Some Applications of Trigonometry: Class 10 Maths · Chapter 9

Find heights and distances you cannot reach, from one length and one angle

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

50 min
24 practice questions
13 MCQs

What you will learn in Some Applications of Trigonometry

Identify the line of sight, the horizontal through the eye, and the angles of elevation and depression in a real situation
Explain why the angle of depression of a point seen from above equals the angle of elevation of the observer seen from that point
Draw a right-triangle model of a heights-and-distances problem and choose the trigonometric ratio that links the known and unknown sides
Find a height, a distance or a slanting length from one right triangle, adding the observer's eye height when needed
Solve problems with two right triangles that share a side, such as two angles seen from one point or one object seen from two points
Give answers in surd form and as decimals using √3 ≈ 1.732, and check that they make sense

Original companion text aligned to NCERT Mathematics Class 10, Chapter 9, pp. 133–143. Not affiliated with or endorsed by NCERT.

9.1 Heights and Distances

How tall is the gopuram of the temple you visit? You could not climb it with a measuring tape, and nobody would let you try. Surveyors, engineers and sailors meet this problem all the time. They measure one length on the ground and one angle, and trigonometry does the rest.

In Chapter 8 you met the trigonometric ratios of an acute angle in a right triangle. Now you will put them to work. Every problem in this chapter follows one plan: draw a right triangle, mark what you know, and pick the ratio that connects it to what you want.

Table 9.1Ratios you will use again and again

	$30^{\circ}$	$45^{\circ}$	$60^{\circ}$
$\sin θ$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$
$\cos θ$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$
$\tan θ$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$
$\tan θ$ (decimal)	$0.577$	$1$	$1.732$

Board papers usually say take $\sqrt{3} = 1.732.$ We do the same throughout this chapter. Keep the surd form, such as $13 \sqrt{3},$ until the last line and only then turn it into a decimal: rounding early makes answers drift.

Line of sight and the angle of elevation

Stand some distance from a tall tower and look at its top. Picture a thread pulled tight from your eye to that top. This thread is your line of sight (दृष्टि रेखा).

Now picture a second thread from your eye, held perfectly level and pointing at the tower. This is the horizontal through your eye. To see the top you tilt your head up, so the line of sight rises above the horizontal. The angle between the two threads is the angle of elevation (उन्नयन कोण) of the top.

4/4The angle $θ$ between them is the angle of elevation. With the vertical from the top, it makes a right triangle.

Fig. 9.1The angle of elevation is measured up from the horizontal through the eye.NCERT Fig. 9.1 · p. 133

A temple gopuram with stepped tiers stands on the left and a person stands on level ground on the right, facing it. Step 1 shows only the scene. Step 2 adds a dashed level line from the person's eye towards the tower, labelled horizontal. Step 3 adds a solid line from the eye to the top of the gopuram, labelled line of sight. Step 4 marks the angle theta between the horizontal and the line of sight at the eye, labels it angle of elevation, and shades the right triangle formed with a dashed vertical line from the top down to eye level.

Definition

Line of sight

The straight line that joins your eye to the point you are looking at.

Definition

Angle of elevation

The angle measured upwards from the horizontal through your eye to your line of sight, when the point you look at is higher than your eye.

ExampleYou measure an angle of elevation when you look up at a kite, a rooftop or the Moon.

Two details trip people up. The angle is measured from the horizontal, not from the ground at the foot of the tower and not from the vertical. And it is measured at the observer's eye, so a tall person and a short person at the same spot see slightly different angles.

The angle of depression

Elevation has a mirror image. When the point you look at is lower than your eye, you tip your head down and the line of sight falls below the horizontal. The angle it falls through is the angle of depression (अवनमन कोण). Picture a lighthouse keeper on the gallery, watching a fishing boat far below: the angle of depression of the boat is measured at her eye.

4/4A horizontal through the boat is parallel to hers. The angle of elevation of the keeper from the boat is the same $θ$ : they are alternate angles.

Fig. 9.2The angle of depression is measured down from the horizontal through the eye.NCERT Fig. 9.3 · p. 134

A striped lighthouse stands on rocks on the right with a keeper on its gallery; a small sailing boat floats on the sea on the left. Step 1 shows the scene. Step 2 adds a dashed horizontal line from the keeper's eye out over the sea and a solid line of sight down to the boat. Step 3 marks the angle theta between the horizontal and the line of sight at the keeper's eye, labelled angle of depression. Step 4 adds a second dashed horizontal through the boat, with small arrowheads on both horizontals to show they are parallel, and marks the equal angle theta at the boat, labelled angle of elevation from the boat.

