Some Applications of Trigonometry

Some Applications of Trigonometry

Some Applications of Trigonometry (heights and distances) carries 5 or 6 marks in the board paper: 6 in the 2026-27 sample paper and 5 in the 2025 and both 2026 board papers.

The marks come either as a 5-mark two-triangle problem (sample paper and Feb 2026) or as a 4-mark case study (a lighthouse in 2025, a ladder between two buildings in May 2026), often with a 1-mark MCQ on a shadow, kite or ladder. Written problems use the angles $30^\circ$, $45^\circ$ and $60^\circ$, with at most two right triangles in one problem; an MCQ may instead give a ratio, such as a kite string with $\tan\theta = \frac{12}{5}$. A common long question sees the same object from two different heights (from the ground and from a roof or deck).

Where marks are usually lost:

• no figure, or a wrong one (the figure carries a mark in a 5-mark answer);
• measuring the angle of depression from the vertical instead of from the horizontal;
• forgetting to add the observer's eye height or the height of a deck, roof or truck;
• adding distances when the points are on the same side, or subtracting when they are on opposite sides.

Revise in 5 Minutes

Words to know

• Line of sight: the line from the observer's eye to the object.
• Angle of elevation: between the horizontal and the line of sight, for an object above the eye.
• Angle of depression: between the horizontal and the line of sight, for an object below the eye.
• The angle of depression of $B$ from $A$ equals the angle of elevation of $A$ from $B$ (alternate angles).

Values you need

Use $\sqrt{3} = 1.73$ and $\sqrt{2} = 1.41$ when decimals are asked.

Which ratio? Height and horizontal distance: use $\tan$. A slant length (string, ladder, cable) with height: $\sin$; with horizontal distance: $\cos$.

Ratio given instead of an angle (e.g. $\tan\theta = \frac{12}{5}$): draw a right triangle with sides 12 and 5, get the hypotenuse 13, then read off $\sin\theta = \frac{12}{13}$ or $\cos\theta = \frac{5}{13}$.

Distance from the foot for height $h$: at $30^\circ$ it is $h\sqrt{3}$; at $45^\circ$ it is $h$; at $60^\circ$ it is $\frac{h}{\sqrt{3}}$.

Method for a 5-marker

1. Draw the figure: vertical lines for towers, a dashed horizontal line through the eye, angles marked at the right place.
2. Mark the unknown as $h$ or $x$.
3. Write one tan (or sin/cos) equation for each right triangle.
4. Solve, rationalise, then put in $\sqrt{3} = 1.73$ only at the end.

Patterns

• Walking towards a tower: distances at the two angles differ by the distance walked.
• Two points on opposite sides of a tower: distances add.
• From a window or deck: draw a horizontal line through the eye; the part below it equals the eye's height.
• Same object from the ground and from a height $h$: the two heights found differ by $h$.

Traps

• Angle of depression is drawn from the horizontal at the observer, not from the vertical.
• Add the height of the hand, eye, deck or truck.
• $\frac{h}{\sqrt{3} - 1}$: multiply by $\frac{\sqrt{3} + 1}{\sqrt{3} + 1}$, giving $\frac{h(\sqrt{3} + 1)}{2}$.
• Doubling the distance does not halve the angle.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

Riya is flying a kite. She holds the string 1.2 m above the ground. The string is 100 m long and makes an angle of $30^\circ$ with the horizontal. Assuming the string is straight, find the height of the kite above the ground.

Answer.

1. Height of the kite above her hand $= 100\sin 30^\circ = 100 \times \frac{1}{2} = 50$ m — 1 mark
2. Height above the ground $= 50 + 1.2 = 51.2$ m — 1 mark

Question 2 (2 marks)

During a storm, a bamboo pole standing vertically on level ground breaks at a point, without separating. The broken part, 8 m long, bends over and its top touches the ground, making an angle of $60^\circ$ with the ground. Find the height of the pole before it broke. (Use $\sqrt{3} = 1.73$)

Answer.

1. Standing part $= 8\sin 60^\circ = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}$ m — 1 mark
2. Original height $= 8 + 4\sqrt{3} = 8 + 6.92 = 14.92$ m — 1 mark

Question 3 (2 marks)

From a point $P$ on level ground, the angle of elevation of the top of a tower is $60^\circ$. Ravi says that from a point $Q$ on the same line, twice as far from the foot of the tower as $P$, the angle of elevation will be $30^\circ$. Is he right? Give a reason.

