Some Applications of Trigonometry - Calculating Heights and Distances

In theoretical trigonometry, ratios such as sine, cosine, and tangent are defined using abstract right-angled triangles. In practical geometry and real-world measurement, these mathematical ratios serve as a powerful indirect measurement tool. Historically developed for astronomy, geography, and celestial navigation, trigonometry allows us to determine the height of towering structures, the depth of valleys, or the distance to distant objects without physically scaling or traversing them.

In Class 10 Mathematics, Chapter 9 (Some Applications of Trigonometry) translates pure trigonometric concepts into practical problem-solving methods. By converting physical contexts into geometric diagrams of right-angled triangles, we can easily calculate unknown lengths and heights using measured angles and known baseline distances.

---

Fundamental Concepts & Terminology

To solve height and distance problems accurately, you must first master the specific terminology used to construct geometric diagrams from textual descriptions.

                  [Object observed]
                        /|
                       / |
       Line of Sight  /  |
                     /   | Height
                    /    |
  Observer's Eye --+-----|  [Foot of object]
    (Angle of     | Base |
    Elevation θ)  +------+
               Horizontal Level

1. Line of Sight

The line of sight is the imaginary straight line drawn from the eye of an observer to the point or object being observed.

2. Horizontal Level

The horizontal level is the straight horizontal line passing through the eye of the observer, parallel to the ground surface.

3. Angle of Elevation

When an object is located above the horizontal level, the observer must raise their head to view it. The angle formed between the line of sight and the horizontal level is defined as the angle of elevation.

$\text{Angle of Elevation } (\theta) = \angle \text{ (Line of Sight, Horizontal Level looking UP)}$

4. Angle of Depression

When an object is located below the horizontal level, the observer must lower their head to view it. The angle formed between the line of sight and the horizontal level is defined as the angle of depression.

$\text{Angle of Depression } (\phi) = \angle \text{ (Line of Sight, Horizontal Level looking DOWN)}$

  Observer's Eye --+----------------- Horizontal Level
                    \  (Angle of      (Parallel to Ground)
                     \  Depression φ)
      Line of Sight   \
                       \
                        * [Object observed below]
  =================================== Ground Level

Crucial Geometric Property: Since the horizontal level through the observer's eye is always parallel to the ground level, the angle of depression from an elevated observer to a ground point is mathematically equal to the angle of elevation from the ground point back to the observer (due to alternate interior angles formed by parallel lines).

$\text{Angle of Depression from top} = \text{Angle of Elevation from bottom}$

---

Standard Trigonometric Ratios Table

To evaluate height and distance equations, you must know the standard trigonometric values for acute angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$).

Useful square root approximations for final calculations:

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

---

Step-by-Step Problem-Solving Methodology

Every heights and distances question can be systematically solved using the following five steps:

1. Read and Conceptualize: Read the question carefully and distinguish between the height of the object, the position of the observer, and the horizontal distances involved.
2. Construct a Right-Angled Diagram: Sketch a clear, labeled 2D diagram. Represent vertical objects (towers, poles, trees, buildings) as vertical line segments perpendicular to the horizontal ground ($90^\circ$).
3. Account for Observer Height: If the height of the observer is explicitly given (e.g., "A 1.5 m tall boy..."), draw the observer as a vertical segment and measure the angle of elevation from the top of the observer's line of sight. If no height is mentioned, treat the observer as a point on the ground.
4. Identify Appropriate Trigonometric Ratios:
   • If working with Perpendicular ($P$) and Base ($B$), use $\tan \theta = \frac{P}{B}$ or $\cot \theta = \frac{B}{P}$.
   • If working with Hypotenuse ($H$) and Perpendicular ($P$), use $\sin \theta = \frac{P}{H}$.
   • If working with Hypotenuse ($H$) and Base ($B$), use $\cos \theta = \frac{B}{H}$.
5. Solve System of Linear/Algebraic Equations: For complex diagrams involving two right triangles sharing a common side, set up simultaneous equations and solve for the target variable. Rationalize surds in denominators where appropriate.

---

Real-World Applications

1. Land Surveying and Civil Engineering

Engineers and land surveyors measure the height of mountains, bridges, and tall infrastructure using an optical instrument called a clinometer (or theodolite). A clinometer measures the angle of elevation from a known baseline position, allowing the height of structures to be computed without physical scaling.

2. Maritime Navigation & Lighthouses

Lighthouse keepers and naval navigators determine a ship's distance from dangerous reefs or shorelines. By measuring the angle of depression of a ship from the top of a cliff or lighthouse of known height, navigators can compute the exact distance to shore using $d = h \cdot \cot \phi$.

3. Aviation & Radar Systems

Air traffic controllers and pilots calculate an aircraft's approach glide path using altitude and ground distance ratios. Trigonometric calculations ensure that an aircraft maintains a safe descent angle (typically $3^\circ$ in commercial aviation) relative to the runway threshold.

---

Step-by-Step Solved Examples

Example 1: Single Right Triangle (Finding Tower Height)

Problem: A tower stands vertically on the ground. From a point on the ground, which is $20\text{ m}$ 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.