Definition

Angle of depression

The angle measured downwards from the horizontal through your eye to your line of sight, when the point you look at is lower than your eye.

ExampleYou measure an angle of depression when you look down from a balcony at a scooter parked on the street.

Elevation or depression?

1 of 4

Riya stands on her first-floor balcony and looks at a scooter parked on the street below. The angle her line of sight makes with the horizontal is

an angle of depressionan angle of elevationan angle measured from the wall

Show answer

Answer (A) an angle of depression

The scooter is below her eye level, so she looks down: an angle of depression. Elevation would mean looking up, and neither angle is measured from the wall.

Planning a heights problem

A neem tree in your school ground is too tall to measure with a tape. Stand where you can see its top. Your eye, the treetop and the point on the trunk level with your eye form a right triangle. To solve it you need two numbers: the angle of elevation of the top and your distance from the trunk. One more number finishes the job: the height of your eye above the ground, because the triangle starts there, not at the ground.

3/3The tree is taller than $x$ by her eye height $e,$ so $H = e + x = e + d \tan θ .$

Fig. 9.3Three measurements give the height: the distance, the angle and the eye height.NCERT p. 135

A large tree stands on the left and a student on the right holds a small clinometer to her eye. Step 1 shows the scene. Step 2 draws the model: a dashed horizontal from her eye to the tree's trunk, a line of sight to the top of the tree, a dashed vertical down from the top with a right-angle mark, the angle theta at her eye, the ground distance d, and the height x of the top above her eye, marked on the left. Step 3 adds her eye height e, marked from the ground to eye level, and the result H equals e plus d tan theta.

The triangle in Fig. 9.3 has its right angle on the trunk, at your eye level. Its side along the trunk, $x,$ is opposite $θ .$ Its level side, $d,$ touches $θ .$ The tangent is the ratio built from exactly these two sides, so $x = d \tan θ .$ Add your eye height and you have the height of the tree.

Height from an angle of elevation

H = e + d \tan θ

(9.1)

where

$H$: height of the object
$e$: height of the observer's eye above the ground (0 if the angle is measured at ground level)
$d$: horizontal distance from the observer to the foot of the object
$θ$: angle of elevation of the top

Not every problem asks for a height. Sometimes the slanting side is known (a ladder, a rope, a kite string), or it is the thing you want. Choose the ratio that contains the side you know and the side you want, and nothing else.

Table 9.2Which ratio to use

You know	You want	Use
horizontal distance (adjacent)	height (opposite)	$\tan θ = \frac{opp}{adj}$
height (opposite)	horizontal distance (adjacent)	$\tan θ$
slanting side (hypotenuse)	height (opposite)	$\sin θ = \frac{opp}{hyp}$
height (opposite)	slanting side (hypotenuse)	$\sin θ$
slanting side (hypotenuse)	horizontal distance (adjacent)	$\cos θ = \frac{adj}{hyp}$

Let's explore

Make a clinometer and measure something tall

You’ll need

a protractor with a small hole at its centre
a drinking straw and sticky tape
20 cm of thread with a small nut tied to one end
a measuring tape
a partner

Predict first

You sight the top of the school flagpole, then take ten steps back and sight it again. The angle you read the second time will be

largersmallerthe same

Tape the straw along the straight edge of the protractor. Pass the thread through the centre hole so the nut hangs freely below the curved scale.
Look at the top of a flagpole or a tree through the straw. Hold still while your partner reads where the thread crosses the scale.
The thread always hangs straight down. So the angle of elevation is the difference between the reading and $90^{\circ}$ : a reading of $60^{\circ}$ or $120^{\circ}$ both mean $30^{\circ} .$
Measure your distance from the foot of the object and the height of your eye above the ground.
Work out the height with (9.1). Then repeat from a second spot.

Table 9.3My measurements

Spot	$d$ (m)	$θ$ (°)	$H$ (m)
First spot
Second spot

Your entries are saved automatically.