Answer.

Model answer:

No, Ravi is not right.

Let the height of the tower be $h$. From $P$, $\tan 60^\circ = \frac{h}{PB}$, where $B$ is the foot, so $PB = \frac{h}{\sqrt{3}}$.

$Q$ is twice as far, so $QB = \frac{2h}{\sqrt{3}}$. The angle of elevation $\theta$ at $Q$ has $\tan\theta = h \div \frac{2h}{\sqrt{3}} = \frac{\sqrt{3}}{2}$.

But $\tan 30^\circ = \frac{1}{\sqrt{3}}$, which is not $\frac{\sqrt{3}}{2}$. So the angle at $Q$ is not $30^\circ$. Doubling the distance does not halve the angle. For an angle of $30^\circ$, the distance would have to be $h\sqrt{3}$, which is 3 times $PB$.

Marking scheme:

1. If the height is $h$, then $P$ is $\frac{h}{\sqrt{3}}$ from the foot and $Q$ is $\frac{2h}{\sqrt{3}}$; the angle at $Q$ has $\tan = \frac{h}{2h/\sqrt{3}} = \frac{\sqrt{3}}{2}$ — 1 mark
2. $\frac{\sqrt{3}}{2} \neq \frac{1}{\sqrt{3}} = \tan 30^\circ$, so Ravi is wrong (for $30^\circ$ the point must be 3 times as far) — 1 mark

Question 4 (2 marks)

A ladder leaning against a vertical wall makes an angle of $60^\circ$ with the level ground, and its top touches the wall at a height of $3\sqrt{3}$ m. Find the length of the ladder and the distance of its foot from the wall.

Answer.

1. $\sin 60^\circ = \frac{3\sqrt{3}}{l}$, so $l = 3\sqrt{3} \times \frac{2}{\sqrt{3}} = 6$ m — 1 mark
2. $\tan 60^\circ = \frac{3\sqrt{3}}{d}$, so $d = \frac{3\sqrt{3}}{\sqrt{3}} = 3$ m — 1 mark

Question 5 (3 marks)

From a window 15 m above the level road, Aman sees a car coming straight towards his house. The angle of depression of the car changes from $30^\circ$ to $60^\circ$. How far did the car travel in this time? (Use $\sqrt{3} = 1.73$)

Answer.

1. Let $A$ be the window, $B$ the foot of the house, $D$ and $C$ the two positions of the car; $\angle ADB = 30^\circ$ and $\angle ACB = 60^\circ$ (alternate angles); $BD = \frac{15}{\tan 30^\circ} = 15\sqrt{3}$ m — 1 mark
2. $BC = \frac{15}{\tan 60^\circ} = \frac{15}{\sqrt{3}} = 5\sqrt{3}$ m — 1 mark
3. Distance travelled $= CD = 15\sqrt{3} - 5\sqrt{3} = 10\sqrt{3} = 17.3$ m — 1 mark

Question 6 (3 marks)

Two vertical poles $AB$ and $CD$ stand on opposite sides of a straight road, with their feet $B$ and $D$ on the road. $AB = 15$ m. From the midpoint $M$ of $BD$, the angles of elevation of the tops $A$ and $C$ are $60^\circ$ and $30^\circ$ respectively. Find the width of the road $BD$ and the height of the pole $CD$.

Answer.

1. In $\triangle ABM$: $\tan 60^\circ = \frac{15}{BM}$, so $BM = \frac{15}{\sqrt{3}} = 5\sqrt{3}$ m — 1 mark
2. $BD = 2BM = 10\sqrt{3}$ m — 1 mark
3. In $\triangle CDM$: $MD = 5\sqrt{3}$ m and $CD = MD\tan 30^\circ = 5\sqrt{3} \times \frac{1}{\sqrt{3}} = 5$ m — 1 mark

Question 7 (3 marks)

Kavya looks out of her window, 12 m above the level ground. The angle of elevation of the top of the building opposite is $30^\circ$ and the angle of depression of its foot is $45^\circ$. Find the height of the opposite building. (Use $\sqrt{3} = 1.73$)

Answer.

