NCERT Solutions Class 10 Math Chapter 9: Some Applications of Trigonometry

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H1 Title: NCERT Solutions for Class 10 Math Chapter 9: Some Applications of Trigonometry (Exhaustive Board Guide)

Navigating through the CBSE Class 10 Mathematics curriculum requires an in-depth understanding of right-triangle geometry, horizontal baselines, line of sight, and the contextual relationship between angles of elevation and depression. Chapter 9 of Class 10 Mathematics, “Some Applications of Trigonometry”, forms the foundation of modern surveying, navigation, astronomical distance calculation, and architectural engineering. It investigates the measurement of heights and distances without physical measurement; explores the translation of real-world descriptive physical situations into precise right-angled geometric diagrams; and details the methodical application of the primary trigonometric ratios—sine, cosine, and tangent—to determine inaccessible dimensions. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.

Every question presented in the official NCERT textbook—ranging from fully detailed walkthroughs of all 7 foundational textbook examples to the complete 16-question set of Exercise 9.1 and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic four-step mathematical workflows (Given Data $\rightarrow$ Formula Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Answer with Units), and practical drawing protocols have been highlighted to ensure students secure maximum marks in their CBSE Board Examinations.

Chapter 9: Some Applications of Trigonometry

Master Chapter Summary & Strategic Examination Blueprint

The study of heights and distances using trigonometry provides an analytical method to determine physical lengths and vertical elevations using angular measurements obtained from instruments such as the theodolite or clinometer. In the rationalised NCERT Class 10 syllabus, this entire practical domain is consolidated within Chapter 9 under a single, high-yield exercise.

Concept / ElementCore Mathematical DefinitionGoverning Formula / Value SetCommon Board PitfallCBSE Weightage Tag
Line of SightStraight visual path from observer eye to objectVector along hypotenuse of reference triangleConfusing line of sight with horizontal distance1 Mark (Objective/MCQ)
Angle of ElevationAngle formed above horizontal level to an elevated object$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$Forgetting to add observer’s height to vertical result2 to 3 Marks
Angle of DepressionAngle formed below horizontal level to an object down below$\theta_{\text{depression}} = \theta_{\text{elevation}}$ (Alternate Interior Angles)Drawing angle against vertical wall instead of horizontal3 to 5 Marks
Standard RatiosKey trigonometric values for $30^\circ, 45^\circ, 60^\circ$$\tan 30^\circ = \frac{1}{\sqrt{3}}$, $\tan 45^\circ = 1$, $\tan 60^\circ = \sqrt{3}$Inverting $\tan 30^\circ$ and $\tan 60^\circ$ during substitutionPrerequisite Skill
Dual-Triangle GeometrySystems involving two right triangles with a shared side$h = x \tan \alpha = (x+d) \tan \beta$Algebraic substitution errors in simultaneous equations5 Marks (Long Answer / CBQ)

🧠 Examiner’s Secret: In the CBSE Class 10 Board Examinations, examiners award up to $1$ full mark solely for an accurately labelled geometric diagram. Even if your ultimate numerical computation suffers an arithmetic slip, a correct diagram featuring right-angle notation, clearly designated angles of elevation/depression, and labeled vertices ($A, B, C$) safeguards partial marks under the official CBSE step-marking guidelines.

Foundational Concepts and Theoretical Architecture

Line of Sight

The line of sight is the straight geometric ray drawn directly from the eye of an observer to the target point on the object being viewed.

[🖼️ Insert Image Here: Geometric representation of Line of Sight, Angle of Elevation, and Horizontal Level]
Image Alt-Text: CBSE Class 10 Maths – Diagram of Line of Sight, Angle of Elevation, and Horizontal Level for Chapter 9

When an observer looks at an object, their eye forms the vertex of an angular reference frame. The straight line extending from the observer’s eye to the exact feature being inspected represents the hypotenuse of an imaginary right-angled triangle formed with the ground plane.

Angle of Elevation

The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal level when the object is positioned above the horizontal plane.

To observe an object placed higher than the observer’s eye level, the observer must raise their head. The angular displacement measured upwards from the horizontal reference line to the line of sight constitutes the angle of elevation.

If point $O$ is the eye of the observer, $P$ is the elevated object, and $X$ is a point on the horizontal line passing through $O$ directly underneath or parallel to the ground, then:

$$\text{Angle of Elevation} = \angle XOP$$

💡 Did You Know?: The primary instrument historically utilized by civil surveyors and geodetic engineers to measure angles of elevation and depression is the theodolite, an instrument dating back to the sixteenth century that relies directly on right-triangle trigonometry.

Angle of Depression

The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal level when the object is located below the horizontal plane.

When an object sits lower than the observer’s eye level, the observer must lower their head. The angular measurement taken downward from the horizontal line to the line of sight is the angle of depression.

Crucially, because the horizontal line drawn at the observer’s eye level is parallel to the ground or water surface, the angle of depression from the observer to the object is strictly equal to the angle of elevation of the observer as measured from the object. This equivalence arises because they form a pair of alternate interior angles between two parallel horizontal lines intersected by the transversal line of sight.

[🖼️ Insert Image Here: Geometric representation of Angle of Depression showing Alternate Interior Angles equivalence]
Image Alt-Text: CBSE Class 10 Maths – Diagram of Angle of Depression and Alternate Interior Angles for Chapter 9

Standard Trigonometric Values Reference Bank

In height and distance problems, angles of $30^\circ$, $45^\circ$, and $60^\circ$ appear across all textbook scenarios. Below is the reference table of values:

Trigonometric Ratio$\theta = 30^\circ$$\theta = 45^\circ$$\theta = 60^\circ$
$\sin \theta$$\frac{1}{2}$$\frac{1}{\sqrt{2}}$$\frac{\sqrt{3}}{2}$
$\cos \theta$$\frac{\sqrt{3}}{2}$$\frac{1}{\sqrt{2}}$$\frac{1}{2}$
$\tan \theta$$\frac{1}{\sqrt{3}}$$1$$\sqrt{3}$

🧠 Examiner’s Secret: Unless the board question paper explicitly specifies $\sqrt{3} = 1.732$ or $\sqrt{2} = 1.414$, the CBSE marking scheme accepts answers left in rationalised radical form (e.g., $10\sqrt{3}\text{ m}$ or $\frac{20\sqrt{3}}{3}\text{ m}$). However, never leave an irrational radical in the denominator; always rationalise denominators fully.

[👉 Also Read: Class 10 Math Chapter 8 Introduction to Trigonometry NCERT Solutions]

Step-by-Step Solutions: NCERT Class 10 Mathematics Chapter 9 Solved Examples

Example 1 (Page 134) [CBSE 2018, 2020 Typology]

A tower stands vertically on the ground. From a point on the ground, which is $15\text{ m}$ away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^\circ$. Find the height of the tower.

Answer:

Let $AB$ represent the vertical tower of height $h\text{ metres}$, standing perpendicular to the ground at point $B$.

Let $C$ represent the observation point on the horizontal ground.

[🖼️ Insert Image Here: Right angled triangle ABC with vertical tower AB and observation point C 15m away]
Image Alt-Text: CBSE Class 10 Maths – Diagram for Example 1 Tower and Observation Point

Step 1: Identify Given Data

  • Distance from the foot of the tower to point $C$, $BC = 15\text{ m}$.
  • Angle of elevation of the top of the tower from point $C$, $\angle ACB = 60^\circ$.
  • Angle of the vertical tower with ground, $\angle ABC = 90^\circ$.

Step 2: Formula and Trigonometric Selection
In right-angled triangle $ABC$, we know the base adjacent to the angle ($BC$) and need to determine the perpendicular opposite to the angle ($AB$).

The relevant trigonometric ratio connecting opposite side and adjacent side is the tangent function:
$$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$$

Step 3: Step-by-Step Substitution
$$\tan 60^\circ = \frac{AB}{BC}$$

Substitute the standard geometric values:
$$\sqrt{3} = \frac{AB}{15}$$

Multiply both sides by $15$:
$$AB = 15\sqrt{3}\text{ m}$$

Final Answer:
The height of the vertical tower is $15\sqrt{3}\text{ m}$ (or approximately $25.98\text{ m}$ if taking $\sqrt{3} \approx 1.732$).