InteractiveElevation explorer

An interactive scene drawn to scale. In the Look up view a tall object such as a gopuram, a lighthouse or a coconut palm stands on the left and an observer stands on the ground to the right. You drag the observer, or use a slider, to change the distance d. The line of sight, the horizontal through the eye and the angle of elevation are drawn, and a panel shows d, the angle, tan of the angle, the height above the eye d tan theta and the total height e plus d tan theta. Buttons move the observer to where the angle is exactly 30, 45 or 60 degrees, a switch adds an eye height of 1.5 m, and a Play button walks the observer towards the object. In the Look down view you drag a boat and see that the angle of depression from the top equals the angle of elevation from the boat. A Challenge mode hides the height and asks you to calculate it from the distance and the angle.

Fig. 9.4Move the observer and watch the angle of elevation and

d \tan θ

change.

Choose the ratio

1 of 3

At Makar Sankranti, Ishaan's kite string is 100 m long and makes

30^{\circ}

with the horizontal at his hand. Taking the string as straight, how high is the kite above his hand?

Show answer

Answer 50 m

The string is the hypotenuse, so

\sin 30^{\circ} = \frac{h}{100}

and

h = 100 \times \frac{1}{2} = 50 m .

Using

\tan

would need the horizontal distance, which you do not know.

Problems with one right triangle

Let us use the plan. In each example, sketch the figure first, then mark the right angle, and only then choose the ratio.

1/2From $P,$ 39 m away, the line of sight rises at $30^{\circ}$ to the top of the palm.

Fig. 9.5Example 9.1: the line of sight from

P

to the top of the palm.

A tall coconut palm with fronds at the top stands on the left of level ground. Point P is on the ground to the right, 39 m from the foot of the palm, marked with a dimension arrow. Step 1 shows the line of sight from P to the top of the palm making 30 degrees with the ground. Step 2 shades the right triangle, adds the right-angle mark at the foot and marks the unknown height h, and shows the result h equals 13 root 3, about 22.52 m.

Example 9.1NCERT Example 1 · p. 135

Height of a coconut palm

A coconut palm stands straight up on level ground. Point $P$ is on the ground, 39 m from its foot. From $P$ the top of the palm is seen at an angle of elevation of $30^{\circ} .$ Find the height of the palm. (Fig. 9.5)

Try it yourself first. Then reveal the solution one step at a time.

2/3The rise is $13.2 - 1.2 = 12$ m. The ladder $L$ is the hypotenuse and $b$ is the horizontal side.

Fig. 9.6Example 9.2: the ladder only has to climb from the deck to the sill.

A fire engine stands on the left of a road and a building with windows stands on the right. The ladder rises from a hinge on the truck deck, 1.2 m above the road, to a window sill 13.2 m above the road; both heights are marked with dimension arrows. Step 1 shows the scene. Step 2 shades the right triangle formed by the ladder L, a dashed horizontal b from the hinge to the wall and a dashed vertical of 12 m, with the 60 degree angle at the hinge. Step 3 shows the answers L equals 8 root 3, about 13.86 m, and b equals 4 root 3, about 6.93 m.

Example 9.2NCERT Example 2 · p. 136

A fire-engine ladder

A fire engine parks beside a building. Its ladder is hinged on the truck's deck, 1.2 m above the road, and must reach a window sill 13.2 m above the road. The ladder has to make $60^{\circ}$ with the horizontal. How long must it be, and how far is the hinge from the wall? Use $\sqrt{3} \approx 1.732.$ (Fig. 9.6)

Try it yourself first. Then reveal the solution one step at a time.

1/2The angle of elevation at her eye is $60^{\circ}$ ; she is 11 m away and her eyes are 1.6 m up.

Fig. 9.7Example 9.3: the right triangle sits at Meera's eye level, not on the ground.

A water tank on four legs stands on the left and a girl stands on the right, 11 m from its foot. Her eye height of 1.6 m is marked with a dimension arrow. A dashed horizontal runs from her eye to the tower and a line of sight rises to the top of the tank at 60 degrees. Step 1 shows these givens. Step 2 shades the right triangle above her eye level, marks the unknown side x on the left, and shows x equals 11 root 3, about 19.05 m, and the total height about 20.65 m.

Example 9.3NCERT Example 3 · p. 137

Counting the eye height

Meera's eyes are 1.6 m above the ground. She stands 11 m from the foot of the water-tank tower in her colony. From there she sees the top of the tank at an angle of elevation of $60^{\circ} .$ How tall is the tower, up to the top of the tank? Use $\sqrt{3} \approx 1.732.$ (Fig. 9.7)

Try it yourself first. Then reveal the solution one step at a time.

