Some Applications of Trigonometry

Stand at the foot of a tall minar and look up at its top: your eyes trace a straight line, tilted up from the horizontal by some angle. Nobody needs to climb the minar with a tape to find its height — the previous chapter's trigonometric ratios, applied to that one tilted line and a distance you can pace out, hand you the answer directly. That is the entire idea of this chapter, dressed up in different everyday scenes: a circus artist's rope tied from a pole to the ground, an electrician's ladder leaned at a fixed angle, a kite string pulled taut against the sky, the shadow a tower throws at different times of day, and a river whose width you need without wading across it. The chapter gives this idea precise names. The line of sight is the line from an observer's eye to the point being viewed; when that point is above the horizontal, the tilt angle is the angle of elevation, and when it is below, it is the angle of depression. Every problem reduces to the same routine — draw a right-triangle diagram from the words, decide which trigonometric ratio connects the known side to the wanted one, and solve. What makes the chapter genuinely test problem-solving skill, rather than rote formula plugging, is that many situations need two right triangles stitched together: a flagstaff on top of a building, two buildings facing each other across a road, or the same tower viewed from two points as an observer walks toward it. By the end of this chapter you will translate a height-and-distance word problem into a labelled diagram without hesitation, pick the correct ratio — tangent almost always, occasionally sine or cotangent — and handle two-triangle problems as simultaneous equations. These techniques resurface in coordinate geometry, in three-dimensional geometry at higher levels, and in any real-world surveying task where direct measurement is impossible but angles and one distance are not.

Key takeaways

The line of sight is the line drawn from an observer's eye to the point on the object being viewed.
The angle of elevation is the angle between the line of sight and the horizontal when the observer raises their head to look at a point above the horizontal level.
The angle of depression is the angle between the line of sight and the horizontal when the observer lowers their head to look at a point below the horizontal level.
Every height-and-distance problem is solved by drawing a right triangle from the description and choosing the trigonometric ratio that links the known side or angle to the unknown one — usually tangent, since it directly relates the vertical (opposite) side to the horizontal (adjacent) side.
When the observer's own eye-level height matters, the total height of the object is the height found from the triangle plus the observer's height, since the line of sight starts at eye level, not at the ground.
Problems with a flagstaff on a building, or two buildings of different heights, need two right triangles sharing a common side, solved as a pair of simultaneous trigonometric equations.
The angle of depression from an elevated point to a ground point equals the angle of elevation from that ground point back up, because the two horizontal lines are parallel and the line of sight is a transversal, making them alternate angles.
When a tower's foot cannot be reached, its height can still be found from two angles of elevation measured from two points a known distance apart on the same line as the foot of the tower.
As an observer walks toward a tall object, the angle of elevation increases; this fact underlies "boy walking towards a building" and "car approaching a tower" problems that ask for a distance walked or a time taken.
Key idea: every height-and-distance problem becomes a right triangle whose angle is the angle of elevation or depression and whose sides are found with tangent, sine, or cotangent — with two right triangles needed whenever two angles are given from the same or different points.

The core insight

Height-and-distance problems always reduce to one or two right triangles built from the angle of elevation or depression, solved with the trigonometric ratio that connects the known side to the unknown one.

What this chapter covers

The chapter opens by naming the two central angles of the whole topic: the angle of elevation, formed when the line of sight rises above the horizontal to view a point overhead, and the angle of depression, formed when it dips below the horizontal to view a point below. It then works through NCERT's own Examples 1 to 7 — a tower found from a single angle (Example 1), a ladder problem needing both sine and cotangent (Example 2), a chimney where the observer's own height matters (Example 3), a flagstaff sitting on a building requiring two right triangles (Example 4), a shadow-length problem solved with two angles of elevation from two points (Example 5), a pair of buildings connected through angles of depression and alternate angles (Example 6), and a river width found from a bridge (Example 7). Across every one of these, the method never changes: draw the triangle, name the sides relative to the known angle, pick tangent, sine, or cotangent, and solve.