Example 2 (Page 135) [CBSE 2016, 2019, 2023 Set-2]

An electrician has to repair an electric fault on a pole of height $5\text{ m}$. She needs to reach a point $1.3\text{ m}$ below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of $60^\circ$ to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take $\sqrt{3} = 1.73$)

Answer:

Let $AD$ be the vertical electric pole of total height $5\text{ m}$.

The electrician must reach point $B$ on the pole such that $AB = 1.3\text{ m}$.

Let $BC$ represent the ladder inclined at an angle of $60^\circ$ to the horizontal ground $DC$.

[🖼️ Insert Image Here: Right triangle BDC showing electric pole AD, repair point B, and inclined ladder BC]
Image Alt-Text: CBSE Class 10 Maths – Diagram for Example 2 Electric Pole and Ladder Setup

Step 1: Identify Given Data and Effective Vertical Height

  • Total height of the pole, $AD = 5\text{ m}$.
  • Distance from top of the pole to repair point, $AB = 1.3\text{ m}$.
  • Therefore, effective working height $BD = AD – AB$:
    $$BD = 5\text{ m} – 1.3\text{ m} = 3.7\text{ m}$$
  • Angle of inclination of the ladder, $\angle BCD = 60^\circ$.
  • Angle at base of the pole, $\angle BDC = 90^\circ$.

Step 2: Calculate Length of the Ladder ($BC$)
In right-angled triangle $BDC$, the side opposite to $\angle BCD$ is $BD$ (perpendicular), and the ladder represents the hypotenuse ($BC$).
$$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BD}{BC}$$

$$\sin 60^\circ = \frac{3.7}{BC}$$

$$\frac{\sqrt{3}}{2} = \frac{3.7}{BC}$$

$$BC = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{\sqrt{3}}$$

Rationalising the denominator:
$$BC = \frac{7.4\sqrt{3}}{3}$$

Substitute $\sqrt{3} = 1.73$:
$$BC = \frac{7.4 \times 1.73}{3} = \frac{12.802}{3} \approx 4.28\text{ m}$$

Step 3: Calculate Distance of Foot of Ladder from Base of Pole ($DC$)
In right-angled triangle $BDC$, the side adjacent to $\angle BCD$ is $DC$ (base).
$$\cot \theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{DC}{BD}$$

$$\cot 60^\circ = \frac{DC}{3.7}$$

$$\frac{1}{\sqrt{3}} = \frac{DC}{3.7}$$

$$DC = \frac{3.7}{\sqrt{3}} = \frac{3.7 \times \sqrt{3}}{3}$$

Substitute $\sqrt{3} = 1.73$:
$$DC = \frac{3.7 \times 1.73}{3} = \frac{6.401}{3} \approx 2.14\text{ m}$$

Final Answer:
The length of the ladder required is $4.28\text{ m}$, and the foot of the ladder must be placed $2.14\text{ m}$ away from the foot of the pole.

Example 3 (Page 136) [CBSE 2017, 2021 Term-2]

An observer $1.5\text{ m}$ tall is $28.5\text{ m}$ away from a chimney. The angle of elevation of the top of the chimney from her eyes is $45^\circ$. What is the height of the chimney?

Answer:

Let $AB$ represent the chimney of height $h$, and $CD$ represent the observer of height $1.5\text{ m}$.

The distance between the foot of the chimney and the observer is $BD = 28.5\text{ m}$.

From the eye of the observer at point $C$, construct a horizontal line parallel to the ground intersecting the chimney $AB$ perpendicularly at point $E$.

[🖼️ Insert Image Here: Diagram of chimney AB, observer CD of 1.5m height, and horizontal eye level CE]
Image Alt-Text: CBSE Class 10 Maths – Diagram for Example 3 Chimney and Observer with Elevated Eye Level

Step 1: Establish Geometric Relations

  • $CD = 1.5\text{ m}$ (Height of observer).
  • $BD = 28.5\text{ m}$ (Distance between observer and chimney).
  • Quadrilateral $CDBE$ forms a rectangle because $\angle CDB = \angle DBE = \angle CEB = 90^\circ$.
  • Therefore:
    $$BE = CD = 1.5\text{ m}$$
    $$CE = BD = 28.5\text{ m}$$
  • The total height of the chimney is $AB = AE + BE = AE + 1.5\text{ m}$.
  • Angle of elevation from observer’s eye, $\angle ACE = 45^\circ$.

Step 2: Solve Triangle $ACE$
In right-angled triangle $ACE$:
$$\tan \angle ACE = \frac{AE}{CE}$$

$$\tan 45^\circ = \frac{AE}{28.5}$$

Since $\tan 45^\circ = 1$:
$$1 = \frac{AE}{28.5} \implies AE = 28.5\text{ m}$$

Step 3: Determine Total Height of Chimney
$$AB = AE + BE = 28.5\text{ m} + 1.5\text{ m} = 30\text{ m}$$

Final Answer:
The total height of the chimney is $30\text{ m}$.

Example 4 (Page 137) [CBSE 2015, 2019 Set-1]

From a point $P$ on the ground the angle of elevation of the top of a $10\text{ m}$ tall building is $30^\circ$. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from $P$ is $45^\circ$. Find the length of the flagstaff and the distance of the building from the point $P$. (You may take $\sqrt{3} = 1.732$)

Answer:

Let $AB$ represent the building of height $10\text{ m}$.

Let $BD$ represent the flagstaff mounted on top of the building of height $x\text{ metres}$.

Let $P$ represent the observation point on the horizontal ground. The total height of building plus flagstaff is $AD = AB + BD = (10 + x)\text{ m}$.

[🖼️ Insert Image Here: Two right triangles sharing base AP showing building AB and flagstaff BD on top]
Image Alt-Text: CBSE Class 10 Maths – Diagram for Example 4 Building and Flagstaff with Point P

Step 1: Identify Given Data

  • Height of the building, $AB = 10\text{ m}$.
  • Angle of elevation of top of building, $\angle APB = 30^\circ$.
  • Angle of elevation of top of flagstaff, $\angle APD = 45^\circ$.
  • $\angle PAB = 90^\circ$.

Step 2: Calculate Distance of Building from Point $P$ ($AP$)
In right-angled triangle $PAB$:
$$\tan 30^\circ = \frac{AB}{AP}$$

$$\frac{1}{\sqrt{3}} = \frac{10}{AP}$$

$$AP = 10\sqrt{3}\text{ m}$$

Substituting $\sqrt{3} = 1.732$:
$$AP = 10 \times 1.732 = 17.32\text{ m}$$

Step 3: Calculate Length of Flagstaff ($BD$)
In right-angled triangle $PAD$:
$$\tan 45^\circ = \frac{AD}{AP}$$

Since $\tan 45^\circ = 1$:
$$1 = \frac{AB + BD}{AP}$$

$$AP = AB + BD$$

Substitute the known values $AP = 10\sqrt{3}$ and $AB = 10$:
$$10\sqrt{3} = 10 + BD$$

$$BD = 10\sqrt{3} – 10 = 10(\sqrt{3} – 1)$$

Substitute $\sqrt{3} = 1.732$:
$$BD = 10(1.732 – 1) = 10(0.732) = 7.32\text{ m}$$

Final Answer:
The length of the flagstaff is $7.32\text{ m}$, and the distance of the building from the observation point $P$ is $17.32\text{ m}$.

Example 5 (Page 138) [CBSE 2014, 2019, 2024 Set-3]

The shadow of a tower standing on a level ground is found to be $40\text{ m}$ longer when the Sun’s altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower.

Answer:

Let $AB$ represent the vertical tower of height $h\text{ metres}$.

Let $BC$ be the length of the shadow when the Sun’s altitude (angle of elevation) is $60^\circ$.

