CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry
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Boards begin February 2027
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Some Applications of Trigonometry is in Trigonometry, which carries 12 of the 80 marks in the 2026-27 CBSE Maths paper
11 NCERT textbook questions from Chapter 9
Exercise 9.1
NCERT Ex 9.1 Q12 marks
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the rope makes an angle of 30° with the ground.
Step 1Right triangle: hypotenuse = rope = 20 m; angle with ground = 30°; opposite side = height h.
Step-by-step solution
Right triangle: hypotenuse = rope = 20 m; angle with ground = 30°; opposite side = height h.
A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree and the point where the top touches the ground is 8 m. Find the height of the tree.
Step 1Let the unbroken part standing upright be BC = h (vertical), and the broken/leaning part CA = ℓ. The broken top touches the ground at A, 8 m from B (foot).
Step-by-step solution
Let the unbroken part standing upright be BC = h (vertical), and the broken/leaning part CA = ℓ. The broken top touches the ground at A, 8 m from B (foot).
Angle at A = 30°; right angle is at B.
tan 30° = BC/BA ⇒ 1/√3 = h/8 ⇒ h = 8/√3 = 8√3/3 m.
cos 30° = BA/CA ⇒ √3/2 = 8/ℓ ⇒ ℓ = 16/√3 = 16√3/3 m.
Total height of tree = h + ℓ = 8√3/3 + 16√3/3 = 24√3/3 = 8√3 m.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m from the foot of the tower, is 30°. Find the height of the tower.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground; the inclination of the string with the ground is 60°. Find the length of the string assuming no slack.
Step 1Let string length = ℓ. Height of kite = 60 m. Angle of inclination with ground = 60°.
Step-by-step solution
Let string length = ℓ. Height of kite = 60 m. Angle of inclination with ground = 60°.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Step 1Height available above eye-level = 30 − 1.5 = 28.5 m.
Step-by-step solution
Height available above eye-level = 30 − 1.5 = 28.5 m.
Let the initial horizontal distance from eye to top of building be x and final be y. Walking forward gives 30° → 60°.
At 30°: tan 30° = 28.5/x ⇒ x = 28.5 √3.
At 60°: tan 60° = 28.5/y ⇒ y = 28.5/√3 = 28.5√3/3 = 9.5√3.
Distance walked = x − y = 28.5√3 − 9.5√3 = 19√3 m.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Step 1Let the building be AB = 20 m and the tower be BC = h on top of it. Let P be the point on the ground at distance d from A. ∠APB = 45° (to top of building) and ∠APC = 60° (to top of tower).
Step-by-step solution
Let the building be AB = 20 m and the tower be BC = h on top of it. Let P be the point on the ground at distance d from A. ∠APB = 45° (to top of building) and ∠APC = 60° (to top of tower).
The angle of elevation of the top of a building from the foot of the tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Step 1Let the building be PQ (height = h) and the tower be RS (height = 50 m). The distance between their feet is d.
Step-by-step solution
Let the building be PQ (height = h) and the tower be RS (height = 50 m). The distance between their feet is d.
From foot of tower (R), angle of elevation of top of building (P) = 30°: tan 30° = h/d ⇒ 1/√3 = h/d ⇒ d = h√3. …(1)
From foot of building (Q), angle of elevation of top of tower (S) = 60°: tan 60° = 50/d ⇒ √3 = 50/d ⇒ d = 50/√3. …(2)
Equate (1) and (2): h√3 = 50/√3 ⇒ 3h = 50 ⇒ h = 50/3.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from her eyes at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.
Step 1Effective vertical height above eye level = 88.2 − 1.2 = 87 m (balloon moves along this horizontal line).
Step-by-step solution
Effective vertical height above eye level = 88.2 − 1.2 = 87 m (balloon moves along this horizontal line).
Let x₁ = horizontal distance from girl's eyes to balloon when angle = 60°; x₂ = horizontal distance when angle = 30°.
A tower stands vertically on the ground. From a point on the ground, 15 m from the foot of the tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower.
Step 1Picture: right triangle with horizontal leg = 15 m (distance from point to tower foot), angle of elevation 60° at the observer's foot, vertical leg = height of tower h.
Step-by-step solution
Picture: right triangle with horizontal leg = 15 m (distance from point to tower foot), angle of elevation 60° at the observer's foot, vertical leg = height of tower h.
From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (Take √3 = 1.732.)
Step 1Let A be the foot of the building, B the top of the building, D the top of the flagstaff; P is on the ground. AB = 10 m (building), BD = x (flagstaff). PA = distance from P to building foot.
Step-by-step solution
Let A be the foot of the building, B the top of the building, D the top of the flagstaff; P is on the ground. AB = 10 m (building), BD = x (flagstaff). PA = distance from P to building foot.
In right ∆PAB: tan 30° = AB/PA ⇒ 1/√3 = 10/PA ⇒ PA = 10√3 m = 10(1.732) = 17.32 m.
In right ∆PAD: tan 45° = AD/PA = (10 + x)/PA. Since tan 45° = 1, AD = PA = 10√3.
10 + x = 10√3 ⇒ x = 10√3 − 10 = 10(√3 − 1) = 10(1.732 − 1) = 10(0.732) = 7.32 m.
So the flagstaff is 7.32 m long and the building is 17.32 m from P.
Answer:Length of flagstaff = 7.32 m; distance of building from P = 17.32 m.
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Questions from the NCERT Class 10 Mathematics textbook; the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.