The core ideas

Line of sight, elevation and depression — the vocabulary

The line of sight is the straight line joining the observer's eye to the point on the object being looked at. Two everyday situations give this line a special name for its tilt above or below the horizontal. When the object is higher than eye level, the observer must raise their head, and the angle between the line of sight and the horizontal is the angle of elevation of that point. When the object is lower than eye level, the observer must lower their head, and the corresponding angle is the angle of depression. Both angles are always measured from a horizontal reference line drawn at the observer's own eye, never from the ground and never from a vertical.

Choosing the right trigonometric ratio

Once the diagram is drawn as a right triangle, the whole problem collapses to picking the correct trigonometric ratio. If the known angle sits at the observer, the side opposite that angle is the vertical height being sought (or the vertical side already known), and the side adjacent to it is the horizontal distance. The ratio that connects opposite and adjacent directly is the tangent (or its reciprocal, cotangent), which is why nearly every problem in this chapter reaches for $\tan\theta$ first.

H = d\tan\theta + h_e \qquad (1)

where: $H$ is the total height of the object above the ground, $d$ is the horizontal distance from the observer to the foot of the object, $\theta$ is the angle of elevation of the top of the object from the observer's eye, and $h_e$ is the height of the observer's own eye above the ground. When the observer's height is negligible or the angle is measured from ground level with an instrument, $h_e = 0$ and the formula reduces to the simpler $H = d\tan\theta$ used whenever a "tower stands on level ground" and the angle is measured from a point on the ground itself.

Figure

A horizontal eye level line with one line of sight rising to a tower top marked angle of elevation and a second line of sight from a height falling to a ground point marked angle of depression

The two-point method and the angle of depression as an alternate angle

Many problems give two angles instead of one — either two angles of elevation measured from two different points on the ground, or an angle of depression paired with an angle of elevation across two structures. Both patterns reduce to solving two right-triangle equations together.

H = \dfrac{d\,\tan\alpha\,\tan\beta}{\tan\beta - \tan\alpha} \qquad (2)

where: $H$ is the height of a tower whose foot cannot be directly reached, $d$ is the distance between two points on the same straight line as the foot, $\alpha$ is the (smaller) angle of elevation from the farther point, and $\beta$ is the (larger) angle of elevation from the nearer point, with $\beta > \alpha$ always, since moving closer to a tall object always increases its angle of elevation. The second recurring pattern uses the fact that the horizontal at an elevated observer and the horizontal at a ground-level point are parallel lines, so the line of sight joining them is a transversal, and the angle of depression at the top exactly equals the angle of elevation measured back up from the ground point, by the property of alternate angles.

Figure

A building with a flagstaff on top and a point on the ground with two lines of sight, one to the foot of the flagstaff at thirty degrees and one to its top at forty five degrees

Two to three worked derivations

Derivation ADeriving the general height formula from a single angle of elevation

NCERT introduces this method in words at the start of the chapter, using a student looking up at a minar, but it deserves to be written out as a clean, reusable formula, because nearly every single-triangle problem in the chapter — a tower, a pole, a chimney, a kite — is this one derivation wearing a different costume. The goal is to derive equation $(1)$ , $H = d\tan\theta + h_e$ , from first principles, starting only from the definition of the angle of elevation and the tangent ratio, and to show precisely why the observer's own height must be added back in.

Step 1. Let the object stand vertically as $ED$ , with foot $E$ on the ground and top $D$ . Let the observer stand at $F$ , a horizontal distance $d$ from $E$ , with eyes at point $A$ , so that $AF = h_e$ is the observer's eye height and $EF = d$ . Draw the horizontal line through $A$ meeting the object at point $B$ , so $AB$ is horizontal and equal to $d$ , and $BD$ is the remaining part of the object above eye level, while $BE = AF = h_e$ (since $ABEF$ is a rectangle).

