Some Applications of Trigonometry Class 10: Heights and Distances

What Some Applications of Trigonometry is about

Weightage: This chapter is part of Unit V, Trigonometry, which carries 12 of the 80 theory marks together with Introduction to Trigonometry.Source: CBSE Mathematics syllabus 2026-27, Unit V Trigonometry

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.7 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.5 marks per paper. See the questions

Key concepts

Line of sight and angle of elevation

NCERT section 9.1 Heights and Distances

The line of sight is the straight line from your eye to the object you are looking at. When the object is above your eye level, you raise your head. The angle between the horizontal line through your eye and the line of sight is the angle of elevation.

Why this matters: the horizontal ground, the vertical height of the object and the line of sight form a right triangle. The right angle sits at the foot of the object. Once you know one side and the angle of elevation, the trigonometric ratios give you every other side.

Always measure the angle from the horizontal, never from the vertical. Mark it at the observer's eye, between the ground line and the slanted line of sight. Use $\tan$ when you link height and horizontal distance, and $\sin$ or $\cos$ when the slanted distance (hypotenuse) is involved.

Worked example

A drone hovers vertically above a point B on level ground at a height of 45 m. From a point A on the same ground, the angle of elevation of the drone is $60^{\circ }$. Find (i) the distance AB and (ii) the straight-line distance from A to the drone. Use $\sqrt{3}=1.732$.

Try it first, then show the solutionSolution

1. Let D be the drone. Then $DB=45$ m, $\angle DBA=90^{\circ }$ and $\angle DAB=60^{\circ }$.
2. In right $△DBA$: $\tan 60^{\circ }=\frac{DB}{AB}$
3. $\sqrt{3}=\frac{45}{AB}$, so $AB=\frac{45}{\sqrt{3}}=\frac{45\sqrt{3}}{3}=15\sqrt{3}$ m
4. $AB=15\times 1.732=25.98$ m
5. Also $\sin 60^{\circ }=\frac{DB}{AD}$, so $\frac{\sqrt{3}}{2}=\frac{45}{AD}$
6. $AD=\frac{90}{\sqrt{3}}=30\sqrt{3}=51.96$ m
7. Check: $A{B}^2+D{B}^2=675+2025=2700=(30\sqrt{3}{)}^2$ ✓

Answer: (i) $AB=15\sqrt{3}\approx 25.98$ m; (ii) distance from A to the drone $=30\sqrt{3}\approx 51.96$ m.

Angle of depression and why it equals the angle of elevation

NCERT section 9.1 Heights and Distances

When the object is below your eye level, you lower your head. The angle between the horizontal line through your eye and the line of sight going down is the angle of depression.

The key fact: the horizontal line through the observer's eye is parallel to the ground. The line of sight cuts both parallel lines, so the angle of depression at the top equals the angle the line of sight makes with the ground at the bottom (alternate interior angles). So the angle of depression of B from A equals the angle of elevation of A from B.

This is why you never put the depression angle inside the triangle at the top. Draw the horizontal at the eye, mark the depression there, then transfer it to the ground point using alternate angles and write the reason. Now it is an ordinary elevation problem.

Worked example

From the top of a vertical cliff 80 m high, the angle of depression of a boat at sea is $30^{\circ }$. Find the distance of the boat from the foot of the cliff and the length of the line of sight from the top of the cliff to the boat. Use $\sqrt{3}=1.732$.

Try it first, then show the solutionSolution

1. Let AB be the cliff with top A, foot B, and boat C on the sea level through B. $AB=80$ m.
2. Draw horizontal AX through A. $\angle XAC=30^{\circ }$ (angle of depression).
3. $AX\parallel BC$, so $\angle ACB=\angle XAC=30^{\circ }$ (alternate angles).
4. In right $△ABC$: $\tan 30^{\circ }=\frac{AB}{BC}$, so $\frac{1}{\sqrt{3}}=\frac{80}{BC}$
5. $BC=80\sqrt{3}=80\times 1.732=138.56$ m
6. $\sin 30^{\circ }=\frac{AB}{AC}$, so $\frac{1}{2}=\frac{80}{AC}$, giving $AC=160$ m

Answer: The boat is $80\sqrt{3}\approx 138.56$ m from the foot of the cliff, and the line of sight is 160 m long.

