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Some Applications of Trigonometry: 3 marks Questions (CBSE Class 10)

20 different 3 marks questions on Some Applications of Trigonometry from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (30)2 marks (3)3 marks (20)4 marks (25)5 marks (73)

Q93 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/1/1

In Figure 2, the angles of elevation of the top of a tower AB of height ‘h’ m, from two points P and Q at a distance of x m and y m from the base of the tower respectively and in the same straight line with it, are $6 0^{\circ}$ and $3 0^{\circ}$ , respectively. Prove that $h^{2} = x y$ .

Diagram for CBSE 2022 Class 10 Maths question 9

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Answer: Proved.

In right $△ B A P$ : $tan 6 0^{\circ} = \frac{A B}{A P} = \frac{h}{x}$ , so $h = 3 x$ ... (1)
In right $△ B A Q$ : $tan 3 0^{\circ} = \frac{A B}{A Q} = \frac{h}{y}$ , so $h = \frac{y}{3}$ ... (2)
Multiplying (1) and (2): $h^{2} = 3 x \times \frac{y}{3} = x y$
Hence $h^{2} = x y$ .

Also asked in: 2022 Basic 430/1/3

Q103 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/1/2

A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is $6 0^{\circ}$ . Find the height of the tower.

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Answer:

153

\approx 25.98

Let AB be the tower and C the point on the ground, with BC = 15 m
$tan 6 0^{\circ} = \frac{A B}{B C}$
$3 = \frac{A B}{15}$
$A B = 153 \approx 25.98$ m

Q83 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/2/1

As observed from the top of a light house 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from $3 0^{\circ}$ to $4 5^{\circ}$ . Determine the distance travelled by the ship during this time.
(Use $3$ = 1.73)

Diagram for CBSE 2022 Class 10 Maths question 8

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Answer: 73 m

Let AB = 100 m be the light house, D and C the initial and final positions of the ship.
$∠ A D B = 3 0^{\circ}$ and $∠ A C B = 4 5^{\circ}$ (alternate angles).
In $△ A B C$ : $tan 4 5^{\circ} = \frac{A B}{B C} \Rightarrow B C = 100$ m.
In $△ A B D$ : $tan 3 0^{\circ} = \frac{A B}{B D} \Rightarrow B D = 1003 = 173$ m.
Distance travelled CD = BD − BC = 173 − 100 = 73 m.

Also asked in: 2022 Basic 430/2/2, 2022 Basic 430/2/3

Q8 (OR) (OR)3 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/2/1Not in current syllabus

At a point on level ground, the angle of elevation of a vertical tower is, found to be $α$ such that $tan α = \frac{1}{3}$ . After walking 100 m towards the tower, the angle of elevation $β$ becomes such that $tan β = \frac{3}{4}$ . Find the height of the tower.

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Answer: 60 m

Let the height be $h$ and the distance of the second point from the foot be $x$ .
$tan β = \frac{h}{x} = \frac{3}{4} \Rightarrow x = \frac{4 h}{3}$ .
$tan α = \frac{h}{x + 100} = \frac{1}{3} \Rightarrow 3 h = x + 100$ .
$3 h = \frac{4 h}{3} + 100 \Rightarrow \frac{5 h}{3} = 100 \Rightarrow h = 60$ .
Height of the tower = 60 m.

Also asked in: 2022 Basic 430/2/2, 2022 Basic 430/2/3

Q83 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/3/1

Two poles of equal heights are standing opposite each other on either side of the road of width 60 m. From a point P between them on the road, the angles of elevation of the top of the poles are $6 0^{\circ}$ and $3 0^{\circ}$ respectively, as shown in Figure 3. Find the height of the poles and distances of the point from the poles.

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Answer: Height of each pole

= 153

\approx 25.98

m; distances of P from the poles are 15 m (from CD) and 45 m (from AB)

In right $△ D C P$ : $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow h = 3 x$
In right $△ B A P$ : $tan 3 0^{\circ} = \frac{h}{y} \Rightarrow y = 3 h = 3 x$
$x + y = 60 \Rightarrow 4 x = 60 \Rightarrow x = 15$ m, $y = 45$ m
$h = 153 \approx 25.98$ m

Q73 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/3/2

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 $4 5^{\circ}$ and $6 0^{\circ}$ respectively as shown in Figure 3. Find the height of the transmission tower.

Diagram for CBSE 2022 Class 10 Maths question 7

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Answer: Height of tower

= 20 (3 - 1)

\approx 14.64

In right $△ B C D$ : $tan 4 5^{\circ} = \frac{B C}{C D} \Rightarrow C D = B C = 20$ m
In right $△ A C D$ : $tan 6 0^{\circ} = \frac{A C}{C D} \Rightarrow A C = 203$ m
Height of tower AB $= A C - B C = 203 - 20 = 20 (3 - 1) \approx 20 \times 0.732 = 14.64$ m

Q83 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/3/3

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ as shown in Figure 3. Determine the height of the tower.