Let $BD$ be the length of the shadow when the Sun’s altitude is $30^\circ$.

According to the question, the shadow lengthens by $40\text{ m}$ as the angle shifts:
$$CD = 40\text{ m} \quad \text{and} \quad BD = BC + CD = BC + 40$$

[🖼️ Insert Image Here: Triangles ABD and ABC sharing vertical tower AB with ground points D and C]
Image Alt-Text: CBSE Class 10 Maths – Diagram for Example 5 Shadow of Tower at Sun Altitudes 30 and 60 Degrees

Step 1: Set up Triangle $ABC$
In right-angled triangle $ABC$, with $\angle ACB = 60^\circ$:
$$\tan 60^\circ = \frac{AB}{BC}$$

$$\sqrt{3} = \frac{h}{BC} \implies BC = \frac{h}{\sqrt{3}} \quad \text{— (Equation 1)}$$

Step 2: Set up Triangle $ABD$
In right-angled triangle $ABD$, with $\angle ADB = 30^\circ$:
$$\tan 30^\circ = \frac{AB}{BD}$$

$$\frac{1}{\sqrt{3}} = \frac{h}{BC + 40}$$

$$BC + 40 = h\sqrt{3} \implies BC = h\sqrt{3} – 40 \quad \text{— (Equation 2)}$$

Step 3: Equate Equations (1) and (2) to solve for $h$
$$\frac{h}{\sqrt{3}} = h\sqrt{3} – 40$$

Multiply the entire equation by $\sqrt{3}$:
$$h = h(\sqrt{3} \times \sqrt{3}) – 40\sqrt{3}$$

$$h = 3h – 40\sqrt{3}$$

$$40\sqrt{3} = 3h – h$$

$$2h = 40\sqrt{3}$$

$$h = \frac{40\sqrt{3}}{2} = 20\sqrt{3}\text{ m}$$

Final Answer:
The height of the vertical tower is $20\sqrt{3}\text{ m}$ (or approximately $34.64\text{ m}$).

Example 6 (Page 139) [CBSE 2017, 2020, 2023 Set-1]

The angles of depression of the top and the bottom of an $8\text{ m}$ tall building from the top of a multi-storeyed building are $30^\circ$ and $45^\circ$, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Answer:

Let $PC$ represent the multi-storeyed building of height $H$.

Let $AB$ represent the building of height $8\text{ m}$.

Let the horizontal distance between the bases of the two buildings be $AC = d$.

From the top of the multi-storeyed building ($P$), horizontal line $PX$ is drawn parallel to the ground.

  • Angle of depression of the top of building $A$: $\angle XPA = 30^\circ$.
  • Angle of depression of the bottom of building $B$: $\angle XPB = 45^\circ$.

Draw $PD \perp AB$ extended or draw horizontal line $AD$ perpendicular to $PC$ from point $A$ meeting $PC$ at $D$.
Then quadrilateral $ACDE$ or $ACDA’$ forms a rectangle where $DC = AB = 8\text{ m}$, and $AD = AC = d$.

Therefore, $PD = PC – DC = H – 8$.

[🖼️ Insert Image Here: Multi-storeyed building PC and 8m building AB with angles of depression 30 and 45 degrees]
Image Alt-Text: CBSE Class 10 Maths – Diagram for Example 6 Multi-storeyed Building and Shorter Building

Step 1: Apply Alternate Interior Angles

  • Horizontal line $PX \parallel AD \implies \angle PAD = \angle XPA = 30^\circ$.
  • Horizontal line $PX \parallel BC \implies \angle PBC = \angle XPB = 45^\circ$.

Step 2: Solve Triangle $PBC$
In right-angled triangle $PBC$, $\angle PCB = 90^\circ$:
$$\tan 45^\circ = \frac{PC}{BC}$$

$$1 = \frac{H}{d} \implies H = d \quad \text{— (Equation 1)}$$
The height of the multi-storeyed building equals the distance between the two buildings.

Step 3: Solve Triangle $PAD$
In right-angled triangle $PAD$, $\angle PDA = 90^\circ$:
$$\tan 30^\circ = \frac{PD}{AD}$$

$$\frac{1}{\sqrt{3}} = \frac{H – 8}{d}$$

Since $d = H$ from Equation 1:
$$\frac{1}{\sqrt{3}} = \frac{H – 8}{H}$$

$$H = \sqrt{3}(H – 8)$$

$$H = H\sqrt{3} – 8\sqrt{3}$$

$$8\sqrt{3} = H\sqrt{3} – H$$

$$H(\sqrt{3} – 1) = 8\sqrt{3}$$

$$H = \frac{8\sqrt{3}}{\sqrt{3} – 1}$$

Step 4: Rationalise the Denominator
$$H = \frac{8\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} – 1)(\sqrt{3} + 1)}$$

$$H = \frac{8(3 + \sqrt{3})}{3 – 1} = \frac{8(3 + \sqrt{3})}{2} = 4(3 + \sqrt{3})\text{ m}$$

Expanding:
$$H = (12 + 4\sqrt{3})\text{ m} = 4(3 + \sqrt{3})\text{ m}$$

Since $d = H$:
$$d = 4(3 + \sqrt{3})\text{ m}$$

Final Answer:
The height of the multi-storeyed building is $4(3 + \sqrt{3})\text{ m}$, and the distance between the two buildings is also $4(3 + \sqrt{3})\text{ m}$.

Example 7 (Page 140) [CBSE 2015, 2018, 2022 Term-2]

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^\circ$ and $45^\circ$, respectively. If the bridge is at a height of $3\text{ m}$ from the banks, find the width of the river.

Answer:

Let $A$ and $B$ represent points on the two opposite banks of the river.

The width of the river is the horizontal line segment $AB$.

Let $P$ represent the observation point on the bridge across the river.

The perpendicular height of the bridge above the river line $AB$ is $PQ = 3\text{ m}$, where $Q$ lies on $AB$.

[🖼️ Insert Image Here: Point P on bridge at height 3m with lines of sight to banks A and B]
Image Alt-Text: CBSE Class 10 Maths – Diagram for Example 7 Bridge Over River and Opposite Banks

Step 1: Geometric Setup and Angular Identification
Horizontal line through $P$ is parallel to $AB$.

  • Angle of depression of bank $A = 30^\circ \implies \angle PAQ = 30^\circ$ (Alternate interior angles).
  • Angle of depression of bank $B = 45^\circ \implies \angle PBQ = 45^\circ$ (Alternate interior angles).
  • $PQ \perp AB \implies \angle PQA = \angle PQB = 90^\circ$.
  • Total river width $AB = AQ + QB$.

Step 2: Solve Triangle $PQA$ for $AQ$
In right-angled triangle $PQA$:
$$\tan 30^\circ = \frac{PQ}{AQ}$$

$$\frac{1}{\sqrt{3}} = \frac{3}{AQ} \implies AQ = 3\sqrt{3}\text{ m}$$

Step 3: Solve Triangle $PQB$ for $QB$
In right-angled triangle $PQB$:
$$\tan 45^\circ = \frac{PQ}{QB}$$

$$1 = \frac{3}{QB} \implies QB = 3\text{ m}$$

Step 4: Compute Total Width $AB$
$$AB = AQ + QB = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}$$

Substitute $\sqrt{3} \approx 1.732$ (optional):
$$AB = 3(1.732 + 1) = 3(2.732) = 8.196\text{ m}$$

Final Answer:
The total width of the river is $3(\sqrt{3} + 1)\text{ m}$ (or approximately $8.2\text{ m}$).

[👉 Also Read: Class 10 Math Chapter 10 Circles NCERT Solutions]

Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 9.1

Question 1 (Page 141) [CBSE 2012, 2019, 2023]

A circus artist is climbing a $20\text{ m}$ long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^\circ$ (see Fig. 9.11).