Solution:

Let $AB$ be the vertical tower of height $h\text{ metres}$. Let $C$ be the point of observation on the ground such that $BC = 20\text{ m}$. The angle of elevation $\angle ACB = 60^\circ$.

  A (Top of tower)
  | \
  |  \
h |   \
  |    \
  | 60° \
  B------C
    20 m

In the right-angled triangle $\triangle ABC$, $\angle B = 90^\circ$:

$\tan(\angle ACB) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$

$\tan 60^\circ = \frac{h}{20}$

Substitute the value $\tan 60^\circ = \sqrt{3}$:

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

$h = 20\sqrt{3}\text{ m}$

Taking $\sqrt{3} \approx 1.732$:

$h = 20 \times 1.732 = 34.64\text{ m}$

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

---

Example 2: Observer Height Included (Electrical Maintenance)

Problem: An electrician needs 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? (Use $\sqrt{3} = 1.73$)

Solution:

Let $AD$ be the electric pole of height $5\text{ m}$. The electrician needs to reach point $B$ such that:

$AB = 1.3\text{ m}$

$BD = AD - AB = 5\text{ m} - 1.3\text{ m} = 3.7\text{ m}$

Let $BC$ represent the ladder of length $L\text{ metres}$, inclined at $\angle BCD = 60^\circ$ to the horizontal ground $CD$.

  A
  | 1.3m
  B
  | \
  |  \
  |   \ L (Ladder)
3.7m   \
  |     \
  |  60° \
  D-------C
      d

Step 1: Finding the length of the ladder ($L$)

In right-angled triangle $\triangle BDC$, $\angle D = 90^\circ$:

$\sin 60^\circ = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BD}{BC}$

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

$L = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{1.73} \approx 4.28\text{ m}$

Step 2: Finding the distance of the foot of the ladder from the pole ($d$)

$\cot 60^\circ = \frac{\text{Base}}{\text{Perpendicular}} = \frac{CD}{BD}$

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

$d = \frac{3.7}{\sqrt{3}} = \frac{3.7}{1.73} \approx 2.14\text{ m}$

Answer:

• Length of the ladder required = $4.28\text{ m}$
• Distance from foot of the pole = $2.14\text{ m}$

---

Example 3: Dual Angles (Elevation and Depression Combined)

Problem: 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.

Solution:

Let $AB$ be the building of height $7\text{ m}$. Let $CD$ be the cable tower of height $H\text{ metres}$. Draw $AE \perp CD$ such that $ABDE$ forms a rectangle where $AE = BD$ and $ED = AB = 7\text{ m}$. Let $CE = h$, so total height of tower $CD = H = h + 7$.

        C (Top of tower)
       /|
      / |
     /  | h
    /   |
  A+----E
  | \ 45|
7m|  \  | 7m
  |45°\ |
  B-----D
     x

From point $A$:

• Angle of elevation of top $C$: $\angle CAE = 60^\circ$
• Angle of depression of foot $D$: $\angle EAD = 45^\circ$

Since horizontal lines $AE$ and $BD$ are parallel:

$\angle ADB = \angle EAD = 45^\circ\quad \text{(Alternate interior angles)}$

Step 1: In right triangle $\triangle ABD$

$\tan 45^\circ = \frac{AB}{BD}$

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

Thus, horizontal distance $AE = BD = 7\text{ m}$.

Step 2: In right triangle $\triangle CEA$

$\tan 60^\circ = \frac{CE}{AE}$

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

Step 3: Calculating total height of tower $CD$

$H = CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1)\text{ m}$

Substituting $\sqrt{3} \approx 1.732$:

$H = 7(1.732 + 1) = 7(2.732) = 19.124\text{ m}$

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

---

Example 4: Variable Shadow Length (Changing Angle of Sun)

Problem: The shadow of a tower standing on 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.

Solution:

Let $AB$ be the vertical tower of height $h\text{ metres}$. Let $C$ and $D$ be the endpoints of the shadows when the Sun's angle of elevation (altitude) is $60^\circ$ and $30^\circ$, respectively. Given: $CD = 40\text{ m}$. Let $BC = x\text{ metres}$. Therefore, $BD = x + 40\text{ metres}$.

      A (Top of tower)
      | \
      |  \
    h |   \
      |    \
      | 60° \ 30°
      B------C-------D
         x     40 m

Step 1: In right triangle $\triangle ABC$

$\tan 60^\circ = \frac{AB}{BC}$

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

Step 2: In right triangle $\triangle ABD$

$\tan 30^\circ = \frac{AB}{BD}$

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

$x + 40 = h\sqrt{3} \quad \text{--- (Equation 2)}$

Step 3: Substitute Equation 1 into Equation 2

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

$40 = h\sqrt{3} - \frac{h}{\sqrt{3}}$

$40 = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right)$

$40 = h \left( \frac{3 - 1}{\sqrt{3}} \right) = h \left( \frac{2}{\sqrt{3}} \right)$

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

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

---

Common Student Mistakes to Avoid

---

Practice Questions for Self-Assessment

Practice Question 1

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.