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Answer: Height of tower

= 7 (3 + 1)

\approx 19.12

ED = 7 m is the building and AB the tower; CE $∥$ BD, so BC = ED = 7 m.
In right $△ B C E$ : $tan 4 5^{\circ} = \frac{B C}{C E} \Rightarrow C E = 7$ m
In right $△ A C E$ : $tan 6 0^{\circ} = \frac{A C}{C E} \Rightarrow A C = 73$ m
AB = AC + BC $= 73 + 7 = 7 (3 + 1) \approx 19.12$ m

Q103 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/4/1

Two poles of heights 25 m and 35 m stand vertically on the ground. The tops of two poles are connected by a wire, which is inclined to the horizontal at an angle of $3 0^{\circ}$ . Find the length of the wire and the distance between the poles.

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Answer: Length of wire = 20 m; distance between poles

= 103

\approx 17.32

Difference in heights $= 35 - 25 = 10$ m; this is the vertical side of a right triangle whose hypotenuse is the wire $l$ .
$sin 3 0^{\circ} = \frac{10}{l} \Rightarrow \frac{1}{2} = \frac{10}{l} \Rightarrow l = 20$ m.
Horizontal distance $d$ : $tan 3 0^{\circ} = \frac{10}{d} \Rightarrow d = 103$ m.
$d \approx 17.32$ m.

Also asked in: 2022 Basic 430/4/2

Q103 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Basic 430/4/3

The string of a kite is 100 metres long and it makes an angle of $6 0^{\circ}$ with the horizontal. Find the height of the kite, assuming that there is no slack in the string. [Use $3 = 1.73$ ]

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Answer:

503

m = 86.5 m

Let the height be $h$ m; the string (100 m) is the hypotenuse.
$sin 6 0^{\circ} = \frac{h}{100} \Rightarrow h = 100 \times \frac{3}{2} = 503$ .
$h = 50 \times 1.73 = 86.5$ m.

Q93 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/1/1

The angle of elevation of the top of a building from the foot of the tower is $3 0^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $6 0^{\circ}$ . If the tower is 50 m high, then find the height of the building.

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Answer:

\frac{50}{3}

= 16 \frac{2}{3}

Let the tower AB = 50 m and the building CD = $h$ m, with feet B and D a distance $d$ apart.
From D: $tan 6 0^{\circ} = \frac{50}{d} \Rightarrow d = \frac{50}{3}$ m.
From B: $tan 3 0^{\circ} = \frac{h}{d} \Rightarrow h = \frac{d}{3} = \frac{50}{3}$ m.
Height of the building $= 16 \frac{2}{3}$ m (about 16.67 m).

Also asked in: 2022 Standard 30/1/2, 2022 Standard 30/1/3

Q9 (OR) (OR)3 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/1/1

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. If the bridge is at a height of 3 m from the banks, then find the width of the river.

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Answer:

3 (1 + 3)

\approx 8.19

Let P be the point on the bridge, 3 m above the banks, and A, B the banks on opposite sides; D is the point directly below P.
Angle of depression $3 0^{\circ}$ to A: $tan 3 0^{\circ} = \frac{3}{A D} \Rightarrow A D = 33$ m.
Angle of depression $4 5^{\circ}$ to B: $tan 4 5^{\circ} = \frac{3}{D B} \Rightarrow D B = 3$ m.
Width AB $= 33 + 3 = 3 (1 + 3)$ m $\approx 8.19$ m.

Also asked in: 2022 Standard 30/1/2, 2022 Standard 30/1/3

Q83 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/2/1

In Fig. 3, AB is tower of height 50 m. A man standing on its top, observes two cars on the opposite sides of the tower with angles of depression $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the distance between the two cars.

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Answer:

50 (3 + 1)

\approx 136.6

Angles of depression equal the angles of elevation from the cars: $∠ A C B = 3 0^{\circ}$ , $∠ A D B = 4 5^{\circ}$ .
In right $△ A B C$ : $tan 3 0^{\circ} = \frac{A B}{B C} \Rightarrow B C = 503$ m.
In right $△ A B D$ : $tan 4 5^{\circ} = \frac{A B}{B D} \Rightarrow B D = 50$ m.
Distance $C D = B C + B D = 503 + 50 = 50 (3 + 1)$ m $\approx 136.6$ m.

Q93 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/2/2

Two men on either side of a cliff 75 m high observe the angles of elevation of the top of the cliff to be $3 0^{\circ}$ and $6 0^{\circ}$ . Find the distance between the two men.

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Answer:

1003

\approx 173.2

In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{A C} \Rightarrow A C = 753$ m.
In right $△ B A D$ : $tan 6 0^{\circ} = \frac{A B}{A D} \Rightarrow A D = \frac{75}{3} = 253$ m.
Distance $C D = A C + A D = 753 + 253 = 1003$ m $\approx 173.2$ m.