One triangle

1 of 3

From a point 25 m from the foot of a flagpole, the angle of elevation of its top, measured at ground level, is

45^{\circ} .

How tall is the flagpole?

Show answer

Answer 25 m

h = 25 \tan 45^{\circ} = 25 \times 1 = 25 m .

45^{\circ}

the height always equals the distance.

Problems with two right triangles

Harder problems give two angles. The figure then holds two right triangles that share a side. That shared side is the key. Write one equation from each triangle, then eliminate the side you do not need.

Watch for three set-ups in the examples below.

One object standing on another, seen from one point (Example 9.4).
One object seen at two different angles along a line (Example 9.5).
Two angles of depression from one high point (Example 9.6 and Example 9.7).

1/3A lighthouse stands at the edge of a 40 m cliff. A boat is out at sea.

Fig. 9.8Example 9.4: the two triangles share the distance

d .

A vertical cliff 40 m high rises on the right with a striped lighthouse at its edge; a boat B floats on the sea on the left. Step 1 shows the scene. Step 2 draws the lower right triangle: the sea-level distance d from the boat to the foot of the cliff, the cliff as the vertical side and a line of sight to the cliff top at 30 degrees, with the result d equals 40 root 3, about 69.28 m. Step 3 adds a dashed line of sight to the top of the lighthouse at 45 degrees and marks the lighthouse height h, with the result h equals 40 root 3 minus 40, about 29.28 m.

Example 9.4NCERT Example 4 · p. 137

A lighthouse on a cliff

A lighthouse stands at the edge of a vertical cliff 40 m high. A boat $B$ is out at sea. From the boat, the top of the cliff (the foot of the lighthouse) is seen at an angle of elevation of $30^{\circ},$ and the lamp at the top at $45^{\circ} .$ How tall is the lighthouse? How far is the boat from the foot of the cliff? Use $\sqrt{3} \approx 1.732.$ (Fig. 9.8)

Try it yourself first. Then reveal the solution one step at a time.

The altitude of the Sun is its angle of elevation seen from the ground. The higher the Sun climbs, the shorter every shadow becomes. The next example turns that into a measurement.

2/3With the Sun higher, at 60°, the shadow is shorter.

Fig. 9.9Example 9.5: as the Sun climbs, the shadow shrinks.

A flag-mast stands on the right of level ground. Step 1 shows the Sun low on the right at 45 degrees, with a ray passing over the top of the mast to the tip of a long shadow on the left. Step 2 shows the Sun higher at 60 degrees, with a dashed ray to the tip of a shorter shadow. Step 3 shows both rays and both shadows together, the 6 m difference between the two tips, the shorter shadow x and the height h, with the result h equals 3 times 3 plus root 3, about 14.20 m.

Example 9.5NCERT Example 5 · p. 138

Shadows of a flag-mast

On a sunny morning the shadow of the flag-mast in a temple courtyard is measured twice. The first time, the altitude of the Sun is $45^{\circ} .$ Later it is $60^{\circ},$ and the shadow has become 6 m shorter. Find the height of the mast. Use $\sqrt{3} \approx 1.732.$ (Fig. 9.9)

Try it yourself first. Then reveal the solution one step at a time.

2/3From $P,$ the line to $A$ dips $30^{\circ}$ and the line to $B$ dips $60^{\circ}$ below the horizontal.

Fig. 9.10Example 9.6: two angles of depression from one roof corner.NCERT Fig. 9.9 · p. 139

A tall apartment block stands on the left and a 10 m house stands on the right, with a road between them. P is the roof corner of the block, A the top of the house and B its foot. Step 1 shows the scene. Step 2 adds a dashed horizontal from P and two lines of sight down to A and B, with the angles of depression 30 and 60 degrees marked at P. Step 3 adds a dashed horizontal from the block to A with a right-angle mark, the equal alternate angles at A and B, the road width d, and the results d equals 5 root 3, about 8.66 m, and h equals 15 m.

Example 9.6NCERT Example 6 · p. 139

Across the road

A 10 m tall house stands across the road from an apartment block. Look from the roof corner $P$ of the block. The top $A$ of the house is $30^{\circ}$ below the horizontal, and its foot $B$ is $60^{\circ}$ below it. Find the width of the road and the height of the block. (Fig. 9.10)

Try it yourself first. Then reveal the solution one step at a time.