[🖼️ Insert Image Here: Right triangle ABC showing vertical pole AB, ground BC, and 20m rope hypotenuse AC]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 1 Circus Artist Climbing Rope

Answer:

Step 1: Identify Given Data

  • Let $AB$ be the vertical pole of height $h\text{ metres}$.
  • Let $AC$ represent the tightly stretched rope of length $20\text{ m}$.
  • Angle formed by the rope with horizontal ground: $\angle ACB = 30^\circ$.
  • Pole is perpendicular to horizontal ground: $\angle ABC = 90^\circ$.

Step 2: Choose Correct Trigonometric Ratio
We need to determine the opposite side ($AB$) given the hypotenuse ($AC$). The ratio connecting perpendicular and hypotenuse is the sine function:
$$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{AC}$$

Step 3: Execution and Substitution
$$\sin 30^\circ = \frac{AB}{AC}$$

$$\frac{1}{2} = \frac{AB}{20}$$

$$AB = \frac{20}{2} = 10\text{ m}$$

Final Answer:
The height of the vertical pole is $10\text{ m}$.

Question 2 (Page 141) [CBSE 2011, 2013, 2017, 2020 Standard]

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^\circ$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8\text{ m}$. Find the height of the tree.

Answer:

Let $AC’$ represent the original unbroken vertical tree standing at point $B$ on the ground.

Suppose the tree breaks at a height corresponding to point $A$. The upper section $AC’$ folds over along point $A$ such that its top tip $C’$ strikes the ground at point $C$.

Thus, length of the broken fallen segment is $AC$, meaning:
$$\text{Original Tree Height } H = AB + AC$$

[🖼️ Insert Image Here: Broken tree geometry showing standing vertical trunk AB, folded broken segment AC, and ground distance BC]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 2 Broken Tree Touching Ground

Step 1: Identify Given Data

  • Distance from foot of tree $B$ to ground contact point $C$: $BC = 8\text{ m}$.
  • Angle formed by the broken part with horizontal ground: $\angle ACB = 30^\circ$.
  • Angle at base: $\angle ABC = 90^\circ$.

Step 2: Find the Standing Trunk Height ($AB$)
In right-angled triangle $ABC$:
$$\tan 30^\circ = \frac{AB}{BC}$$

$$\frac{1}{\sqrt{3}} = \frac{AB}{8}$$

$$AB = \frac{8}{\sqrt{3}}\text{ m}$$

Step 3: Find the Broken Section Length ($AC$)
In right-angled triangle $ABC$:
$$\cos 30^\circ = \frac{BC}{AC}$$

$$\frac{\sqrt{3}}{2} = \frac{8}{AC}$$

$$AC\sqrt{3} = 16 \implies AC = \frac{16}{\sqrt{3}}\text{ m}$$

Step 4: Calculate Total Original Height ($H$)
$$H = AB + AC = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}}\text{ m}$$

Rationalise the denominator:
$$H = \frac{24\sqrt{3}}{3} = 8\sqrt{3}\text{ m}$$

Final Answer:
The total original height of the tree is $8\sqrt{3}\text{ m}$ (or approximately $13.86\text{ m}$).

Question 3 (Page 141) [CBSE 2014, 2018]

A contractor plans to install two slides for the children to play in a park. For the children below the age of $5\text{ years}$, she prefers to have a slide whose top is at a height of $1.5\text{ m}$, and is inclined at an angle of $30^\circ$ to the ground, whereas for elder children, she wants to have a steep slide at a height of $3\text{ m}$, and inclined at an angle of $60^\circ$ to the ground. What should be the length of the slide in each case?

Answer:

Case I: Slide for Children Below the Age of 5 Years

Let $AB$ be the vertical support ladder of the slide, and $AC$ be the slide itself.

[🖼️ Insert Image Here: Two separate right-angled triangles showing Case 1 slide and Case 2 slide]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 3 Two Playground Slides

  • Height of the slide: $AB = 1.5\text{ m}$.
  • Angle of inclination with the ground: $\angle ACB = 30^\circ$.
  • $\angle ABC = 90^\circ$.

In right-angled triangle $ABC$:
$$\sin 30^\circ = \frac{AB}{AC}$$

$$\frac{1}{2} = \frac{1.5}{AC}$$

$$AC = 1.5 \times 2 = 3\text{ m}$$

Case II: Slide for Elder Children

Let $PQ$ be the vertical support of the steeper slide, and $PR$ be the slide itself.

  • Height of the slide: $PQ = 3\text{ m}$.
  • Angle of inclination with the ground: $\angle PRQ = 60^\circ$.
  • $\angle PQR = 90^\circ$.

In right-angled triangle $PQR$:
$$\sin 60^\circ = \frac{PQ}{PR}$$

$$\frac{\sqrt{3}}{2} = \frac{3}{PR}$$

$$PR\sqrt{3} = 6 \implies PR = \frac{6}{\sqrt{3}}$$

Rationalise the denominator:
$$PR = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\text{ m}$$

Final Answer:
The length of the slide for children below $5\text{ years}$ is $3\text{ m}$, and the length of the slide for elder children is $2\sqrt{3}\text{ m}$ (or approximately $3.46\text{ m}$).

Question 4 (Page 141) [CBSE 2013, 2019 Set-1]

The angle of elevation of the top of a tower from a point on the ground, which is $30\text{ m}$ away from the foot of the tower, is $30^\circ$. Find the height of the tower.

Answer:

Let $AB$ represent the vertical tower of height $h\text{ metres}$.

Let $C$ represent the observation point situated on the horizontal ground.

[🖼️ Insert Image Here: Right triangle showing tower AB, ground line BC of 30m, and angle of elevation 30 degrees]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 4 Tower Height Calculation

Step 1: Identify Given Data

  • Distance from foot of tower to point $C$: $BC = 30\text{ m}$.
  • Angle of elevation: $\angle ACB = 30^\circ$.
  • $\angle ABC = 90^\circ$.

Step 2: Apply Tangent Ratio
In right-angled triangle $ABC$:
$$\tan 30^\circ = \frac{AB}{BC}$$

$$\frac{1}{\sqrt{3}} = \frac{h}{30}$$

$$h = \frac{30}{\sqrt{3}}$$

Step 3: Rationalise Denominator
$$h = \frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}\text{ m}$$

Final Answer:
The height of the vertical tower is $10\sqrt{3}\text{ m}$ (or approximately $17.32\text{ m}$).

Question 5 (Page 141) [CBSE 2012, 2016, 2023]

A kite is flying at a height of $60\text{ m}$ above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^\circ$. Find the length of the string, assuming that there is no slack in the string.

Answer:

Let $K$ represent the position of the kite in the sky.

Let $P$ represent the ground anchor point where the taut string is tied.

Let $KL$ be the vertical distance from the kite directly down to the horizontal ground plane.

[🖼️ Insert Image Here: Right triangle showing flying kite K at vertical height 60m and taut string KP inclined at 60 degrees]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 5 Kite Flying Height and String Length

Step 1: Identify Given Data

  • Vertical altitude of the kite: $KL = 60\text{ m}$.
  • Angle of inclination of the string with ground: $\angle KPL = 60^\circ$.
  • Hypotenuse $KP$ represents the length of the string ($L$).
  • $\angle KLP = 90^\circ$.

Step 2: Apply Sine Ratio
In right-angled triangle $KLP$:
$$\sin 60^\circ = \frac{KL}{KP}$$

$$\frac{\sqrt{3}}{2} = \frac{60}{KP}$$

$$KP \times \sqrt{3} = 120$$

$$KP = \frac{120}{\sqrt{3}}$$

Step 3: Rationalise Denominator
$$KP = \frac{120\sqrt{3}}{3} = 40\sqrt{3}\text{ m}$$

Final Answer:
The length of the kite string is $40\sqrt{3}\text{ m}$ (or approximately $69.28\text{ m}$).

Question 6 (Page 141) [CBSE 2015, 2017, 2020 Standard]

A $1.5\text{ m}$ tall boy is standing at some distance from a $30\text{ m}$ tall building. The angle of elevation from his eyes to the top of the building increases from $30^\circ$ to $60^\circ$ as he walks towards the building. Find the distance he walked towards the building.