Step 2. The line of sight from the observer's eye $A$ to the top $D$ makes the angle of elevation $\theta$ with the horizontal $AB$ , so triangle $ABD$ is right-angled at $B$ with the known angle $\theta$ at $A$ . The side $BD$ is opposite $\theta$ and the side $AB = d$ is adjacent to it, so the ratio connecting them directly is the tangent: $\tan\theta = \dfrac{BD}{AB} = \dfrac{BD}{d}$ .

Step 3. Solve for the unknown side: $BD = d\tan\theta$ . This is the portion of the object's height that lies above the observer's eye level, found purely from the angle and the horizontal distance.

Step 4. The total height of the object is $H = ED = EB + BD = h_e + d\tan\theta$ , which is exactly equation $(1)$ . When the angle is measured by an instrument placed directly on the ground, or when the observer's height is small enough to ignore, $h_e = 0$ and this collapses to the familiar $H = d\tan\theta$ , the formula behind NCERT's Example 1, where a $15$ m distance and a $60°$ angle of elevation give a tower height of $15\sqrt{3}$ m.

Go deeper · Derivation B · 1 minDeriving the two-point formula for the height of an inaccessible tower

Derivation BDeriving the two-point formula for the height of an inaccessible tower

NCERT's Example 5 — a tower's shadow is $40$ m longer when the sun's altitude is $30°$ than when it is $60°$ — is solved from scratch using two separate right triangles and a pair of simultaneous equations. This derivation carries out the same idea in full generality, producing equation $(2)$ , $H = \dfrac{d\tan\alpha\tan\beta}{\tan\beta - \tan\alpha}$ , so that any "two angles of elevation from two points a known distance apart" problem can be solved by substitution into a single formula instead of re-deriving it every time.

Step 1. Let the tower have foot $O$ and top $T$ , with height $H = OT$ . Let $A$ be the farther observation point, where the angle of elevation of $T$ is the smaller angle $\alpha$ , and let $B$ be the nearer point, on segment $OA$ , where the angle of elevation is the larger angle $\beta$ , with the given distance $AB = d$ between them.

Step 2. In right triangle $OAT$ , $\tan\alpha = \dfrac{OT}{OA} = \dfrac{H}{OA}$ , so $OA = H\cot\alpha$ . In right triangle $OBT$ , $\tan\beta = \dfrac{OT}{OB} = \dfrac{H}{OB}$ , so $OB = H\cot\beta$ .

Step 3. Since $B$ lies between $A$ and $O$ , the given distance is the difference of these two distances: $AB = OA - OB = H\cot\alpha - H\cot\beta = H(\cot\alpha - \cot\beta) = d$ . Rewrite $\cot\alpha - \cot\beta$ using $\cot\theta = \dfrac{1}{\tan\theta}$ : $\cot\alpha - \cot\beta = \dfrac{1}{\tan\alpha} - \dfrac{1}{\tan\beta} = \dfrac{\tan\beta - \tan\alpha}{\tan\alpha\tan\beta}$ .

Step 4. Substitute and solve for $H$ : $d = H\cdot\dfrac{\tan\beta - \tan\alpha}{\tan\alpha\tan\beta}$ , so $H = \dfrac{d\tan\alpha\tan\beta}{\tan\beta - \tan\alpha}$ , exactly equation $(2)$ . Checking it against Example 5, with $\alpha = 30°$ , $\beta = 60°$ , $d = 40$ : $H = \dfrac{40\cdot\frac{1}{\sqrt3}\cdot\sqrt3}{\sqrt3 - \frac{1}{\sqrt3}} = \dfrac{40}{\frac{2}{\sqrt3}} = \dfrac{40\sqrt3}{2} = 20\sqrt3$ m, matching NCERT's own answer exactly, which confirms the formula.

