Question 1

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is $30^\circ$. The height of the tower is: ```svg ```

Accepted Answer

**Answer: (a)** $10\sqrt3$ m $	an30^\circ=\dfrac{h}{30}\Rightarrow h=30	an30^\circ=\dfrac{30}{\sqrt3}=10\sqrt3$ m.

Question 2

If the length of the shadow of a vertical pole is equal to the height of the pole, then the angle of elevation of the Sun is:

Accepted Answer

**Answer: (b)** $45^\circ$ If height $=$ shadow $=h$, then $	an	heta=\dfrac{h}{h}=1\Rightarrow	heta=45^\circ$.

Question 3

A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the ground, then the height of the wall is:

Accepted Answer

**Answer: (a)** $\dfrac{15\sqrt3}{2}$ m The ladder is the hypotenuse, so $\sin60^\circ=\dfrac{	ext{height}}{15}\Rightarrow	ext{height}=15\sin60^\circ=15	imes\dfrac{\sqrt3}{2}=\dfrac{15\sqrt3}{2}$ m.

Question 4

A tower stands vertically on the ground. From a point 50 m away from its foot, the top of the $50\sqrt3$ m high tower is observed. The angle of elevation of the top is:

Accepted Answer

**Answer: (c)** $60^\circ$ $	an	heta=\dfrac{	ext{height}}{	ext{distance}}=\dfrac{50\sqrt3}{50}=\sqrt3\Rightarrow	heta=60^\circ$.

Question 5

**Assertion (A):** For a tower of fixed height, the length of its shadow increases as the Sun's altitude (angle of elevation) decreases. **Reason (R):** The length of the shadow of a tower of height $h$ is $h\cot	heta$, and $\cot	heta$ increases as $	heta$ decreases from $90^\circ$ towards $0^\circ$.

Accepted Answer

**Answer: (a)** Both A and R are true and R is the correct explanation of A. If the tower has height $h$ and the Sun's altitude is $	heta$, then $	an	heta=\dfrac{h}{	ext{shadow}}$, so shadow $=h\cot	heta$. As $	heta$ decreases, $\cot	heta$ increases, hence the shadow lengthens. R correctly explains A.

Question 6

The angle of elevation of the top of a 15 m high tower at a point on the ground is $45^\circ$. Find the distance of the point from the foot of the tower. ```svg ```

Accepted Answer

Let the required distance be $d$. Then $	an45^\circ=\dfrac{15}{d}$. Since $	an45^\circ=1$, we get $1=\dfrac{15}{d}\Rightarrow d=15$ m. The point is 15 m from the foot of the tower.

Question 7

A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied to a point on the ground and is taut (no slack). If the string makes an angle of $60^\circ$ with the ground, find the length of the string. ```svg ```

Accepted Answer

The string is the hypotenuse of the right triangle, and 60 m is the side opposite the $60^\circ$ angle. Let the length of the string be $L$ .

$\sin60^\circ=\dfrac{60}{L}\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{60}{L}\Rightarrow L=\dfrac{120}{\sqrt3}=40\sqrt3$ m.

The length of the string is $40\sqrt3\approx69.2$ m.

Question 8

The shadow of a tower standing on level ground is found to be 40 m longer when the Sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower. (Take $\sqrt3=1.73$.) ```svg ```

Accepted Answer

Let the height of the tower be $h$ and the shorter shadow (at $60^\circ$ ) be $x$ . The longer shadow (at $30^\circ$ ) is then $x+40$ .

At $60^\circ$ : $\tan60^\circ=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x$ .

At $30^\circ$ : $\tan30^\circ=\dfrac{h}{x+40}\Rightarrow h=\dfrac{x+40}{\sqrt3}$ .

Equating: $\sqrt3\,x=\dfrac{x+40}{\sqrt3}\Rightarrow 3x=x+40\Rightarrow 2x=40\Rightarrow x=20$ m.

Therefore $h=\sqrt3\times20=20\sqrt3=20\times1.73=34.6$ m.

The height of the tower is $20\sqrt3\approx34.6$ m.

Question 9

A tree is broken by the wind. Its top strikes the ground at an angle of $30^\circ$ and at a distance of 8 m from the foot of the tree. Find the original height of the tree. (Take $\sqrt3=1.73$.) ```svg ```

Accepted Answer

Let $B$ be the foot of the tree, $D$ the point where it broke, and $A$ the point on the ground where the top strikes, with $AB=8$ m and $\angle A=30^\circ$ . The broken part $DA$ (now leaning) equals the upper part of the tree.

Standing part: $\tan30^\circ=\dfrac{BD}{AB}\Rightarrow BD=8\tan30^\circ=\dfrac{8}{\sqrt3}$ m.

Broken (leaning) part: $\cos30^\circ=\dfrac{AB}{DA}\Rightarrow DA=\dfrac{8}{\cos30^\circ}=\dfrac{8}{\sqrt3/2}=\dfrac{16}{\sqrt3}$ m.

Original height $=BD+DA=\dfrac{8}{\sqrt3}+\dfrac{16}{\sqrt3}=\dfrac{24}{\sqrt3}=8\sqrt3=8\times1.73=13.84$ m.

The tree was $8\sqrt3\approx13.84$ m tall.