Answer:

Let $PQ$ represent the vertical building of total height $30\text{ m}$.

Let the initial position of the boy be represented by line segment $AB$ of height $1.5\text{ m}$, and his second position closer to the building be represented by segment $CD$ of height $1.5\text{ m}$.

From the eye of the boy at $A$, draw a horizontal line parallel to ground $BQ$, intersecting the building $PQ$ perpendicularly at point $E$.

[🖼️ Insert Image Here: Geometry of 1.5m boy walking towards 30m building with eye level line and elevation angles 30 and 60]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 6 Observer Approaching Building

Step 1: Compute Effective Height ($PE$)

  • Total building height: $PQ = 30\text{ m}$.
  • Height of boy: $AB = CD = EQ = 1.5\text{ m}$.
  • Effective perpendicular height above horizontal eye line:
    $$PE = PQ – EQ = 30\text{ m} – 1.5\text{ m} = 28.5\text{ m} = \frac{57}{2}\text{ m}$$

Step 2: Angular Assignments

  • Initial eye position: point $A$, angle of elevation $\angle PAE = 30^\circ$.
  • Final eye position: point $C$, angle of elevation $\angle PCE = 60^\circ$.
  • Distance walked towards the building: $d = AC = BD$.

Step 3: Triangle $PCE$ (Final Position)
In right-angled triangle $PCE$:
$$\tan 60^\circ = \frac{PE}{CE}$$

$$\sqrt{3} = \frac{28.5}{CE} \implies CE = \frac{28.5}{\sqrt{3}}\text{ m}$$

Step 4: Triangle $PAE$ (Initial Position)
In right-angled triangle $PAE$:
$$\tan 30^\circ = \frac{PE}{AE}$$

$$\frac{1}{\sqrt{3}} = \frac{28.5}{AE} \implies AE = 28.5\sqrt{3}\text{ m}$$

Step 5: Compute Distance Walked ($AC$)
$$AC = AE – CE$$

$$AC = 28.5\sqrt{3} – \frac{28.5}{\sqrt{3}}$$

Factor out $28.5$:
$$AC = 28.5\left(\sqrt{3} – \frac{1}{\sqrt{3}}\right) = 28.5\left(\frac{3 – 1}{\sqrt{3}}\right) = 28.5 \times \frac{2}{\sqrt{3}} = \frac{57}{\sqrt{3}}$$

Rationalise the denominator:
$$AC = \frac{57\sqrt{3}}{3} = 19\sqrt{3}\text{ m}$$

Final Answer:
The distance the boy walked towards the building is $19\sqrt{3}\text{ m}$ (or approximately $32.91\text{ m}$).

Question 7 (Page 142) [CBSE 2016, 2019, 2024 Set-1]

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20\text{ m}$ high building are $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.

Answer:

Let $BC$ represent the vertical building of height $20\text{ m}$.

Let $AB$ represent the transmission tower of height $h\text{ metres}$ fixed vertically on top of building $BC$.

Let $D$ represent the observation point on the ground.

Total vertical height of point $A$ from ground is $AC = AB + BC = (h + 20)\text{ m}$.

[🖼️ Insert Image Here: Building BC with transmission tower AB on top and observation point D on ground]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 7 Transmission Tower on Building

Step 1: Identify Given Data

  • Building height: $BC = 20\text{ m}$.
  • Angle of elevation of the bottom of tower (point $B$): $\angle BDC = 45^\circ$.
  • Angle of elevation of the top of tower (point $A$): $\angle ADC = 60^\circ$.
  • $\angle BCD = 90^\circ$.

Step 2: Solve Triangle $BCD$
In right-angled triangle $BCD$:
$$\tan 45^\circ = \frac{BC}{CD}$$

$$1 = \frac{20}{CD} \implies CD = 20\text{ m}$$

Step 3: Solve Triangle $ACD$
In right-angled triangle $ACD$:
$$\tan 60^\circ = \frac{AC}{CD}$$

$$\sqrt{3} = \frac{h + 20}{CD}$$

Substitute $CD = 20\text{ m}$:
$$\sqrt{3} = \frac{h + 20}{20}$$

$$20\sqrt{3} = h + 20$$

$$h = 20\sqrt{3} – 20 = 20(\sqrt{3} – 1)\text{ m}$$

Substitute $\sqrt{3} \approx 1.732$:
$$h = 20(1.732 – 1) = 20(0.732) = 14.64\text{ m}$$

Final Answer:
The height of the transmission tower is $20(\sqrt{3} – 1)\text{ m}$ (or approximately $14.64\text{ m}$).

Question 8 (Page 142) [CBSE 2015, 2019, 2023 Set-3]

A statue, $1.6\text{ m}$ tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ$. Find the height of the pedestal.

Answer:

Let $BC$ represent the vertical pedestal of height $h\text{ metres}$.

Let $AB$ represent the statue of height $1.6\text{ m}$ situated on top of the pedestal.

Let $D$ represent the point of observation on the horizontal ground.

Total vertical height $AC = AB + BC = 1.6 + h$.

[🖼️ Insert Image Here: Pedestal BC with statue AB of height 1.6m and observation point D on horizontal ground]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 8 Statue on Pedestal

Step 1: Identify Given Data

  • Height of statue: $AB = 1.6\text{ m}$.
  • Angle of elevation of top of pedestal: $\angle BDC = 45^\circ$.
  • Angle of elevation of top of statue: $\angle ADC = 60^\circ$.
  • $\angle ACD = 90^\circ$.

Step 2: Solve Triangle $BCD$
In right-angled triangle $BCD$:
$$\tan 45^\circ = \frac{BC}{CD}$$

$$1 = \frac{h}{CD} \implies CD = h \quad \text{— (Equation 1)}$$

Step 3: Solve Triangle $ACD$
In right-angled triangle $ACD$:
$$\tan 60^\circ = \frac{AC}{CD}$$

$$\sqrt{3} = \frac{h + 1.6}{CD}$$

Substitute $CD = h$ from Equation 1:
$$\sqrt{3} = \frac{h + 1.6}{h}$$

$$h\sqrt{3} = h + 1.6$$

$$h\sqrt{3} – h = 1.6$$

$$h(\sqrt{3} – 1) = 1.6$$

$$h = \frac{1.6}{\sqrt{3} – 1}$$

Step 4: Rationalise the Denominator
$$h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} – 1)(\sqrt{3} + 1)}$$

$$h = \frac{1.6(\sqrt{3} + 1)}{3 – 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1)\text{ m}$$

Substitute $\sqrt{3} \approx 1.732$:
$$h = 0.8(1.732 + 1) = 0.8(2.732) = 2.1856\text{ m} \approx 2.19\text{ m}$$

Final Answer:
The height of the pedestal is $0.8(\sqrt{3} + 1)\text{ m}$ (or approximately $2.19\text{ m}$).

Question 9 (Page 142) [CBSE 2011, 2018, 2020 Set-2]

The angle of elevation of the top of a building from the foot of the tower is $30^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^\circ$. If the tower is $50\text{ m}$ high, find the height of the building.

Answer:

Let $AB$ represent the vertical tower of height $50\text{ m}$.

Let $CD$ represent the vertical building of height $h\text{ metres}$.

Let $BD$ represent the horizontal ground distance between the bases of the tower and the building.

[🖼️ Insert Image Here: Dual right triangles on common base BD between tower AB of 50m and building CD]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 9 Tower and Building Mutual Elevation Angles

Step 1: Identify Given Data

  • Height of tower: $AB = 50\text{ m}$.
  • Angle of elevation of top of tower from foot of building ($D$): $\angle ADB = 60^\circ$.
  • Angle of elevation of top of building from foot of tower ($B$): $\angle CBD = 30^\circ$.
  • $\angle ABD = \angle CDB = 90^\circ$.