Go deeper · Derivation C · 2 minDeriving the alternate-angle rule and solving a two-building depression system

Derivation CDeriving the alternate-angle rule and solving a two-building depression system

NCERT's Example 6 involves an $8$ m building observed from the top of a taller, multi-storeyed building, with the angles of depression to the smaller building's top and bottom given as $30°$ and $45°$ . Before that system can be set up, one geometric fact needs justifying: that an angle of depression at the top equals the corresponding angle of elevation measured back up from the ground. This derivation proves that fact and solves the full system.

Step 1. At the elevated observer $P$ , draw the horizontal line $PQ'$ . At the ground point $A$ being viewed, draw the horizontal line through $A$ . Because "horizontal" means the same fixed direction everywhere on a level plane, these two horizontal lines are parallel. The line of sight $PA$ crosses both of them, making it a transversal.

Step 2. A transversal crossing two parallel lines creates equal alternate interior angles on opposite sides of the transversal. The angle of depression at $P$ (between the horizontal at $P$ and $PA$ ) and the angle of elevation at $A$ (between the horizontal at $A$ and $AP$ ) are exactly this pair of alternate interior angles, so they are equal. This justifies writing, for instance, $\angle PAC = 45°$ whenever the angle of depression of $A$ from $P$ is given as $45°$ .

Step 3. Now set up Example 6's system. Let $PC$ be the multi-storeyed building of height $H$ , and $AB = 8$ m the smaller building, with $AC = BD = x$ the horizontal distance between them ( $ABDC$ forms a rectangle since both buildings stand on level ground). The angle of depression of the bottom $A$ is $45°$ , so by Step 2, $\angle PAC = 45°$ in right triangle $PAC$ : $\tan 45° = \dfrac{PC}{AC} = \dfrac{H}{x} = 1$ , giving $H = x$ . The angle of depression of the top $B$ is $30°$ , so $\angle PBD = 30°$ in right triangle $PBD$ , where $PD = PC - DC = H - 8$ and $BD = x$ : $\tan 30° = \dfrac{PD}{BD} = \dfrac{H-8}{x} = \dfrac{1}{\sqrt3}$ .

Step 4. Substitute $H = x$ from Step 3 into the second equation: $\sqrt3(x - 8) = x$ , so $\sqrt3 x - 8\sqrt3 = x$ , giving $x(\sqrt3 - 1) = 8\sqrt3$ , so $x = \dfrac{8\sqrt3}{\sqrt3 - 1} = \dfrac{8\sqrt3(\sqrt3+1)}{2} = 4\sqrt3(\sqrt3+1) = 12 + 4\sqrt3$ . Since $H = x$ , both the height of the multi-storeyed building and the distance between the two buildings equal $12 + 4\sqrt3 \approx 18.93$ m, matching NCERT's own solution.

Worked examples

Board level - Height of a tower from a single angle of elevation (Example 1 pattern)

A tower stands vertically on the ground. From a point on the ground $15$ m away 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 1. Draw the right triangle $ACB$ , right-angled at $B$ , where $AB$ is the tower, $CB = 15$ m is the horizontal distance, and $\angle ACB = 60°$ is the angle of elevation.

Step 2. The side $AB$ (height, unknown) is opposite the known angle, and $CB = 15$ m (adjacent) is known, so use the tangent ratio: $\tan 60° = \dfrac{AB}{CB}$ .

Step 3. Substitute $\tan 60° = \sqrt3$ : $\sqrt3 = \dfrac{AB}{15}$ , so $AB = 15\sqrt3$ .

Step 4. This uses equation $(1)$ with $h_e = 0$ , since the angle is measured directly from ground level. Answer: the height of the tower is 15√3 m, approximately 25.98 m.

Board level - Ladder length and foot distance using two ratios (Example 2 pattern)

An electrician has to repair a fault on a pole of height $5$ m. She needs to reach a point $1.3$ m below the top of the pole, using a ladder inclined at $60°$ to the horizontal. Find the length of the ladder and the distance of its foot from the pole. Take $\sqrt3 = 1.73$ .