Choosing the right ratio: tan, sin or cos

NCERT section 9.1 Heights and Distances

Every heights and distances triangle has three sides: the vertical height (opposite the angle), the horizontal distance (adjacent to the angle) and the slanted length such as a rope, ladder, string or line of sight (the hypotenuse).

Before writing any ratio, ask: which side do I know, and which side do I want?

• Height and horizontal distance: use $\tan$.
• Height and slanted length: use $\sin$.
• Horizontal distance and slanted length: use $\cos$.

Most tower and building questions only involve height and ground distance, so $\tan$ is the most used ratio. Ladders, ropes, kite strings and slides bring in the hypotenuse, so $\sin$ or $\cos$ appears there. Picking the ratio that contains the known side and the unknown side means one equation, one unknown, and no extra work.

Worked example

A rope is tied from the top of a vertical pole to a peg on level ground 8 m from the foot of the pole. The rope is taut and makes an angle of $60^{\circ }$ with the ground. Find the height of the pole and the length of the rope.

Try it first, then show the solutionSolution

1. Let AB be the pole (A top, B foot) and C the peg. $BC=8$ m, $\angle ACB=60^{\circ }$, $\angle ABC=90^{\circ }$.
2. Height and ground distance are linked by tan: $\tan 60^{\circ }=\frac{AB}{BC}$
3. $\sqrt{3}=\frac{AB}{8}$, so $AB=8\sqrt{3}$ m
4. Ground distance and rope are linked by cos: $\cos 60^{\circ }=\frac{BC}{AC}$
5. $\frac{1}{2}=\frac{8}{AC}$, so $AC=16$ m
6. Check: $8^2+(8\sqrt{3}{)}^2=64+192=256=16^2$ ✓

Answer: Height of the pole $=8\sqrt{3}\approx 13.86$ m; length of the rope $=16$ m.

Finding the angle: shadows and the sun's altitude

NCERT section 9.1 Heights and Distances

Sometimes you know two sides and must find the angle. Work out the ratio, then match it with a standard value: $\tan \theta =\frac{1}{\sqrt{3}}$ gives $30^{\circ }$, $\tan \theta =1$ gives $45^{\circ }$, $\tan \theta =\sqrt{3}$ gives $60^{\circ }$.

Shadow questions are the classic case. The sun's rays touch the top of a vertical object and end at the tip of its shadow on the ground. The angle these rays make with the ground is the altitude of the sun, which is just the angle of elevation of the sun from the shadow's tip. So $\tan (\text{altitude})=\frac{\text{height}}{\text{shadow}}$.

This also explains a common reasoning question: as the sun rises higher, the altitude increases, $\tan$ increases, so for a fixed height the shadow gets shorter. At $45^{\circ }$ the shadow equals the height.

Worked example

A vertical tower 30 m tall casts a shadow $10\sqrt{3}$ m long on level ground at a certain time. (i) Find the altitude of the sun at that time. (ii) Later the altitude becomes $30^{\circ }$. By how much does the shadow lengthen?

Try it first, then show the solutionSolution

1. Let AB be the tower (height 30 m) and BC the first shadow, $BC=10\sqrt{3}$ m, $\angle B=90^{\circ }$.
2. $\tan \theta =\frac{AB}{BC}=\frac{30}{10\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$
3. $\tan \theta =\sqrt{3}=\tan 60^{\circ }$, so $\theta =60^{\circ }$
4. Let BD be the new shadow. $\tan 30^{\circ }=\frac{AB}{BD}$, so $\frac{1}{\sqrt{3}}=\frac{30}{BD}$
5. $BD=30\sqrt{3}$ m
6. Increase $=BD-BC=30\sqrt{3}-10\sqrt{3}=20\sqrt{3}$ m $\approx 34.64$ m

Answer: (i) Altitude of the sun $=60^{\circ }$; (ii) the shadow becomes $20\sqrt{3}\approx 34.64$ m longer.

Drawing the figure and modelling a word problem (including the observer's height)

NCERT section 9.1 Heights and Distances

Most lost marks start with a wrong figure. Use the same routine every time:

• Draw the ground as a horizontal line and every tower, pole or building as a vertical line.
• Mark the observer's eye. If the observer has a height, the horizontal line of sight starts at eye level, not at ground level.
• Mark the angle between the horizontal and the line of sight.
• Write known lengths and call the unknowns $h$ and $x$.