Step 2: Solve Triangle $ABD$ to Find Ground Separation ($BD$)
In right-angled triangle $ABD$:
$$\tan 60^\circ = \frac{AB}{BD}$$

$$\sqrt{3} = \frac{50}{BD}$$

$$BD = \frac{50}{\sqrt{3}}\text{ m} \quad \text{— (Equation 1)}$$

Step 3: Solve Triangle $CDB$ to Find Building Height ($h$)
In right-angled triangle $CDB$:
$$\tan 30^\circ = \frac{CD}{BD}$$

$$\frac{1}{\sqrt{3}} = \frac{h}{BD} \implies h = \frac{BD}{\sqrt{3}}$$

Substitute Equation 1 into this expression:
$$h = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}} = \frac{50}{3} = 16\frac{2}{3}\text{ m} \approx 16.67\text{ m}$$

Final Answer:
The height of the building is $\frac{50}{3}\text{ m}$ (or $16\frac{2}{3}\text{ m}$ / $16.67\text{ m}$).

Question 10 (Page 142) [CBSE 2014, 2017, 2019, 2024 Standard]

Two poles of equal heights are standing opposite each other on either side of the road, which is $80\text{ m}$ wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^\circ$ and $30^\circ$, respectively. Find the height of the poles and the distances of the point from the poles.

Answer:

Let $AB$ and $CD$ represent two vertical poles of equal height $h$, such that $AB = CD = h$.

Let $BD$ represent the width of the road, where $BD = 80\text{ m}$.

Let $P$ represent the observation point on the road between the two poles.

  • Let the distance from pole $AB$ to point $P$ be $BP = x\text{ metres}$.
  • Then the distance from point $P$ to pole $CD$ is $PD = (80 – x)\text{ metres}$.

[🖼️ Insert Image Here: Two equal poles AB and CD on road BD of 80m with intermediate point P]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 10 Two Equal Poles on Opposite Sides of Road

Step 1: Assign Angles of Elevation

  • Angle of elevation of top of pole $AB$ from $P$: $\angle APB = 60^\circ$.
  • Angle of elevation of top of pole $CD$ from $P$: $\angle CPD = 30^\circ$.
  • $\angle ABP = \angle CDP = 90^\circ$.

Step 2: Solve Triangle $ABP$
In right-angled triangle $ABP$:
$$\tan 60^\circ = \frac{AB}{BP}$$

$$\sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \text{— (Equation 1)}$$

Step 3: Solve Triangle $CDP$
In right-angled triangle $CDP$:
$$\tan 30^\circ = \frac{CD}{PD}$$

$$\frac{1}{\sqrt{3}} = \frac{h}{80 – x} \implies h = \frac{80 – x}{\sqrt{3}} \quad \text{— (Equation 2)}$$

Step 4: Equate Expressions for $h$ to Solve for $x$
$$x\sqrt{3} = \frac{80 – x}{\sqrt{3}}$$

Multiply both sides by $\sqrt{3}$:
$$3x = 80 – x$$

$$4x = 80 \implies x = 20\text{ m}$$

Thus:

  • $BP = 20\text{ m}$
  • $PD = 80 – x = 80 – 20 = 60\text{ m}$

Step 5: Compute Pole Height ($h$)
Substitute $x = 20$ into Equation 1:
$$h = 20\sqrt{3}\text{ m}$$

Substitute $\sqrt{3} \approx 1.732$:
$$h = 20 \times 1.732 = 34.64\text{ m}$$

Final Answer:
The height of each pole is $20\sqrt{3}\text{ m}$ (or $34.64\text{ m}$), and the distances of the observation point from the poles are $20\text{ m}$ and $60\text{ m}$.

Question 11 (Page 142) [CBSE 2012, 2017, 2020 Standard]

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^\circ$. From another point $20\text{ m}$ away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $30^\circ$ (see Fig. 9.12). Find the height of the tower and the width of the canal.

[🖼️ Insert Image Here: TV tower AB on canal bank BC with collinear points C and D at 20m separation]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 11 TV Tower and Canal Bank

Answer:

Let $AB$ represent the vertical TV tower of height $h\text{ metres}$.

Let $BC$ represent the width of the canal, denoted by $x\text{ metres}$.

Let $D$ be the point $20\text{ m}$ further away from point $C$ collinear with $B$, such that $CD = 20\text{ m}$ and $BD = x + 20$.

Step 1: Identify Given Data

  • $\angle ACB = 60^\circ$ (Angle of elevation from opposite bank).
  • $\angle ADB = 30^\circ$ (Angle of elevation from point $20\text{ m}$ further away).
  • $\angle ABC = 90^\circ$.

Step 2: Solve Triangle $ABC$
In right-angled triangle $ABC$:
$$\tan 60^\circ = \frac{AB}{BC}$$

$$\sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \text{— (Equation 1)}$$

Step 3: Solve Triangle $ABD$
In right-angled triangle $ABD$:
$$\tan 30^\circ = \frac{AB}{BD}$$

$$\frac{1}{\sqrt{3}} = \frac{h}{x + 20}$$

$$x + 20 = h\sqrt{3} \quad \text{— (Equation 2)}$$

Step 4: Substitute Equation 1 into Equation 2
$$x + 20 = (x\sqrt{3})\sqrt{3}$$

$$x + 20 = 3x$$

$$20 = 3x – x$$

$$2x = 20 \implies x = 10\text{ m}$$

Step 5: Compute Tower Height ($h$)
Substitute $x = 10$ into Equation 1:
$$h = 10\sqrt{3}\text{ m}$$

Final Answer:
The height of the TV tower is $10\sqrt{3}\text{ m}$ (or approximately $17.32\text{ m}$), and the width of the canal is $10\text{ m}$.

Question 12 (Page 143) [CBSE 2013, 2018, 2023 Set-2]

From the top of a $7\text{ m}$ high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Determine the height of the tower.

Answer:

Let $AB$ represent the vertical building of height $7\text{ m}$.

Let $CD$ represent the vertical cable tower of height $H$.

Let $BD$ represent the horizontal ground distance separating the building and the tower.

From the observation point at the top of the building ($A$), construct horizontal line $AE$ parallel to $BD$, meeting the tower $CD$ at point $E$.

[🖼️ Insert Image Here: Building AB of 7m and cable tower CD with eye level line AE, elevation 60 and depression 45]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 12 Building and Cable Tower Heights

Step 1: Establish Geometric Relationships

  • $ABED$ forms a rectangle because $AB \perp BD$, $AE \parallel BD$, and $CD \perp BD$.
  • Therefore:
    $$ED = AB = 7\text{ m}$$
    $$AE = BD$$
  • Total height of the tower: $CD = CE + ED = CE + 7$.
  • Angle of elevation of top of tower $C$: $\angle CAE = 60^\circ$.
  • Angle of depression of foot of tower $D$: $\angle EAD = 45^\circ$.
  • Consequently, alternate interior angle $\angle ADB = \angle EAD = 45^\circ$.

Step 2: Solve Triangle $ABD$ to Find Separation ($BD$)
In right-angled triangle $ABD$:
$$\tan 45^\circ = \frac{AB}{BD}$$

$$1 = \frac{7}{BD} \implies BD = 7\text{ m}$$

Since $AE = BD$:
$$AE = 7\text{ m}$$

Step 3: Solve Triangle $CEA$ to Find Upper Tower Segment ($CE$)
In right-angled triangle $CEA$:
$$\tan 60^\circ = \frac{CE}{AE}$$

$$\sqrt{3} = \frac{CE}{7} \implies CE = 7\sqrt{3}\text{ m}$$

Step 4: Compute Total Tower Height ($CD$)
$$CD = CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1)\text{ m}$$

Substitute $\sqrt{3} \approx 1.732$:
$$CD = 7(1.732 + 1) = 7(2.732) = 19.124\text{ m}$$

Final Answer:
The height of the cable tower is $7(\sqrt{3} + 1)\text{ m}$ (or approximately $19.12\text{ m}$).

Question 13 (Page 143) [CBSE 2014, 2019 Set-3, 2024 Set-2]

As observed from the top of a $75\text{ m}$ high lighthouse from the sea-level, the angles of depression of two ships are $30^\circ$ and $45^\circ$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer:

Let $AB$ represent the vertical lighthouse of height $75\text{ m}$ standing on sea level at $B$.