Step 1. The point to be reached is $BD = 5 - 1.3 = 3.7$ m up the pole $AD$ . The ladder $BC$ is the hypotenuse of right triangle $BDC$ , right-angled at $D$ , with $\angle BCD = 60°$ .

Step 2. To find the hypotenuse $BC$ from the opposite side $BD$ , use sine: $\sin 60° = \dfrac{BD}{BC} = \dfrac{3.7}{BC}$ , so $BC = \dfrac{3.7}{\sin 60°} = \dfrac{3.7 \times 2}{\sqrt3} = \dfrac{7.4}{1.73} \approx 4.28$ m.

Step 3. To find the foot distance $DC$ from the same opposite side $BD$ , use cotangent, which connects adjacent and opposite directly: $\cot 60° = \dfrac{DC}{BD}$ , so $DC = BD\cot 60° = \dfrac{3.7}{\sqrt3} = \dfrac{3.7}{1.73} \approx 2.14$ m.

Step 4. Two different ratios were needed here because the two unknowns, hypotenuse and adjacent side, are related to the known opposite side $BD$ by two different pairs of ratios. Answer: the ladder should be about 4.28 m long, with its foot about 2.14 m from the pole.

Board HOTS / foundation JEE-NEET - Twin buildings from two angles of depression (Example 6 pattern)

The angles of depression of the top and bottom of an $8$ m tall building from the top of a multi-storeyed building are $30°$ and $45°$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Step 1. Let $PC$ be the multi-storeyed building of height $H$ and $AB = 8$ m the smaller building, with common horizontal distance $AC = BD = x$ . By the alternate-angle rule proved in the third derivation, the angle of depression of the bottom $A$ , $45°$ , equals $\angle PAC$ , and the angle of depression of the top $B$ , $30°$ , equals $\angle PBD$ .

Step 2. In right triangle $PAC$ : $\tan 45° = \dfrac{H}{x} = 1$ , so $H = x$ .

Step 3. In right triangle $PBD$ , where $PD = H - 8$ : $\tan 30° = \dfrac{H-8}{x} = \dfrac{1}{\sqrt3}$ , so $\sqrt3(x-8) = x$ , giving $x(\sqrt3 - 1) = 8\sqrt3$ and $x = \dfrac{8\sqrt3}{\sqrt3-1} = 4\sqrt3(\sqrt3+1) = 12 + 4\sqrt3$ .

Step 4. Since $H = x$ , both quantities equal $12 + 4\sqrt3 \approx 18.93$ m. Answer: the multi-storeyed building is 12 + 4√3 m (about 18.93 m) tall, and the two buildings are also 12 + 4√3 m (about 18.93 m) apart. Timing note: budget about three to four minutes for a two-triangle depression problem like this. The fastest route is to spot immediately that the $45°$ triangle gives a one-line relation ( $H = x$ , since $\tan 45° = 1$ ), substitute it straight into the $30°$ equation, and rationalise the surd denominator only once, at the very end, rather than at every intermediate step.

Common traps

The most common error is measuring the angle of elevation or depression from the wrong reference line — both are always measured from the horizontal at the observer's eye, never from the ground and never from the vertical side of the triangle. A closely related mistake is forgetting to add the observer's own eye height back onto the triangle's answer in problems like the chimney (Example 3) or the balloon (Exercise 9.1), where the observer is a person of given height rather than an instrument on the ground. Students also frequently pick the wrong ratio: if the two known or wanted sides are the hypotenuse and one leg, sine or cosine is needed, not tangent, and mixing this up is the single most common source of a wrong final answer. In two-triangle problems, another frequent slip is assuming the horizontal distance is the same in both triangles without checking that the two structures genuinely stand on the same level ground (what actually makes $AC = BD$ true in the twin-building problem). Finally, many students leave a surd like $\dfrac{1}{\sqrt3}$ un-rationalised or substitute an approximate value of $\sqrt3$ too early, piling up rounding errors through a multi-step problem.