When the observer is a person of height $p$, the right triangle you solve has vertical side (object height $-p$), because the horizontal through the eye cuts the object at height $p$. You add $p$ back at the end if the question asks for the full height. Forgetting this shift is a very common error. A figure also shows which distances belong to which triangle, which is vital once there are two triangles.

Worked example

A girl whose eyes are 1.6 m above the ground stands on level ground in front of a building 25.6 m tall. The angle of elevation of the top of the building from her eyes is $30^{\circ }$. She walks straight towards the building until the angle of elevation becomes $60^{\circ }$. Find the distance she walked. Use $\sqrt{3}=1.732$.

Try it first, then show the solutionSolution

1. Let AB be the building (A top). The horizontal line through her eyes meets AB at E, so $EB=1.6$ m and $AE=25.6-1.6=24$ m.
2. Let her eye be at P first and Q later, both on this horizontal line, with $\angle APE=30^{\circ }$ and $\angle AQE=60^{\circ }$.
3. In right $△AEP$: $\tan 30^{\circ }=\frac{AE}{PE}$, so $PE=24\sqrt{3}$ m
4. In right $△AEQ$: $\tan 60^{\circ }=\frac{AE}{QE}$, so $QE=\frac{24}{\sqrt{3}}=8\sqrt{3}$ m
5. Distance walked $=PQ=PE-QE=24\sqrt{3}-8\sqrt{3}=16\sqrt{3}$ m
6. $16\times 1.732=27.712$ m

Answer: She walked $16\sqrt{3}\approx 27.71$ m towards the building.

Two right triangles on the same side (observer moves)

NCERT section 9.1 Heights and Distances

Here the observer sees the top of a tower from two points on the same straight line through its foot. The two triangles share the tower as a common vertical side.

The method: let the height be $h$ and the distance from the nearer point to the foot be $x$. The farther point is then at distance $x+d$, where $d$ is the distance between the two points. Write one $\tan$ equation for each triangle, so you get two equations in two unknowns. Express $h$ from both and equate, or substitute one into the other.

Why it works: you cannot solve either triangle alone, because each has only an angle and an unknown side. Together they give enough information. Note that the larger angle belongs to the nearer point: moving closer makes you look up more steeply. Use this as a quick check on your figure.

Worked example

From a point A on level ground, the angle of elevation of the top of a tower is $30^{\circ }$. After walking 40 m straight towards the tower to point B, the angle of elevation becomes $60^{\circ }$. Find the height of the tower and the distance of B from its foot. Use $\sqrt{3}=1.732$.

Try it first, then show the solutionSolution

1. Let PQ be the tower (P top, Q foot), $PQ=h$, $BQ=x$, so $AQ=x+40$.
2. In right $△PQB$: $\tan 60^{\circ }=\frac{h}{x}$, so $h=\sqrt{3}x$ ... (1)
3. In right $△PQA$: $\tan 30^{\circ }=\frac{h}{x+40}$, so $\sqrt{3}h=x+40$ ... (2)
4. Substitute (1) in (2): $\sqrt{3}(\sqrt{3}x)=x+40$, so $3x=x+40$
5. $2x=40$, so $x=20$ m
6. From (1): $h=20\sqrt{3}=34.64$ m
7. Check in (2): $\sqrt{3}\times 20\sqrt{3}=60=20+40$ ✓

Answer: Height of the tower $=20\sqrt{3}\approx 34.64$ m; B is 20 m from the foot of the tower.

Two right triangles on opposite sides

NCERT section 9.1 Heights and Distances

In this type, two objects (cars, boats, people) lie on opposite sides of a tower, in one straight line with its foot. Or one observer stands between two poles. The vertical line is shared, and the two horizontal distances add up to the total distance.

Method: solve each triangle separately using $\tan$, find each horizontal distance in terms of the height, then add. If the total distance is given instead and the height is unknown, write both distances in terms of $h$, add them, and set the sum equal to the given total.

When angles of depression are given from the top of the tower, first transfer each one to the ground using alternate angles. Then each car sees the top at an angle of elevation equal to the given depression. Do not subtract distances here: subtraction is for points on the same side.