2/3From $D,$ post $A$ is $45^{\circ}$ and post $B$ is $60^{\circ}$ below the horizontal.

Fig. 9.11Example 9.7: the two parts of AB come from two separate triangles.NCERT Fig. 9.10 · p. 140

A small drone D hovers above a level path with post A on the left and post B on the right. A dashed vertical of 60 m runs from the drone to the point directly below it. Step 1 shows the scene. Step 2 adds a dashed horizontal through the drone and lines of sight to A and B, with angles of depression of 45 and 60 degrees. Step 3 marks the equal alternate angles at A and B and the two ground distances, 60 m on the left and 20 root 3 m on the right, with the total AB about 94.64 m.

Example 9.7NCERT Example 7 · p. 140

A drone over a canal path

A survey drone hovers 60 m above a straight, level canal path. Two marker posts $A$ and $B$ stand on the path, on opposite sides of the drone. Their angles of depression from the drone are $45^{\circ}$ and $60^{\circ} .$ How far apart are the posts? Use $\sqrt{3} \approx 1.732.$ (Fig. 9.11)

Try it yourself first. Then reveal the solution one step at a time.

Two triangles

1 of 3

From the top of a tower, the angles of depression of two cars on the same side are

30^{\circ}

and

60^{\circ} .

Which car is nearer the tower?

the car at

30^{\circ}

the car at

60^{\circ}

both are equally far

Show answer

Answer (B) the car at $60^{\circ}$

A steeper line of sight lands closer to the foot: the distance is

\frac{h}{\tan θ},

which is smaller for the larger angle.