1. Let $A$ be the window, $B$ the foot of Kavya's building (so $AB = 12$ m), $CD$ the opposite building and $AE$ horizontal with $E$ on $CD$; from the $45^\circ$ depression, $BD = AE = AB = 12$ m — 1 mark
2. In $\triangle AEC$: $CE = AE\tan 30^\circ = \frac{12}{\sqrt{3}} = 4\sqrt{3}$ m — 1 mark
3. Height $CD = 12 + 4\sqrt{3} = 12 + 6.92 = 18.92$ m — 1 mark

Question 8 (3 marks)

A drone hovers 30 m directly above a point on the ground between the two banks of a river. The angles of depression of the two banks, on opposite sides of the drone and in line with it, are $45^\circ$ and $60^\circ$. Find the width of the river. (Use $\sqrt{3} = 1.73$)

Answer.

1. Let the drone be $P$, 30 m above $Q$, with banks $A$ and $B$; $\angle PAQ = 45^\circ$ gives $AQ = 30$ m — 1 mark
2. $\angle PBQ = 60^\circ$ gives $QB = \frac{30}{\sqrt{3}} = 10\sqrt{3}$ m — 1 mark
3. Width $AB = 30 + 10\sqrt{3} = 30 + 17.3 = 47.3$ m — 1 mark

Question 9 (3 marks)

Two vertical poles, 8 m and 20 m high, stand on level ground. A straight wire joins their tops, and the angle of elevation of the top of the taller pole from the top of the shorter pole is $30^\circ$. Find the length of the wire and the distance between the poles. (Use $\sqrt{3} = 1.73$)

Answer.

1. Let the poles be $AB = 8$ m and $CD = 20$ m, and draw $AE \parallel BD$ meeting $CD$ at $E$; then $ED = 8$ m and $CE = 20 - 8 = 12$ m — 1 mark
2. In $\triangle AEC$: $\sin 30^\circ = \frac{CE}{AC}$, so the wire $AC = 12 \times 2 = 24$ m — 1 mark
3. $\tan 30^\circ = \frac{CE}{AE}$, so $BD = AE = 12\sqrt{3} = 20.76$ m — 1 mark

Long Answer and Case-Based Questions

Question 10 (5 marks)

From the top of a 45 m high hotel, the angles of depression of the top and the foot of a temple standing on the same level ground are $30^\circ$ and $60^\circ$ respectively. Find the height of the temple and the distance between the hotel and the temple. (Use $\sqrt{3} = 1.73$)

Answer.

Model answer:

Let $AB = 45$ m be the hotel and $CD$ the temple. Draw $CE$ parallel to the ground, meeting $AB$ at $E$. The angles of depression from $A$ are equal to the alternate angles $\angle ACE = 30^\circ$ and $\angle ADB = 60^\circ$.

In right $\triangle ABD$: $\tan 60^\circ = \frac{AB}{BD}$, so $\sqrt{3} = \frac{45}{BD}$ and $BD = \frac{45}{\sqrt{3}} = 15\sqrt{3}$ m.

$CE = BD = 15\sqrt{3}$ m. In right $\triangle AEC$: $\tan 30^\circ = \frac{AE}{CE}$, so $AE = 15\sqrt{3} \times \frac{1}{\sqrt{3}} = 15$ m.

Height of the temple $CD = BE = AB - AE = 45 - 15 = 30$ m.

Distance between the hotel and the temple $= BD = 15\sqrt{3} = 15 \times 1.73 = 25.95$ m.

Marking scheme:

1. Correct figure: hotel $AB = 45$ m, temple $CD$, $CE$ horizontal with $E$ on $AB$; $\angle ADB = 60^\circ$, $\angle ACE = 30^\circ$ — 1 mark
2. In $\triangle ABD$: $\tan 60^\circ = \frac{45}{BD}$, so $BD = \frac{45}{\sqrt{3}} = 15\sqrt{3} = 25.95$ m — 2 marks
3. In $\triangle AEC$: $CE = BD = 15\sqrt{3}$ m and $AE = CE\tan 30^\circ = 15\sqrt{3} \times \frac{1}{\sqrt{3}} = 15$ m — 1 mark
4. Height of the temple $CD = BE = 45 - 15 = 30$ m; distance between them $= 25.95$ m — 1 mark

Question 11 (5 marks)

During a flood in Assam, a rescue helicopter hovers in still air. Rahul, standing at the foot of a 40 m high water tower, sees the helicopter at an angle of elevation of $60^\circ$. At the same moment Priya, standing on top of the tower, sees it at an angle of elevation of $30^\circ$. Find the height of the helicopter above the ground, its horizontal distance from the tower, and its distance from Priya. (Use $\sqrt{3} = 1.73$)

Answer.