Worked example

From the top of a tower 60 m high, the angles of depression of two cars on opposite sides of the tower are $30^{\circ }$ and $45^{\circ }$. The cars and the foot of the tower are in one straight line. Find the distance between the cars. Use $\sqrt{3}=1.732$.

Try it first, then show the solutionSolution

1. Let AB be the tower (A top), $AB=60$ m, with cars C and D on opposite sides of B.
2. By alternate angles, $\angle ACB=30^{\circ }$ and $\angle ADB=45^{\circ }$.
3. In right $△ABC$: $\tan 30^{\circ }=\frac{60}{BC}$, so $BC=60\sqrt{3}$ m
4. In right $△ABD$: $\tan 45^{\circ }=\frac{60}{BD}$, so $BD=60$ m
5. Cars are on opposite sides, so $CD=BC+BD=60\sqrt{3}+60=60(\sqrt{3}+1)$ m
6. $60\times 2.732=163.92$ m

Answer: The distance between the cars is $60(\sqrt{3}+1)\approx 163.92$ m.

Elevation and depression from the top of a building

NCERT section 9.1 Heights and Distances

An observer on top of a building looks up at the top of a taller tower and down at its foot. Both angles are measured from the same horizontal line through the observer's eye.

That horizontal line splits the tower into two parts. The lower part equals the building's height (both are the vertical gap between the horizontal and the ground). The upper part is the extra height above the observer. The horizontal distance between the building and the tower is common to both triangles.

Method: use the angle of depression with the building's height to find the common horizontal distance. Then use the angle of elevation with that distance to find the extra height. Add the two parts. The same idea works in reverse when you look down at both the top and the foot of a shorter object: then you subtract.

Worked example

From the top of a building 15 m high, the angle of elevation of the top of a tower is $60^{\circ }$ and the angle of depression of its foot is $30^{\circ }$. Find the height of the tower and the distance between the building and the tower. Use $\sqrt{3}=1.732$.

Try it first, then show the solutionSolution

1. Let AB be the building (A top, $AB=15$ m) and CD the tower (C top, D foot). Draw horizontal AE meeting CD at E, so $ED=AB=15$ m and $AE=BD$.
2. Angle of depression of D is $30^{\circ }$, so $\angle ADB=30^{\circ }$ (alternate angles).
3. In right $△ABD$: $\tan 30^{\circ }=\frac{15}{BD}$, so $BD=15\sqrt{3}$ m, hence $AE=15\sqrt{3}$ m
4. In right $△AEC$: $\tan 60^{\circ }=\frac{CE}{AE}$, so $CE=\sqrt{3}\times 15\sqrt{3}=45$ m
5. Height of tower $CD=CE+ED=45+15=60$ m
6. Distance $BD=15\times 1.732=25.98$ m

Answer: Height of the tower $=60$ m; distance between them $=15\sqrt{3}\approx 25.98$ m.

Maths in real life: Some Applications of Trigonometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Line of Sight and Angle of Elevation
• Angle of Depression
• Heights and Distances: One Right Triangle
• Problems with Two Right Triangles
• Drawing Figures and Solving Word Problems

Some Applications of Trigonometry formula sheet

Tangent ratio

$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{\text{height}}{\text{horizontal distance}}$

The main tool. Use whenever the question links a vertical height with a ground distance.

Sine ratio

$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}$

Use when the height and a slanted length (string, rope, ladder, line of sight) are involved.

Cosine ratio

$\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$

Use when the ground distance and a slanted length are involved.

Values at 30°

$\sin 30^{\circ }=\frac{1}{2},\cos 30^{\circ }=\frac{\sqrt{3}}{2},\tan 30^{\circ }=\frac{1}{\sqrt{3}}$

Small angle, small tan: the height is less than the distance.

Values at 45°

$\sin 45^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}},\tan 45^{\circ }=1$

At 45° the height equals the horizontal distance.

Values at 60°

$\sin 60^{\circ }=\frac{\sqrt{3}}{2},\cos 60^{\circ }=\frac{1}{2},\tan 60^{\circ }=\sqrt{3}$

Large angle, large tan: the height is more than the distance.

Angle of depression equals angle of elevation

$\text{depression of }B\text{ seen from }A=\text{elevation of }A\text{ seen from }B$

Because the horizontal at the eye is parallel to the ground (alternate angles). Write this reason in the answer.