Common trapAngle of depression is not measured at the ground point and is not automatically the same number as any angle inside the triangle without justification

The single biggest misconception in this chapter is treating the angle of depression as though it were simply "the angle at the bottom of the diagram." It is not — the angle of depression is always the angle between the horizontal at the elevated observer and the line of sight, drawn at the top of the figure, not at the ground point being viewed. It can be transferred down to the ground point, and used inside the ground-level right triangle, only because of the alternate-angle argument in this chapter's third derivation: the horizontal at the observer and the horizontal at the ground point are parallel, and the line of sight is a transversal, so the angle of depression above equals the angle of elevation below. Skipping this justification and simply writing the given angle into the ground triangle without saying why gives the right numeric answer but earns no credit in a board answer, since the "why" is exactly what the alternate-angle property is testing.

You should be able to…

Define the line of sight, angle of elevation, and angle of depression precisely, and identify which one applies in a given word problem.
Translate a height-and-distance word problem into a correctly labelled right-triangle diagram before writing any trigonometric equation.
Choose the correct trigonometric ratio, tangent, sine, cosine, or cotangent, based on which two sides relative to the known angle are involved.
Add or subtract the observer's own eye height correctly, depending on whether the angle is measured from the ground or a person's eye level.
Justify, using the alternate-angle property of parallel lines and a transversal, why depression at an elevated point equals elevation at the ground point.
Set up and solve two simultaneous right-triangle equations for a flagstaff on a building, two buildings, or two angles of elevation from two points.
Apply the two-point formula to find the height of a tower whose foot cannot be reached, given two angles of elevation and a distance between points.
Recognise that the angle of elevation increases as the observer moves closer to a tall object, and use this in "distance walked" or "time taken" problems.
Simplify and rationalise surds such as $\dfrac{1}{\sqrt3}$ , and substitute given decimal approximations only at the final step of a calculation.

Ask the tutor: Want to work through one more two-triangle problem together, step by step, using your own choice of numbers?

What exactly is the difference between the angle of elevation and the angle of depression?

Both are angles between a line of sight and a horizontal line drawn at the observer's own eye level, and the only difference is the direction of tilt. If the point is higher than the observer, the head is raised, and that upward tilt is the angle of elevation. If the point is lower, the head is lowered, and that downward tilt is the angle of depression. Neither angle is ever measured from the ground or a vertical line — always from the horizontal at eye level, which is the detail students most often forget under exam pressure.

Why do some problems add the observer's height to the triangle's answer while others do not?

It depends on where the angle is measured from. If a problem says an instrument or a point on the ground gives the angle of elevation, the triangle already reaches from the ground to the top of the object, so nothing extra is added. But if a person of given height is looking up (like the $1.5$ m tall observer viewing the chimney in Example 3, or the $1.2$ m tall girl watching the balloon), the triangle only covers the part above that person's eye level, and the person's own height must be added back on, exactly as equation $(1)$ shows.

How do I decide whether to use tangent, sine, or cotangent in a given problem?

Identify which two sides of the right triangle — opposite, adjacent, or hypotenuse relative to the known angle — are involved, one already known and one wanted. If those two sides are opposite and adjacent, use tangent (or cotangent, its reciprocal). If one of them is the hypotenuse, use sine (opposite and hypotenuse) or cosine (adjacent and hypotenuse). The ladder problem is a good example: the hypotenuse (ladder length) needed sine, while the adjacent side (foot distance) needed cotangent, both from the same known opposite side.

Why do twin-building or flagstaff problems always need two equations instead of one?

Because there are two unknowns, and a single right triangle with one known angle only ever gives one equation relating two sides. Whenever a problem gives two separate angles, such as the $30°$ and $45°$ depression angles in the twin-building problem, or the $30°$ and $45°$ elevation angles to a building and its flagstaff, each angle sits inside its own right triangle, and writing the tangent equation for both gives exactly two equations for the two unknowns, solved together like any pair of simultaneous equations.