Practice · 16 questions

Practice 9.1: questions with answers

NumericalEasyNCERT Exercise 9.1, Q1
A steel guy wire runs tight from the top of a vertical radio mast to a peg in the ground. The wire is 28 m long and makes $60^{\circ}$ with the ground. How tall is the mast? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $14 \sqrt{3} \approx 24.25 m$
$\sin 60^{\circ} = \frac{h}{28},$ so $h = 28 \times \frac{\sqrt{3}}{2} .$
$h = 14 \sqrt{3}$ $\approx 24.25 m .$
NumericalMediumNCERT Exercise 9.1, Q2
A storm snaps a bamboo pole. The top part bends over without coming off, and its tip touches the ground 6 m from the foot, making $60^{\circ}$ with the ground. How tall was the pole before it broke? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $12 + 6 \sqrt{3} \approx 22.39 m$
Standing part $a$ : opposite $60^{\circ},$ with 6 m adjacent. $a = 6 \tan 60^{\circ} = 6 \sqrt{3} .$
Broken part $b$ : the hypotenuse. $\cos 60^{\circ} = \frac{6}{b},$ so $b = 12.$
Height $= a + b$ $= 12 + 6 \sqrt{3}$ $\approx 22.39 m .$
3
NumericalEasyNCERT Exercise 9.1, Q3
A godown has two ramps.
1. A ramp for hand-carts rises to a 1.2 m loading platform at $30^{\circ}$ to the ground. How long is it?
  m
  Show solution
  $\sin 30^{\circ} = \frac{1.2}{l},$ so $l = 1.2 \times 2 = 2.4 m .$
2. A chute for sacks runs down from a 2.4 m loft at $60^{\circ}$ to the floor. How long is it? Use $\sqrt{3} \approx 1.732.$
  m
  Show solution
  $\sin 60^{\circ} = \frac{2.4}{l},$ so $l = \frac{4.8}{\sqrt{3}}$ $= 1.6 \sqrt{3}$ $\approx 2.77 m .$
NumericalEasyNCERT Exercise 9.1, Q4
Standing 20 m from the base of a brick-kiln chimney, Sunil sees its top $60^{\circ}$ above the horizontal (measured at ground level). How tall is the chimney? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $20 \sqrt{3} \approx 34.64 m$
$\tan 60^{\circ} = \frac{h}{20},$ so $h = 20 \sqrt{3}$ $\approx 34.64 m .$
NumericalEasyNCERT Exercise 9.1, Q5
A weather balloon is held 75 m above the ground by a straight cable tied to a hook on the ground. The cable makes $60^{\circ}$ with the ground. How long is the cable? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $50 \sqrt{3} \approx 86.60 m$
$\sin 60^{\circ} = \frac{75}{l},$ so $l = \frac{150}{\sqrt{3}}$ $= 50 \sqrt{3}$ $\approx 86.60 m .$
NumericalMediumNCERT Exercise 9.1, Q6
Kavya walks straight towards a temple tower 43.4 m tall. When she starts, she sees its top $30^{\circ}$ above the horizontal through her eyes; when she stops, it is $45^{\circ}$ above it. Her eyes are 1.4 m above the ground. How far did she walk? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $42 (\sqrt{3} - 1) \approx 30.74 m$
Height above her eyes $= 43.4 - 1.4 = 42 m .$
At $30^{\circ}$ : distance $= \frac{42}{\tan 30^{\circ}} = 42 \sqrt{3} .$ At $45^{\circ}$ : distance $= 42.$
Walked $= 42 \sqrt{3} - 42$ $= 42 (\sqrt{3} - 1)$ $\approx 30.74 m .$
NumericalMediumNCERT Exercise 9.1, Q7
A flag mast rises from the top of an 18 m bastion of an old fort. Rahul, on level ground, measures $45^{\circ}$ up to the foot of the mast and $60^{\circ}$ up to its tip. How long is the mast? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $18 (\sqrt{3} - 1) \approx 13.18 m$
$\tan 45^{\circ} = \frac{18}{d},$ so $d = 18.$
$\tan 60^{\circ} = \frac{18 + m}{18},$ so $18 + m = 18 \sqrt{3} .$
$m = 18 (\sqrt{3} - 1)$ $\approx 13.18 m .$
NumericalMediumNCERT Exercise 9.1, Q8
A 2.4 m bronze statue of a freedom fighter is mounted on a pedestal in a town square. Measured from one spot on the ground, the pedestal's top is $30^{\circ}$ above the horizontal and the statue's head is $45^{\circ}$ above it. How tall is the pedestal? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $1.2 (\sqrt{3} + 1) \approx 3.28 m$
$\tan 30^{\circ} = \frac{p}{d}$ gives $d = \sqrt{3} p .$
$\tan 45^{\circ} = \frac{p + 2.4}{d}$ gives $p + 2.4 = \sqrt{3} p .$
So $p (\sqrt{3} - 1) = 2.4,$ and $p = \frac{2.4}{\sqrt{3} - 1}$ $= 1.2 (\sqrt{3} + 1)$ $\approx 3.28 m .$
NumericalMediumNCERT Exercise 9.1, Q9
A 54 m mobile tower and a hostel stand on level ground. Measured at ground level from the foot of the tower, the hostel's roof is $30^{\circ}$ above the horizontal. From the foot of the hostel, the tower's top is $60^{\circ}$ above it. How tall is the hostel?
m
Show solution
Answer $18 m$
