CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar Solutions Class 10 Maths Chapter 8 Some Applications of Trigonometry
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Boards begin February 2027
Pre-boards are in December
Some Applications of Trigonometry is in Trigonometry, which carries 12 of the 80 marks in the 2026-27 CBSE Maths paper
10 NCERT Exemplar questions from Chapter 8
Exercise 8.2
Exemplar Ch8 Ex 8.2 Q72 marks
State true or false with justification: 'If the length of the shadow of a tower is increasing, then the angle of elevation of the Sun is also increasing.'
Step 1Let the tower have height h and shadow length s. Then tan(angle of elevation) = h/s.
Step-by-step solution
Let the tower have height h and shadow length s. Then tan(angle of elevation) = h/s.
If s increases (with h fixed), h/s decreases, so tan(elevation) decreases, so the angle decreases.
Hence the statement is FALSE — as the shadow lengthens, the angle of elevation of the Sun DECREASES (think of early morning vs noon).
Answer:False — longer shadow ⇒ smaller angle of elevation (since tan θ = h/s decreases).
State true or false with justification: 'The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, the angle of elevation will also double.'
Step 1If original height is h and distance is d, then tan 30° = h/d ⇒ d = h√3.
Step-by-step solution
If original height is h and distance is d, then tan 30° = h/d ⇒ d = h√3.
After doubling height: new tan(angle) = (2h)/d = 2h/(h√3) = 2/√3.
tan 60° = √3 ≈ 1.732, while 2/√3 ≈ 1.155 ≠ √3.
So the new angle is arctan(2/√3) ≈ 49.1°, not 60°.
Hence the statement is FALSE — doubling the height does NOT double the angle.
Answer:False — doubling h gives tan(new) = 2/√3, not √3 (60°). The angle does not double.
State true or false with justification: 'If both the height of a tower and the distance of the observation point from its foot are increased by 10%, then the angle of elevation of its top remains unchanged.'
Step 1Originally: tan θ = h/d.
Step-by-step solution
Originally: tan θ = h/d.
After 10% increase in both: new tan = (1.1 h)/(1.1 d) = h/d = tan θ.
So the new angle equals the old angle.
Hence the statement is TRUE.
Answer:True — proportional scaling of both legs leaves the ratio (and hence the angle) unchanged.
The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation increases by 15°. Find the height of the tower.
Step 1Let the tower be AB with height h, B the foot. Let the first point be P, the second be Q, with QB nearer.
Step-by-step solution
Let the tower be AB with height h, B the foot. Let the first point be P, the second be Q, with QB nearer.
At P: tan 30° = h / PB ⇒ PB = h/tan 30° = h√3.
At Q (after moving 20 m closer): angle is 30° + 15° = 45°. tan 45° = h/QB ⇒ QB = h.
The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60°, and the angle of elevation of the top of the second tower from the foot of the first is 30°. Find the distance between the two towers and the height of the second tower.
Step 1[1 mark] Diagram: two towers on the same horizontal plane — Tower 1 of height 30 m, Tower 2 of height h, distance between their feet = d.
Step-by-step solution
[1 mark] Diagram: two towers on the same horizontal plane — Tower 1 of height 30 m, Tower 2 of height h, distance between their feet = d.
[1 mark] From the foot of Tower 2, the elevation of the top of Tower 1 is 60°: tan 60° = 30/d ⇒ √3 = 30/d.
[1 mark] So d = 30/√3 = 10√3 m ≈ 17.32 m.
[1 mark] From the foot of Tower 1, the elevation of the top of Tower 2 is 30°: tan 30° = h/d ⇒ 1/√3 = h/(10√3).
[1 mark] h = 10√3/√3 = 10 m. Hence the distance between the towers is 10√3 m and the height of the second tower is 10 m.
Answer:Distance between towers = 10√3 m ≈ 17.32 m; height of second tower = 10 m.
The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.
Step 1Let the tower be AB with height h, foot at B. Let P be the ground point and Q be the point 10 m vertically above P.
Step-by-step solution
Let the tower be AB with height h, foot at B. Let P be the ground point and Q be the point 10 m vertically above P.
Let horizontal distance from P (and from Q, since Q is directly above P) to B = d.
From P: tan 60° = h/d ⇒ √3 = h/d ⇒ d = h/√3. …(1)
From Q: the height of the tower as seen from Q is (h − 10) and horizontal distance is still d. tan 45° = (h − 10)/d ⇒ 1 = (h − 10)/d ⇒ d = h − 10. …(2)
Equate (1) and (2): h/√3 = h − 10 ⇒ h = √3(h − 10) ⇒ h = h√3 − 10√3 ⇒ h(√3 − 1) = 10√3 ⇒ h = 10√3/(√3 − 1).
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Case-based4 marks
The lower window of a house is 2 m above the ground and the upper window is 4 m vertically above the lower window. At a certain instant the angles of elevation of a balloon from these two windows are 60° and 30° respectively.
(a) Set up the geometry: let the balloon be at height H above the ground; let horizontal distance from house to balloon = d.
(b) Write the tan equations for both windows.
(c) Find the height of the balloon above the ground.
Step 1Lower window height = 2 m; upper window height = 2 + 4 = 6 m above the ground.
Step-by-step solution
Lower window height = 2 m; upper window height = 2 + 4 = 6 m above the ground.
From the lower window the balloon is at vertical distance (H − 2) and horizontal distance d; angle = 60°. tan 60° = (H − 2)/d ⇒ √3 d = H − 2. …(1)
From the upper window the balloon is at vertical distance (H − 6) and horizontal distance d; angle = 30°. tan 30° = (H − 6)/d ⇒ d = √3 (H − 6). …(2)
Substitute (2) in (1): √3 · √3 (H − 6) = H − 2 ⇒ 3(H − 6) = H − 2 ⇒ 3H − 18 = H − 2 ⇒ 2H = 16 ⇒ H = 8.
So the balloon is 8 m above the ground.
Answer:Height of the balloon above the ground = 8 m.
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Questions from the NCERT Class 10 Mathematics Exemplar Problems (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.