Let $C$ and $D$ represent the positions of the two ships, with ship $D$ situated behind ship $C$ collinear with base $B$.

Draw horizontal reference line $AX$ through the summit of the lighthouse parallel to the sea level line $BD$.

[🖼️ Insert Image Here: Lighthouse AB of 75m observing two ships C and D at angles of depression 45 and 30]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 13 Lighthouse and Two Ships at Sea

Step 1: Identify Given Data and Angular Equivalence

  • Height of lighthouse: $AB = 75\text{ m}$.
  • Angle of depression of closer ship $C$: $\angle XAC = 45^\circ \implies \angle ACB = 45^\circ$ (Alternate interior angles).
  • Angle of depression of farther ship $D$: $\angle XAD = 30^\circ \implies \angle ADB = 30^\circ$ (Alternate interior angles).
  • $\angle ABD = 90^\circ$.
  • Distance between the two ships: $CD = BD – BC$.

Step 2: Solve Triangle $ABC$ to Find $BC$
In right-angled triangle $ABC$:
$$\tan 45^\circ = \frac{AB}{BC}$$

$$1 = \frac{75}{BC} \implies BC = 75\text{ m}$$

Step 3: Solve Triangle $ABD$ to Find $BD$
In right-angled triangle $ABD$:
$$\tan 30^\circ = \frac{AB}{BD}$$

$$\frac{1}{\sqrt{3}} = \frac{75}{BD} \implies BD = 75\sqrt{3}\text{ m}$$

Step 4: Compute Distance Between the Ships ($CD$)
$$CD = BD – BC = 75\sqrt{3} – 75 = 75(\sqrt{3} – 1)\text{ m}$$

Substitute $\sqrt{3} \approx 1.732$:
$$CD = 75(1.732 – 1) = 75(0.732) = 54.9\text{ m}$$

Final Answer:
The distance between the two ships is $75(\sqrt{3} – 1)\text{ m}$ (or approximately $54.9\text{ m}$).

Question 14 (Page 143) [CBSE 2016, 2020 Standard]

A $1.2\text{ m}$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2\text{ m}$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60^\circ$. After some time, the angle of elevation reduces to $30^\circ$ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

[🖼️ Insert Image Here: Girl of height 1.2m observing balloon moving horizontally at 88.2m total height]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 14 Girl Observing Horizontally Drifting Balloon

Answer:

Let $A$ represent the eye of the girl at height $1.2\text{ m}$ above the ground plane.

Let $P$ and $Q$ be the initial and final positions of the balloon floating along a horizontal ceiling line.

Let $PE$ and $QF$ be perpendicular lines dropped from the balloon positions to the horizontal eye-level line drawn from point $A$.

Step 1: Compute Effective Height ($h$)

  • Total height of the balloon above the ground: $88.2\text{ m}$.
  • Height of the girl: $1.2\text{ m}$.
  • Effective perpendicular height above eye level:
    $$h = PE = QF = 88.2\text{ m} – 1.2\text{ m} = 87\text{ m}$$

Step 2: Angular Assignments

  • Initial angle of elevation: $\angle PAE = 60^\circ$.
  • Final angle of elevation: $\angle QAF = 30^\circ$.
  • $\angle PEA = \angle QFA = 90^\circ$.
  • The horizontal distance travelled by the balloon is $PQ = EF = AF – AE$.

Step 3: Solve Triangle $PAE$ to Find $AE$
In right-angled triangle $PAE$:
$$\tan 60^\circ = \frac{PE}{AE}$$

$$\sqrt{3} = \frac{87}{AE} \implies AE = \frac{87}{\sqrt{3}}\text{ m}$$

Step 4: Solve Triangle $QAF$ to Find $AF$
In right-angled triangle $QAF$:
$$\tan 30^\circ = \frac{QF}{AF}$$

$$\frac{1}{\sqrt{3}} = \frac{87}{AF} \implies AF = 87\sqrt{3}\text{ m}$$

Step 5: Compute Horizontal Distance Travelled ($EF$)
$$EF = AF – AE = 87\sqrt{3} – \frac{87}{\sqrt{3}}$$

Factor out $87$:
$$EF = 87\left(\sqrt{3} – \frac{1}{\sqrt{3}}\right) = 87\left(\frac{3 – 1}{\sqrt{3}}\right) = 87 \times \frac{2}{\sqrt{3}} = \frac{174}{\sqrt{3}}$$

Rationalise the denominator:
$$EF = \frac{174\sqrt{3}}{3} = 58\sqrt{3}\text{ m}$$

Substitute $\sqrt{3} \approx 1.732$:
$$EF = 58 \times 1.732 = 100.456\text{ m}$$

Final Answer:
The distance travelled by the balloon during the interval is $58\sqrt{3}\text{ m}$ (or approximately $100.46\text{ m}$).

Question 15 (Page 143) [CBSE 2015, 2017, 2019, 2023 Standard]

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^\circ$, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be $60^\circ$. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

Let $AB$ represent the vertical tower of height $h$ standing at the highway base $B$.

Let the initial position of the car on the highway be point $C$, where the angle of depression from top $A$ is $30^\circ$.

Let the second position of the car, $6\text{ seconds}$ later, be point $D$, where the angle of depression is $60^\circ$.

Let the car travel at a uniform speed of $v\text{ metres per second}$.

[🖼️ Insert Image Here: Tower AB overlooking straight highway with car moving from point C to D towards B]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 15 Approaching Car on Highway

Step 1: Translate Kinematics to Geometry

  • $\text{Distance} = \text{Speed} \times \text{Time}$.
  • Distance travelled from $C$ to $D$ in $6\text{ seconds}$:
    $$CD = 6v\text{ metres}$$
  • Let the time taken to travel from point $D$ to the foot of the tower $B$ be $t\text{ seconds}$.
  • Distance from $D$ to $B$:
    $$DB = vt\text{ metres}$$
  • Total distance $CB = CD + DB = 6v + vt = v(6 + t)\text{ metres}$.

Step 2: Assign Angles Using Alternate Interior Angles

  • Angle of depression at $C = 30^\circ \implies \angle ACB = 30^\circ$.
  • Angle of depression at $D = 60^\circ \implies \angle ADB = 60^\circ$.
  • $\angle ABD = 90^\circ$.

Step 3: Solve Triangle $ABD$
In right-angled triangle $ABD$:
$$\tan 60^\circ = \frac{AB}{DB}$$

$$\sqrt{3} = \frac{h}{vt} \implies h = vt\sqrt{3} \quad \text{— (Equation 1)}$$

Step 4: Solve Triangle $ABC$
In right-angled triangle $ABC$:
$$\tan 30^\circ = \frac{AB}{CB}$$

$$\frac{1}{\sqrt{3}} = \frac{h}{v(6 + t)}$$

$$h\sqrt{3} = v(6 + t) \quad \text{— (Equation 2)}$$

Step 5: Substitute Equation 1 into Equation 2
$$(vt\sqrt{3})\sqrt{3} = v(6 + t)$$

$$3vt = 6v + vt$$

Since uniform speed $v > 0$, divide both sides by $v$:
$$3t = 6 + t$$

$$3t – t = 6$$

$$2t = 6 \implies t = 3\text{ seconds}$$

Final Answer:
The time taken by the car to reach the foot of the tower from point $D$ is $3\text{ seconds}$.

Question 16 (Page 144) [CBSE 2011, 2014, 2017, 2020 Standard]

The angles of elevation of the top of a tower from two points at a distance of $4\text{ m}$ and $9\text{ m}$ from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $6\text{ m}$.

Answer:

Let $AB$ represent the vertical tower of height $h\text{ metres}$.

Let $C$ and $D$ be two points situated on the same straight line passing through base $B$.

  • Distance of first point: $BC = 4\text{ m}$.
  • Distance of second point: $BD = 9\text{ m}$.