Question 10

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Find the height of the cable tower. (Take $\sqrt3=1.73$.) ```svg ```

Accepted Answer

Let the horizontal distance between the building and the tower be $d$ . From the level of the building top, the foot of the tower is 7 m below and the line of sight to it is depressed $45^\circ$ .

Depression to foot: $\tan45^\circ=\dfrac{7}{d}\Rightarrow d=7$ m.

Elevation to top: let the part of the tower above the building-top level be $p$ . Then $\tan60^\circ=\dfrac{p}{d}\Rightarrow p=d\tan60^\circ=7\sqrt3$ m.

Total height of tower $=7+p=7+7\sqrt3=7(1+\sqrt3)=7(1+1.73)=7\times2.73=19.11$ m.

The cable tower is $7(1+\sqrt3)\approx19.11$ m high.

Question 11

Two poles of equal height stand on either side of a road 80 m wide. From a point between the poles on the road, the angles of elevation of the tops of the poles are $60^\circ$ and $30^\circ$ respectively. Find the height of the poles and the distances of the point from the two poles. ```svg ```

Accepted Answer

Let the common height be $h$ and let the point $P$ be $x$ m from the foot of the pole seen at $60^\circ$ . Then $P$ is $(80-x)$ m from the other pole.

From the $60^\circ$ pole: $\tan60^\circ=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x$ .

From the $30^\circ$ pole: $\tan30^\circ=\dfrac{h}{80-x}\Rightarrow h=\dfrac{80-x}{\sqrt3}$ .

Equating the two expressions for $h$ :

$\sqrt3\,x=\dfrac{80-x}{\sqrt3}\Rightarrow 3x=80-x\Rightarrow 4x=80\Rightarrow x=20$ m.

So the point is 20 m from one pole and $80-20=60$ m from the other.

Height $h=\sqrt3\times20=20\sqrt3\approx34.64$ m.

The poles are $20\sqrt3\approx34.64$ m high, and the point is 20 m and 60 m from the two poles.

Question 12

From the top of a lighthouse 75 m high, the angles of depression of two ships, sailing towards it in the same straight line and on the same side, are $30^\circ$ and $45^\circ$. Find the distance between the two ships. (Take $\sqrt3=1.73$.) ```svg ```

Accepted Answer

Let the lighthouse be $LM$ with $L$ at the top, $LM=75$ m. Let the nearer ship (depression $45^\circ$ ) be at distance $d_1$ and the farther ship (depression $30^\circ$ ) at distance $d_2$ from the foot $M$ . The angles of depression equal the corresponding angles of elevation at the ships.

Nearer ship: $\tan45^\circ=\dfrac{75}{d_1}\Rightarrow d_1=75$ m.

Farther ship: $\tan30^\circ=\dfrac{75}{d_2}\Rightarrow d_2=75\sqrt3$ m.

Distance between the ships $=d_2-d_1=75\sqrt3-75=75(\sqrt3-1)=75(1.73-1)=75\times0.73=54.75$ m.

The two ships are $75(\sqrt3-1)\approx54.75$ m apart.

Question 13

During a field trip, some students stand on a straight road leading to a temple. From a point $A$ on the road the angle of elevation of the top $T$ of the temple is $30^\circ$ . On walking 40 m towards the temple to a point $B$ , the angle of elevation becomes $60^\circ$ . Let the height of the temple be $h$ and its foot be $O$ .

(i) Express $OA$ and $OB$ in terms of $h$ .

(ii) Using $OA-OB=40$ , find the distance $OB$ .

(iii) Find the height $h$ of the temple. (Take $\sqrt3=1.73$ .)

Accepted Answer

(i) From $A$: $	an30^\circ=\dfrac{h}{OA}\Rightarrow OA=\dfrac{h}{	an30^\circ}=h\sqrt3$. From $B$: $	an60^\circ=\dfrac{h}{OB}\Rightarrow OB=\dfrac{h}{	an60^\circ}=\dfrac{h}{\sqrt3}$. (ii) Since $A$ is farther, $OA-OB=40$: $h\sqrt3-\dfrac{h}{\sqrt3}=40\Rightarrow \dfrac{3h-h}{\sqrt3}=40\Rightarrow \dfrac{2h}{\sqrt3}=40\Rightarrow h=20\sqrt3$. Then $OB=\dfrac{h}{\sqrt3}=\dfrac{20\sqrt3}{\sqrt3}=20$ m. (iii) $h=20\sqrt3=20	imes1.73=34.6$ m. The temple is $20\sqrt3\approx34.6$ m high (and $B$ is 20 m, $A$ is 60 m from its foot).

Question 14

Are these Some Applications of Trigonometry important questions free?

Accepted Answer

Yes. All 13 CBSE Class 10 Maths important questions for Some Applications of Trigonometry are free, with full model answers and no login required.

Question 15

Do these Some Applications of Trigonometry questions follow the latest CBSE syllabus?

Accepted Answer

Yes — they are aligned to the NCERT 2026–27 syllabus for CBSE Class 10 Maths, so nothing here is outside the current course.

Question 16

How should I practise the Some Applications of Trigonometry important questions?

Accepted Answer

Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.

Question 17

What types of questions are covered for Some Applications of Trigonometry?

Accepted Answer

A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the CBSE paper is covered.

Question type	Count	Marks
MCQ	4	1
Assertion–Reason	1	1
Very Short	2	2
Short Answer	3	3
Long Answer	2	5
Case-based	1	4

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