Altitude of the sun

$\tan (\text{altitude})=\frac{\text{height of object}}{\text{length of shadow}}$

Higher sun means larger altitude and a shorter shadow.

Observer with eye height

$H=p+d\tan \theta$

H is the object's height, p the eye height, d the horizontal distance. Solve the triangle above eye level, then add p.

Same side, two angles (check formula)

$h=\frac{d}{\cot \alpha -\cot \beta },\alpha <\beta$

d is the distance between the two observation points. Use only to check; in the exam, write both tan equations and solve.

Same side, 30° and 60°

$h=\frac{\sqrt{3}}{2}d$

Quick check: points d apart on the same side, angles 30° and 60°. The nearer point is d/2 from the foot.

Opposite sides

$\text{total distance}=h(\cot \alpha +\cot \beta )$

Objects on opposite sides of a tower: add the two horizontal distances.

Side ratios of special triangles

$30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }:1:\sqrt{3}:2,45^{\circ }\text{-}45^{\circ }\text{-}90^{\circ }:1:1:\sqrt{2}$

Lets you check answers fast. The shortest side is opposite the 30° angle.

Rationalising surd denominators

$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3},\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{2}$

Needed in 30°/45° two-triangle problems. Multiply by the conjugate.

Approximate values

$\sqrt{3}\approx 1.732,\sqrt{2}\approx 1.414$

Use only when the question gives these values or asks for a decimal answer. Otherwise leave the surd form.

Pythagoras theorem

$(\text{hypotenuse}{)}^2=(\text{height}{)}^2+(\text{base}{)}^2$

Use to check a slanted length after finding the other two sides.

How the board asks Some Applications of Trigonometry

The question patterns to expect, and the method that works for each.

Tricks to remember Some Applications of Trigonometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

• TOA first

  Tan = Opposite over Adjacent = height over ground distance. If no slanted length is mentioned, the answer is almost always a tan equation.

  Use it when: Choosing the ratio in a few seconds.

• Tan climbs with the angle

  $\tan 30^{\circ },\tan 45^{\circ },\tan 60^{\circ }=\frac{1}{\sqrt{3}},1,\sqrt{3}$: each value is $\sqrt{3}$ times the one before. So the order is small, equal, big: at $30^{\circ }$ the height is less than the distance, at $45^{\circ }$ they are equal, and at $60^{\circ }$ the height is more than the distance.

  Use it when: Spotting a swapped value: if a 60° answer gives a height smaller than the distance, you used the wrong value.

• Depression drops to the ground

  Depression is drawn at the top from the horizontal, then it 'drops' to the ground point by alternate angles. After the drop, treat it exactly like elevation.

  Use it when: Avoiding the wrong angle placement in depression questions.

• Bigger angle, nearer point

  In any same-side figure, the steeper angle sits at the point closer to the tower.

  Use it when: Checking your figure before writing equations.

• The 30-60 half rule

  If points d apart on the same side see the top at 30° and 60°, the nearer point is $\frac{d}{2}$ from the foot and $h=\frac{\sqrt{3}}{2}d$.

  Use it when: Checking a two-triangle answer in seconds.

• 1 : √3 : 2 check

  In a 30-60-90 triangle the sides opposite 30°, 60°, 90° are in the ratio $1:\sqrt{3}:2$. The hypotenuse is double the shortest side.

  Use it when: Verifying a slanted length or height without redoing the working.

Some Applications of Trigonometry: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Some Applications of Trigonometry

The slips examiners see most often, and what to write instead.

• ✗Mistake: Placing the angle of depression inside the triangle at the top

  Students mark the given depression angle between the vertical tower and the line of sight, instead of between the horizontal and the line of sight.

  Wrong working

  Tower $AB=50$ m, depression of car C is $30^{\circ }$. $\tan 30^{\circ }=\frac{BC}{AB}$, so $BC=\frac{50}{\sqrt{3}}$ m

  Right working

  $\angle ACB=30^{\circ }$ (alternate angles). $\tan 30^{\circ }=\frac{AB}{BC}$, so $BC=50\sqrt{3}$ m

  ✓Do this: Always draw the horizontal through the observer's eye, mark the depression there, then transfer it to the ground point and write 'alternate angles'.