Q83 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/2/3

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $3 0^{\circ}$ and $4 5^{\circ}$ . If the bridge is at a height of 8 m from the banks, then find the width of the river.

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Answer:

8 (1 + 3)

\approx 21.86

Let A be the point on the bridge, AC = 8 m, and B, D the banks with $∠ A B C = 4 5^{\circ}$ , $∠ A D C = 3 0^{\circ}$ (alternate angles).
In right $△ A C B$ : $tan 4 5^{\circ} = \frac{A C}{B C} \Rightarrow B C = 8$ m.
In right $△ A C D$ : $tan 3 0^{\circ} = \frac{A C}{C D} \Rightarrow C D = 83$ m.
Width $B D = 8 + 83 = 8 (1 + 3)$ m $\approx 21.86$ m.

Q83 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/3/1

The tops of two poles of heights 20 m and 28 m are connected with a wire. The wire is inclined to the horizontal at an angle of $3 0^{\circ}$ . Find the length of the wire and the distance between the two poles.

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Answer: Length of wire = 16 m; distance between poles =

83

m ≈ 13.86 m

Difference in heights $= 28 - 20 = 8$ m; this is the vertical side of a right triangle with the wire as hypotenuse.
$sin 3 0^{\circ} = \frac{8}{L} \Rightarrow \frac{1}{2} = \frac{8}{L} \Rightarrow L = 16$ m.
$tan 3 0^{\circ} = \frac{8}{d} \Rightarrow \frac{1}{3} = \frac{8}{d} \Rightarrow d = 83$ m ≈ 13.86 m.

Q73 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/3/2

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of $3 0^{\circ}$ with it. The height of the breaking point from the ground is 2 m. Find the total height of the tree.

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Answer: 6 m

Let the broken part (hypotenuse) be L m; the standing part is 2 m.
$sin 3 0^{\circ} = \frac{2}{L} \Rightarrow \frac{1}{2} = \frac{2}{L} \Rightarrow L = 4$ m.
Total height $= 2 + 4 = 6$ m.

Q73 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/3/3

Two boats are sailing in the sea 80 m apart from each other towards a cliff AB. The angles of depression of the boats from the top of the cliff are $3 0^{\circ}$ and $4 5^{\circ}$ respectively, as shown in Figure 3. Find the height of the cliff.

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Answer:

40 (3 + 1)

m ≈ 109.28 m

Let $A B = h$ . The angle of depression equals the angle of elevation from the boat.
In $△ A B C$ : $tan 4 5^{\circ} = \frac{h}{B C} \Rightarrow B C = h$ .
In $△ A B D$ : $tan 3 0^{\circ} = \frac{h}{B D} \Rightarrow B D = h 3$ .
$C D = B D - B C = h (3 - 1) = 80 \Rightarrow h = \frac{80}{3 - 1} = 40 (3 + 1)$ .
$h \approx 40 \times 2.732 = 109.28$ m.

Q93 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/4/1

An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Find the distance between the two planes at that instant.

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Answer: 6250 m

Let the point on the ground be at horizontal distance $d$ from the foot of the vertical line through the planes.
Lower plane (height 3125 m) has angle of elevation $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{3125}{d} \Rightarrow d = 31253$ m.
Higher plane (height $H$ ) has angle of elevation $6 0^{\circ}$ : $H = d tan 6 0^{\circ} = 31253 \times 3 = 9375$ m.
Distance between the planes $= 9375 - 3125 = 6250$ m.

Q93 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/4/2

There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are points directly opposite to each other on two banks and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively $3 0^{\circ}$ and $4 5^{\circ}$ , find the height of the tree. (Use $3 = 1.732$ )

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Answer: 36.6 m

Let the height of the tree be $h$ m and its foot be at distances $x$ from P and $100 - x$ from Q.
From P: $tan 3 0^{\circ} = \frac{h}{x} \Rightarrow x = h 3$ .
From Q: $tan 4 5^{\circ} = \frac{h}{100 - x} \Rightarrow 100 - x = h$ .
$h 3 + h = 100 \Rightarrow h = \frac{100}{3 + 1} = 50 (3 - 1)$
$h = 50 \times 0.732 = 36.6$ m

Q73 marksShort AnswerSome Applications of Trigonometry CBSE 2022 · Standard 30/4/3

An aeroplane at an altitude of 200 metres observes the angles of depression of opposite points on the two banks of a river to be $4 5^{\circ}$ and $6 0^{\circ}$ . Find the width of the river. (Use $3 = 1.732$ )

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Answer: 315.47 m (approx.)

Let the aeroplane be at A, 200 m vertically above point D on the line joining the banks B and C.
For the bank with depression $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{200}{B D} \Rightarrow B D = 200$ m.
For the bank with depression $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{200}{D C} \Rightarrow D C = \frac{200}{3} = \frac{200 3}{3} = \frac{346.4}{3} \approx 115.47$ m.
Width of river $= B D + D C = 200 + 115.47 = 315.47$ m.

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