[🖼️ Insert Image Here: Tower AB with collinear base points C and D at 4m and 9m having complementary elevation angles]
Image Alt-Text: CBSE Class 10 Maths – Exercise 9.1 Question 16 Complementary Angles of Elevation Proof

Step 1: Define Complementary Angles
Two angles are complementary if their sum equals $90^\circ$.

  • Let the angle of elevation from point $C$ be $\angle ACB = \theta$.
  • Then the angle of elevation from point $D$ is $\angle ADB = 90^\circ – \theta$.
  • Tower is vertical: $\angle ABC = 90^\circ$.

Step 2: Solve Triangle $ABC$
In right-angled triangle $ABC$:
$$\tan \theta = \frac{AB}{BC}$$

$$\tan \theta = \frac{h}{4} \quad \text{— (Equation 1)}$$

Step 3: Solve Triangle $ABD$
In right-angled triangle $ABD$:
$$\tan(90^\circ – \theta) = \frac{AB}{BD}$$

Using the complementary angle identity $\tan(90^\circ – \theta) = \cot \theta$:
$$\cot \theta = \frac{h}{9} \quad \text{— (Equation 2)}$$

Step 4: Multiply Equations (1) and (2)
$$\tan \theta \times \cot \theta = \left(\frac{h}{4}\right) \times \left(\frac{h}{9}\right)$$

Since $\cot \theta = \frac{1}{\tan \theta}$, their product is identity $1$:
$$1 = \frac{h^2}{36}$$

$$h^2 = 36$$

Taking the positive square root (since physical height $h > 0$):
$$h = \sqrt{36} = 6\text{ m}$$

Conclusion:
Hence proved, the height of the tower is strictly $6\text{ m}$.

🧠 Examiner’s Secret: In complementary angle questions, students often attempt to solve for angle $\theta$ numerically. The direct algebraic proof requires multiplying Equation 1 ($\tan \theta$) by Equation 2 ($\cot \theta$). Because $\tan \theta \cdot \cot \theta = 1$, the trigonometric functions cancel out immediately, leaving a simple equation: $h = \sqrt{a \cdot b}$.

[👉 Also Read: Class 10 Math Chapter 11 Areas Related to Circles NCERT Solutions]

Master High-Yield Board FAQs (Rank Math Schema Ready)

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

The angle of elevation is measured upward from the observer’s horizontal line of sight to an elevated object located higher than their eye level. In contrast, the angle of depression is measured downward from the horizontal eye line to an object situated below. Geometrically, because any horizontal observer line is parallel to the horizontal ground plane, the angle of depression to a target matches the angle of elevation measured from the target back to the observer as alternate interior angles.

Are students required to substitute decimal values for square roots in CBSE board exams?

If a question explicitly states values such as $\sqrt{3} = 1.732$ or $\sqrt{2} = 1.414$, students must substitute them and compute the decimal value to avoid losing marks. If no decimal value is specified in the question paper, leaving the final answer in simplest radical form with a rationalised denominator (for example, $10\sqrt{3}\text{ m}$) is accepted under CBSE step-marking schemes.

How much credit is awarded for the diagram in heights and distances questions?

Under standard CBSE marking schemes for $3$-mark and $5$-mark questions, an accurate, clearly labelled geometric diagram earns $1$ mark. The diagram must show vertices labeled ($A, B, C$), right angles marked ($90^\circ$), horizontal baseline indicated, and angles of elevation/depression positioned relative to the horizontal. An answer written without a diagram typically receives a deduction or zero credit if calculations cannot be validated geometrically.

Why do angles of depression have to be drawn from a horizontal line rather than the vertical wall?

An angle of depression is defined strictly with respect to a horizontal baseline. A common error is measuring the angle between the vertical structure (like a lighthouse or building) and the line of sight. Drawing an angle against the vertical wall produces the complement of the true angle ($90^\circ – \theta$), which reverses the sine and cosine ratios and leads to incorrect lengths.

When should observer height be included in the calculation?

Observer height must be accounted for whenever the problem statement explicitly provides the observer’s stature (such as a $1.5\text{ m}$ tall boy or a $1.2\text{ m}$ tall girl). In those situations, the observer’s eye level creates a horizontal reference plane elevated above the ground, meaning the effective vertical leg of the right-angled triangle equals the total structure height minus the observer’s height. If no observer height is mentioned, the observer is treated as a point on the ground plane.

What is the most common trigonometric ratio applied in Chapter 9 problems?

The tangent ratio ($\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$) is the most widely used ratio in heights and distances because survey measurements typically provide or require vertical heights (perpendicular) and horizontal ground separations (base). The sine ratio is used when dealing with direct lengths along a line of sight, such as ropes, slides, or kite strings (hypotenuse).

What happens if an irrational radical remains in the denominator of the final answer?

Leaving an irrational radical in the denominator (for instance, $\frac{24}{\sqrt{3}}$ instead of $8\sqrt{3}$) typically results in a half-mark deduction under the mathematical simplification criteria of the CBSE evaluation scheme. Students must rationalise denominators by multiplying both the numerator and the denominator by the conjugate or corresponding root term before writing their final result.

How do you identify whether to use sine, cosine, or tangent in a word problem?

Identify the two sides of the right-angled triangle relevant to the problem: the side provided and the side you need to find. If the problem involves the opposite side and hypotenuse, use the sine ratio. If it involves the adjacent side and hypotenuse, use cosine. When working with the opposite side and adjacent side, apply the tangent ratio.

Can the angle of elevation ever exceed 90 degrees?

No. In the context of right-angled triangle trigonometry and real-world observations within Chapter 9, an angle of elevation is bounded strictly within the open acute interval $0^\circ < \theta < 90^\circ$. An angle of $0^\circ$ corresponds to a purely horizontal view, while $90^\circ$ represents looking straight up along a vertical line, which does not produce a right-angled triangle.

How do complementary angle problems work in heights and distances?

Two acute angles are complementary when their sum is $90^\circ$. When two observation points produce complementary angles of elevation to the top of a tower of height $h$, one angle is $\theta$ and the other is $90^\circ – \theta$. By expressing the two triangles using $\tan \theta = \frac{h}{a}$ and $\tan(90^\circ – \theta) = \cot \theta = \frac{h}{b}$, their product yields $\tan \theta \cdot \cot \theta = 1 = \frac{h^2}{ab}$, which simplifies to the standard formula $h = \sqrt{ab}$.

Is it necessary to write units with the final answer?

Yes. Omitting physical units (such as metres, centimetres, or seconds) in the final statement typically results in a half-mark penalty under CBSE step-marking criteria. Every physical quantity computed must end with its designated unit stated clearly.

How does uniform speed relate to distance in heights and distances problems?

When moving objects like cars, boats, or pedestrians approach an observation point, the distance travelled along the ground equals the uniform speed multiplied by the time taken ($d = v \cdot t$). By framing horizontal line segments in terms of $v$ and $t$, the speed variable $v$ cancels out when setting up ratios between simultaneous right-angled triangles, isolating time $t$.

How should a student tackle case-study questions based on Chapter 9?

Case-study questions break down descriptive practical scenarios into multiple parts. First, translate the written description into a clean geometric line diagram. Second, identify which right-angled triangles share common boundary lines or heights. Third, solve the sub-questions sequentially, as later questions often rely on values calculated in earlier parts.

What is the Clinometer mentioned in textbook side-notes?

A clinometer is a practical surveying device consisting of a graduated protractor card, a sighting tube or straw, and a weighted plumb-line suspended from the central origin point. When an observer sights an elevated object through the tube, the plumb-line hangs vertically under gravity, directly indicating the angle of inclination or elevation relative to the horizontal.

What are the key steps to guarantee maximum marks in 5-mark heights and distances questions?

To earn full marks on 5-mark questions, follow these four structured steps: First, draw a neat, labelled diagram showing all angles, right-angle markers, and side lengths. Second, write an initial “Given Data” statement defining each variable and line segment. Third, write the general trigonometric formula before substituting values. Fourth, show all algebraic steps cleanly and state the final numerical answer with its correct unit in a concluding sentence.

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