
Opposite over adjacent.
2. tan 45?
Show answer and reason
1 - height equals distance.
The 45-degree right triangle is isosceles.
3. Given the slant, use?
Show answer and reason
sin for height, cos for ground distance.
The slant is the hypotenuse.
4. Angle of depression from the top equals?
Show answer and reason
The angle of elevation from below.
Alternate angles on parallel horizontals.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Kite with 75 m of string at 60 degrees - height.
- Shadow of a 10 m pole with the sun at 45 degrees.
- Distance from a 12 m building whose top shows 30 degrees.
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Questions about this practice page
- What is the difference between an angle of elevation and an angle of depression?
- Both are measured from the horizontal line through the observer's eye. Looking up at an object gives the angle of elevation; looking down from a height gives the angle of depression. By alternate angles, one observer's depression equals the other's elevation.
- Why is tan used most often in heights-and-distances problems?
- Most setups give or ask for the height and the ground distance - the two legs of the right triangle. tan links the legs directly without the hypotenuse, which usually cannot be measured on the ground.
- When do I use sin or cos instead?
- When the slant length enters the problem - a kite string, a ladder, a guy rope. The slant is the hypotenuse, and sin and cos are the ratios that involve the hypotenuse.
- Which standard values should I know for this chapter?
- tan 30 = 1/sqrt(3), tan 45 = 1, tan 60 = sqrt(3), plus sin and cos of the same three angles from Chapter 8. Keep sqrt(3) symbolic until the last step to avoid rounding errors.
- Are these the NCERT exercise questions for Some Applications of Trigonometry (Class 10)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.