Model answer:

Let $BC = 40$ m be the tower with foot $B$ and top $C$. Let the helicopter be at $H$, directly above the point $G$ on the ground. Draw $CE$ horizontal, meeting $HG$ at $E$. Then $EG = BC = 40$ m and $CE = BG$. Let $BG = x$ m.

In right $\triangle HGB$: $\tan 60^\circ = \frac{HG}{BG}$, so $HG = x\sqrt{3}$.

In right $\triangle HEC$: $\tan 30^\circ = \frac{HE}{CE}$, so $HE = \frac{x}{\sqrt{3}}$.

Since $HG = HE + EG$: $x\sqrt{3} = \frac{x}{\sqrt{3}} + 40$, that is, $\frac{3x - x}{\sqrt{3}} = 40$, so $x = 20\sqrt{3}$ m.

Height of the helicopter $HG = 20\sqrt{3} \times \sqrt{3} = 60$ m.

Horizontal distance from the tower $= 20\sqrt{3} = 20 \times 1.73 = 34.6$ m.

Distance from Priya: $\cos 30^\circ = \frac{CE}{CH}$, so $CH = 20\sqrt{3} \times \frac{2}{\sqrt{3}} = 40$ m.

Marking scheme:

1. Correct figure: tower $BC = 40$ m ($B$ on the ground), helicopter $H$ above the ground point $G$, $CE$ horizontal with $E$ on $HG$; $\angle HBG = 60^\circ$, $\angle HCE = 30^\circ$ — 1 mark
2. Let $BG = CE = x$; in $\triangle HGB$: $HG = x\tan 60^\circ = x\sqrt{3}$ — 1 mark
3. In $\triangle HEC$: $HE = x\tan 30^\circ = \frac{x}{\sqrt{3}}$, and $HG - HE = 40$, so $x\sqrt{3} - \frac{x}{\sqrt{3}} = 40$ — 1 mark
4. $\frac{2x}{\sqrt{3}} = 40$ gives $x = 20\sqrt{3} = 34.6$ m and $HG = 20\sqrt{3} \times \sqrt{3} = 60$ m — 1 mark
5. Distance from Priya $CH = \frac{CE}{\cos 30^\circ} = 20\sqrt{3} \times \frac{2}{\sqrt{3}} = 40$ m — 1 mark

Question 12 (5 marks)

Deepa's eyes are 1.5 m above the level ground. From a point $A$, the angle of elevation of the top of a mobile tower from her eyes is $45^\circ$. She walks 20 m straight towards the tower to a point $B$, and the angle of elevation becomes $60^\circ$. Find the height of the tower and the distance of $B$ from the foot of the tower. (Use $\sqrt{3} = 1.73$)

Answer.

Model answer:

Let $PQ$ be the tower with foot $Q$. Let $A'$ and $B'$ be Deepa's eye positions, 1.5 m above $A$ and $B$. The line $A'B'$ is horizontal; produce it to meet $PQ$ at $M$, so $MQ = 1.5$ m. Let $PM = x$ m.

In right $\triangle PMA'$: $\tan 45^\circ = \frac{x}{A'M}$, so $A'M = x$.

In right $\triangle PMB'$: $\tan 60^\circ = \frac{x}{B'M}$, so $B'M = \frac{x}{\sqrt{3}}$.

$A'M - B'M = A'B' = 20$: $x - \frac{x}{\sqrt{3}} = 20$, so $x(\sqrt{3} - 1) = 20\sqrt{3}$ and

$x = \frac{20\sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = 10\sqrt{3}(\sqrt{3} + 1) = 30 + 10\sqrt{3} = 47.3$ m.

Height of the tower $PQ = PM + MQ = 47.3 + 1.5 = 48.8$ m.

Distance of $B$ from the foot $= B'M = \frac{x}{\sqrt{3}} = 10 + 10\sqrt{3} = 27.3$ m.