From the hostel's foot: $\tan 60^{\circ} = \frac{54}{d},$ so $d = \frac{54}{\sqrt{3}} = 18 \sqrt{3} .$
From the tower's foot: hostel $= d \tan 30^{\circ}$ $= 18 \sqrt{3} \times \frac{1}{\sqrt{3}} = 18 m .$
NumericalMediumNCERT Exercise 9.1, Q10
A sports ground is 48 m across, with a floodlight tower at each end of a line straight across it. The towers are equally tall. Standing on that line between them, Divya sees the top of one tower at $60^{\circ}$ and the other at $30^{\circ}$ above the horizontal (ignore her height). How tall are the towers, and how far is she from each? Use $\sqrt{3} \approx 1.732.$
Show solution
AnswerTowers $12 \sqrt{3} \approx 20.78 m;$ she is 12 m and 36 m from them
$h = x \tan 60^{\circ} = \sqrt{3} x$ and $h = (48 - x) \tan 30^{\circ} = \frac{48 - x}{\sqrt{3}} .$
So $3 x = 48 - x,$ giving $x = 12.$ The other distance is $36 m .$
$h = 12 \sqrt{3}$ $\approx 20.78 m .$
NumericalHardNCERT Exercise 9.1, Q11
Neha stands at the edge of a canal, exactly opposite a clock tower on the far bank, and sees its top $60^{\circ}$ above the horizontal. She walks 10 m straight back from the water, away from the tower, and the top is now $45^{\circ}$ above the horizontal. Ignoring her height, how tall is the tower and how wide is the canal? Use $\sqrt{3} \approx 1.732.$
Show solution
AnswerTower $5 (3 + \sqrt{3}) \approx 23.66 m;$ canal $\approx 13.66 m$ wide
At the edge: $h = w \tan 60^{\circ}$ $= \sqrt{3} w .$ After stepping back: $h = (w + 10) \tan 45^{\circ}$ $= w + 10.$
So $w (\sqrt{3} - 1) = 10,$ and $w = 5 (\sqrt{3} + 1)$ $\approx 13.66 m .$
$h = w + 10$ $\approx 23.66 m .$
NumericalMediumNCERT Exercise 9.1, Q12
Standing on a hostel terrace 15 m above the ground, Farhan looks up at the top of a mobile tower, $60^{\circ}$ above the horizontal, and down at its foot, $30^{\circ}$ below it. How tall is the tower?
m
Show solution
Answer $60 m$
Foot: by alternate angles, $\tan 30^{\circ} = \frac{15}{d},$ so $d = 15 \sqrt{3} .$
Top: the part above the terrace is $d \tan 60^{\circ}$ $= 15 \sqrt{3} \times \sqrt{3} = 45.$
Tower $= 15 + 45 = 60 m .$
NumericalMediumNCERT Exercise 9.1, Q13
Two boats sit on a lake in a straight line, one behind the other, as seen from a viewpoint 90 m above the water. The nearer boat is $60^{\circ}$ below the horizontal and the farther one $45^{\circ}$ below it. How far apart are the boats? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $90 - 30 \sqrt{3} \approx 38.04 m$
Nearer boat: $\frac{90}{\tan 60^{\circ}} = \frac{90}{\sqrt{3}} = 30 \sqrt{3} .$ Farther boat: $\frac{90}{\tan 45^{\circ}} = 90.$
Gap $= 90 - 30 \sqrt{3}$ $\approx 90 - 51.96$ $= 38.04 m .$
NumericalMediumNCERT Exercise 9.1, Q14
A drone flies level, 61.5 m above a playground, moving away from Aman. When he first looks up, his line of sight to the drone is $60^{\circ}$ above the horizontal; a little later it is only $45^{\circ} .$ His eyes are 1.5 m above the ground. How far did the drone fly between the two sightings? Use $\sqrt{3} \approx 1.732.$
m
Show solution
Answer $60 - 20 \sqrt{3} \approx 25.36 m$
Height above his eyes $= 61.5 - 1.5 = 60 m .$
Horizontal distances: $\frac{60}{\sqrt{3}} = 20 \sqrt{3}$ at $60^{\circ},$ and $60$ at $45^{\circ} .$
Distance travelled $= 60 - 20 \sqrt{3}$ $\approx 25.36 m .$
NumericalHardNCERT Exercise 9.1, Q15
A guard on top of a watchtower watches a cyclist pedalling steadily along a straight road that leads to the tower's foot. When she first looks, the cyclist is $30^{\circ}$ below her horizontal; 10 seconds later, $45^{\circ}$ below it. How many more seconds until the cyclist reaches the foot of the tower? Use $\sqrt{3} \approx 1.732.$
s
Show solution
Answer $5 (\sqrt{3} + 1) \approx 13.66 s$
Let the tower be $h$ m. The distances are $h \sqrt{3}$ at $30^{\circ}$ and $h$ at $45^{\circ} .$
In 10 s the cyclist covers $h (\sqrt{3} - 1),$ so the last $h$ metres take $\frac{10 h}{h (\sqrt{3} - 1)}$ seconds.
Time $= \frac{10}{\sqrt{3} - 1}$ $= 5 (\sqrt{3} + 1)$ $\approx 13.66 s .$
Quick answerEasy
A pole 9 m tall casts a shadow $3 \sqrt{3}$ m long. The altitude of the Sun is
$30^{\circ}$ $45^{\circ}$ $60^{\circ}$ $90^{\circ}$
Show answer
Answer (C) $60^{\circ}$
$\tan θ = \frac{9}{3 \sqrt{3}} = \sqrt{3},$ so $θ = 60^{\circ} .$ $30^{\circ}$ would need the shadow to be $\sqrt{3}$ times the height.