  Usually costs: Usually most of the question, since every later step uses the wrong distance.

• ✗Mistake: Using sin when the given side is the horizontal distance

  Students grab sin because it is the first ratio they learnt, without checking which sides are involved.

  Wrong working

  Distance from foot $=20$ m, elevation $60^{\circ }$. $\sin 60^{\circ }=\frac{h}{20}$, so $h=10\sqrt{3}$ m

  Right working

  $\tan 60^{\circ }=\frac{h}{20}$, so $h=20\sqrt{3}$ m

  ✓Do this: Name the sides first: opposite, adjacent, hypotenuse. Pick the ratio that uses exactly the known side and the wanted side.

  Usually costs: 1 to 2 marks in a 2- or 3-mark question.

• ✗Mistake: Ignoring the observer's height

  Students treat the line of sight as starting from the ground even when the person's eye height is given.

  Wrong working

  Eye height 1.6 m, building 25.6 m, elevation $30^{\circ }$. $\tan 30^{\circ }=\frac{25.6}{x}$

  Right working

  Height above eye level $=25.6-1.6=24$ m. $\tan 30^{\circ }=\frac{24}{x}$, so $x=24\sqrt{3}$ m

  ✓Do this: Draw the horizontal at eye level and subtract the eye height from the object's height before using any ratio.

  Usually costs: 1 to 2 marks, as the method mark may survive but the answer is wrong.

• ✗Mistake: Swapping the values of tan 30° and tan 60°

  Rushed recall under exam pressure; both involve $\sqrt{3}$.

  Wrong working

  Distance 30 m, elevation $30^{\circ }$. $h=30\tan 30^{\circ }=30\sqrt{3}$ m

  Right working

  $h=30\tan 30^{\circ }=\frac{30}{\sqrt{3}}=10\sqrt{3}$ m

  ✓Do this: Remember that at 30° the height must be less than the distance. Check your answer against that before moving on.

  Usually costs: ½ to 1 mark in short answers; can wreck a two-triangle question.

• ✗Mistake: Using the distance walked as the distance from the foot

  In moving-observer questions students attach the given distance to the wrong triangle.

  Wrong working

  Walks 40 m towards a tower; angle changes from $30^{\circ }$ to $60^{\circ }$. $\tan 60^{\circ }=\frac{h}{40}$, so $h=40\sqrt{3}$ m

  Right working

  Let nearer distance $=x$. $h=\sqrt{3}x$ and $\sqrt{3}h=x+40$, so $3x=x+40$, $x=20$, $h=20\sqrt{3}$ m

  ✓Do this: Introduce an unknown $x$ for the distance from the foot. The given 40 m is the gap between the two points, not a side of either triangle alone.

  Usually costs: Usually 3 or more marks of a 5-mark question.

• ✗Mistake: Leaving a surd in the denominator or skipping units and the decimal value

  Students stop as soon as they have an expression, forgetting the question asked for a simplified or decimal answer.

  Wrong working

  $h=\frac{60}{\sqrt{3}-1}$

  Right working

  $h=\frac{60(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=30(\sqrt{3}+1)$ m $=30\times 2.732=81.96$ m

  ✓Do this: Rationalise, write the unit, and if $\sqrt{3}=1.732$ is given, give the decimal value too.

  Usually costs: ½ to 1 mark for the final answer.

• ✗Mistake: No figure, or an unlabelled figure

  Students think the figure is optional and go straight to equations.

  Wrong working

  $\tan 45^{\circ }=\frac{h}{x}$ written with no figure showing what $h$ and $x$ are.

  Right working

  A labelled figure: tower AB with $AB=h$, point C on the ground with $BC=x$, right angle at B, $\angle ACB=45^{\circ }$; then $\tan 45^{\circ }=\frac{AB}{BC}$.

  ✓Do this: Spend 30 seconds on a clean figure with every point, length and angle labelled. Many marking schemes give a mark for it.

  Usually costs: ½ to 1 mark in 3- and 5-mark questions.

NCERT exercise guide: Some Applications of Trigonometry

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Some Applications of Trigonometry: previous year questions

• Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.7 marks per paper; 43 questions in all, counting OR alternatives. In 4 of 4 sample papers.
• Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.5 marks per paper; 40 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Some Applications of Trigonometry

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.