Marking scheme:

1. Correct figure: tower $PQ$, eye positions $A'$ and $B'$ 1.5 m above $A$ and $B$, $A'B'$ produced meets $PQ$ at $M$; $\angle PA'M = 45^\circ$, $\angle PB'M = 60^\circ$, $A'B' = 20$ m — 1 mark
2. Let $PM = x$; in $\triangle PMA'$: $A'M = x$ (as $\tan 45^\circ = 1$); in $\triangle PMB'$: $B'M = \frac{x}{\sqrt{3}}$ — 1 mark
3. $A'M - B'M = 20$: $x - \frac{x}{\sqrt{3}} = 20$, so $x = \frac{20\sqrt{3}}{\sqrt{3} - 1} = 10\sqrt{3}(\sqrt{3} + 1) = 30 + 10\sqrt{3}$ — 1 mark
4. $x = 30 + 17.3 = 47.3$ m, so the tower is $47.3 + 1.5 = 48.8$ m high — 1 mark
5. Distance of $B$ from the foot $= B'M = \frac{x}{\sqrt{3}} = 10\sqrt{3} + 10 = 27.3$ m — 1 mark

Question 13 (4 marks)

At an adventure camp near Rishikesh, a zipline platform $OA$ stands 30 m high on level ground. Cable $AB$ runs straight from the top $A$ to a point $B$ on the ground and makes an angle of $30^\circ$ with the ground. A second cable $AD$ runs from $A$ to a point $D$ on the other side of the platform and makes an angle of $60^\circ$ with the ground, as shown in the figure.

(i) Find the length of the cable $AB$. (1 mark)

Answer.

1. $\sin 30^\circ = \frac{30}{AB}$, so $AB = 60$ m — 1 mark

(ii) How far is $B$ from the foot $O$ of the platform? (1 mark)

Answer.

1. $\tan 30^\circ = \frac{30}{OB}$, so $OB = 30\sqrt{3}$ m — 1 mark

(iii) Find the length of the cable $AD$ and the distance $BD$. (Use $\sqrt{3} = 1.73$) (2 marks)

Answer.

1. $AD = \frac{30}{\sin 60^\circ} = \frac{60}{\sqrt{3}} = 20\sqrt{3} = 34.6$ m — 1 mark
2. $OD = \frac{30}{\tan 60^\circ} = 10\sqrt{3}$ m; $BD = 30\sqrt{3} + 10\sqrt{3} = 40\sqrt{3} = 69.2$ m — 1 mark

OR

(iii) A rider on cable $AB$ stops at a point $P$ that is 20 m from $A$ along the cable. How high is $P$ above the ground, and how far is it horizontally from the platform? (Use $\sqrt{3} = 1.73$) (2 marks)

Answer.

1. Drop below $A$: $20\sin 30^\circ = 10$ m, so the height of $P$ is $30 - 10 = 20$ m — 1 mark
2. Horizontal distance $= 20\cos 30^\circ = 10\sqrt{3} = 17.3$ m — 1 mark

Question 14 (4 marks)

A fire engine reaches a building in Surat. The foot $L$ of its ladder is fixed on the truck, 2 m above the ground. The ladder $LT$ is 20 m long. It is raised to make an angle of $60^\circ$ with the horizontal, and its top $T$ rests against the vertical wall of the building, as shown in the figure.

(i) How high above the ground is the top $T$? (Use $\sqrt{3} = 1.73$) (1 mark)

Answer.

1. $2 + 20\sin 60^\circ = 2 + 10\sqrt{3} = 2 + 17.3 = 19.3$ m — 1 mark

(ii) What is the horizontal distance of $L$ from the wall? (1 mark)

Answer.

1. $20\cos 60^\circ = 20 \times \frac{1}{2} = 10$ m — 1 mark

(iii) A person is trapped at a window 12 m above the ground. The ladder, with the same length, is lowered to make $30^\circ$ with the horizontal. Show that its top now reaches the window's height, and find how far back the truck must move so that the top rests on the wall. (Use $\sqrt{3} = 1.73$) (2 marks)

Answer.

1. Height of the top $= 2 + 20\sin 30^\circ = 2 + 10 = 12$ m, which is the window's height — 1 mark
2. New horizontal distance $= 20\cos 30^\circ = 10\sqrt{3} = 17.3$ m; the truck moves back $17.3 - 10 = 7.3$ m — 1 mark

OR

(iii) Suppose the ladder is kept at $45^\circ$ to the horizontal instead. What length of ladder is needed for its top to reach a window 16 m above the ground? (Use $\sqrt{2} = 1.41$) (2 marks)

Answer.

1. Height to be covered above $L$ $= 16 - 2 = 14$ m, and $\sin 45^\circ = \frac{14}{l}$ — 1 mark
2. $l = 14\sqrt{2} = 14 \times 1.41 = 19.74$ m — 1 mark