9.2 Summary

Recap

Some Applications of Trigonometry: key points and summary

The line of sight joins the observer's eye to the point being looked at (Line of sight and the angle of elevation).
The angle of elevation is measured up from the horizontal through the eye, when the point is above eye level (Fig. 9.1).
The angle of depression is measured down from the horizontal through the eye, when the point is below eye level (The angle of depression).
The angle of depression of $B$ from $A$ equals the angle of elevation of $A$ from $B$ : they are alternate angles (the key idea).
Height from an angle of elevation: $H = e + d \tan θ,$ with $e$ the eye height ((9.1)).
Use $\tan$ for height and horizontal distance; use $\sin$ or $\cos$ when a ladder, rope or string (the hypotenuse) is involved (Table 9.2).
With two angles, find the side two right triangles share and write one equation per triangle (Problems with two right triangles).
Keep surds such as $40 \sqrt{3}$ to the end, then use $\sqrt{3} \approx 1.732$ (Table 9.1).

1/3Height and horizontal distance: opposite and adjacent, so use $\tan θ .$

Fig. 9.12Which ratio? Look at the sides you know and want.

A recap in three steps. Step 1 shows a right triangle with the height and the horizontal distance highlighted, the angle theta at the bottom left and the words use tan, opposite and adjacent. Step 2 shows the same triangle with the slanting side highlighted and labelled ladder, rope, string, and the words use sin or cos. Step 3 shows two right triangles drawn from two observers to the top of one tall vertical side, which is highlighted as the shared side, with angles alpha and beta and the words one equation each.

Flash cards · 5

Quick recall

1 / 5

Board-exam practice

These questions follow the formats of the CBSE board paper: multiple choice, assertion–reason, short and long answers, and a case study.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 9.1, Q1	Practice Q1
NCERT Exercise 9.1, Q2	Practice Q2
NCERT Exercise 9.1, Q3	Practice Q3
NCERT Exercise 9.1, Q4	Practice Q4
NCERT Exercise 9.1, Q5	Practice Q5
NCERT Exercise 9.1, Q6	Practice Q6
NCERT Exercise 9.1, Q7	Practice Q7
NCERT Exercise 9.1, Q8	Practice Q8
NCERT Exercise 9.1, Q9	Practice Q9
NCERT Exercise 9.1, Q10	Practice Q10
NCERT Exercise 9.1, Q11	Practice Q11
NCERT Exercise 9.1, Q12	Practice Q12
NCERT Exercise 9.1, Q13	Practice Q13
NCERT Exercise 9.1, Q14	Practice Q14
NCERT Exercise 9.1, Q15	Practice Q15

Quick check: MCQs on Some Applications of Trigonometry

Elevation or depression?4 questions
Choose the ratio3 questions
One triangle3 questions
Two triangles3 questions

Glossary · 8

Key terms (English–Hindi)

alternate anglesएकांतर कोणekantar kon: Two angles on opposite sides of a line that cuts two other lines, between them. They are equal when the two lines are parallel.First seen in the chapter →
altitude of the Sunसूर्य का उन्नतांशsurya ka unnatansh: The angle of elevation of the Sun seen from the ground. The higher the Sun, the shorter the shadows.First seen in the chapter →
angle of depressionअवनमन कोणavanaman kon: The angle between the line of sight and the horizontal when the point looked at is below eye level.First seen in the chapter →
angle of elevationउन्नयन कोणunnayan kon: The angle between the line of sight and the horizontal when the point looked at is above eye level.First seen in the chapter →

clinometer: A simple instrument for measuring angles of elevation or depression, such as a protractor with a straw and a hanging weight.First seen in the chapter →

horizontalक्षैतिजkshaitij: The level line through the observer's eye, parallel to level ground. Angles of elevation and depression are measured from it.First seen in the chapter →

line of sightदृष्टि रेखाdrishti rekha: The straight line from an observer's eye to the point being looked at.First seen in the chapter →

trigonometric ratioत्रिकोणमितीय अनुपातtrikonmitiya anupat: A ratio of two sides of a right triangle, tied to one of its acute angles, such as $\tan θ = \frac{opposite}{adjacent} .$ First seen in the chapter →

Some Applications of Trigonometry: Class 10 Maths · Chapter 9

What you will learn in Some Applications of Trigonometry

9.1 Heights and Distances

Line of sight and the angle of elevation

The angle of depression

Planning a heights problem

Problems with one right triangle

Problems with two right triangles

Practice 9.1: questions with answers

9.2 Summary

Some Applications of Trigonometry: key points and summary

Board-exam practice

Practice questions with answers (NCERT “Board-exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Some Applications of Trigonometry

Key terms (English–Hindi)

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