Detailed Solution:

Let $AB$ and $CD$ be two poles of equal height $h$ standing on opposite sides of road $BD = 80\text{ m}$. Let $P$ be a point on $BD$. Let $BP = x\text{ m}$, then $PD = (80 - x)\text{ m}$. Given: $\angle APB = 60^\circ$ and $\angle CPD = 30^\circ$.

  A                 C
  |                 |
  |                 |
h |                 | h
  | 60°         30° |
  B-----+-----------D
     x     80 - x

In $\triangle ABP$:

$\tan 60^\circ = \frac{AB}{BP} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \text{--- (Eq. 1)}$

In $\triangle CDP$:

$\tan 30^\circ = \frac{CD}{PD} \implies \frac{1}{\sqrt{3}} = \frac{h}{80 - x} \implies h = \frac{80 - x}{\sqrt{3}} \quad \text{--- (Eq. 2)}$

Equating Eq. 1 and Eq. 2:

$x\sqrt{3} = \frac{80 - x}{\sqrt{3}}$

$3x = 80 - x \implies 4x = 80 \implies x = 20\text{ m}$

Substitute $x = 20$ into Eq. 1:

$h = 20\sqrt{3}\text{ m}$

Distances of the point from the poles:

• Distance from first pole ($BP$) = $20\text{ m}$
• Distance from second pole ($PD$) = $80 - 20 = 60\text{ m}$

Final Answer:

• Height of each pole = $20\sqrt{3}\text{ m}$ (or $34.64\text{ m}$)
• Distances of the point from the poles = $20\text{ m}$ and $60\text{ m}$

---

Practice Question 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.

Detailed Solution:

Let $AB$ be the lighthouse of height $75\text{ m}$. Let $C$ and $D$ be the positions of the two ships along the same straight line from the foot $B$. Angle of depression to ship $C$ (closer ship) = $45^\circ \implies \angle ACB = 45^\circ$. Angle of depression to ship $D$ (farther ship) = $30^\circ \implies \angle ADB = 30^\circ$.

  A (Top)
  |\
  | \
75|  \
  |   \
  | 45°\ 30°
  B----C-------D
    y     y_diff

In $\triangle ABC$:

$\tan 45^\circ = \frac{AB}{BC} \implies 1 = \frac{75}{BC} \implies BC = 75\text{ m}$

In $\triangle ABD$:

$\tan 30^\circ = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{75}{BD} \implies BD = 75\sqrt{3}\text{ m}$

Distance between the two ships $CD = BD - BC$:

$CD = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1)\text{ m}$

Substituting $\sqrt{3} \approx 1.732$:

$CD = 75(1.732 - 1) = 75 \times 0.732 = 54.9\text{ m}$

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

---

Practice Question 3

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 $30^\circ$. Find the distance traveled by the balloon during the interval.

Detailed Solution:

Height of balloon above ground = $88.2\text{ m}$. Height of girl = $1.2\text{ m}$. Height of balloon above girl's eye level ($h$) = $88.2 - 1.2 = 87\text{ m}$.

Let $A$ and $B$ be initial and final positions of the balloon. Let $E$ be the eye of the girl.

  A----------B  (Balloon line)
  |          |
 87m        87m
  | 60°  30° |
  C----------D
  |          | 1.2m
  E----------F (Eye line)

In right triangle $\triangle ACE$:

$\tan 60^\circ = \frac{AC}{EC} \implies \sqrt{3} = \frac{87}{EC} \implies EC = \frac{87}{\sqrt{3}} = 29\sqrt{3}\text{ m}$

In right triangle $\triangle BDE$:

$\tan 30^\circ = \frac{BD}{ED} \implies \frac{1}{\sqrt{3}} = \frac{87}{ED} \implies ED = 87\sqrt{3}\text{ m}$

Distance traveled by balloon = $CD = ED - EC$:

$CD = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\text{ m}$

Substituting $\sqrt{3} \approx 1.732$:

$CD = 58 \times 1.732 = 100.456\text{ m}$

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

---

Practice Question 4

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.

Detailed Solution:

Let $AB = h$ be the height of the tower. Let $C$ be the initial position of the car at $t = 0$ with angle of depression $30^\circ$ ($\angle ACB = 30^\circ$). Let $D$ be the position of the car at $t = 6\text{ seconds}$ with angle of depression $60^\circ$ ($\angle ADB = 60^\circ$). Let the uniform speed of the car be $v\text{ units/second}$.

  A (Top)
  |\
  | \
h |  \
  |   \
  | 60°\ 30°
  B----D-------C
       <-6 sec->

In right triangle $\triangle ABD$:

$\tan 60^\circ = \frac{AB}{BD} \implies \sqrt{3} = \frac{h}{BD} \implies BD = \frac{h}{\sqrt{3}}$

In right triangle $\triangle ABC$:

$\tan 30^\circ = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{h}{BC} \implies BC = h\sqrt{3}$

Distance covered in 6 seconds ($CD$):

$CD = BC - BD = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left( \frac{3 - 1}{\sqrt{3}} \right) = \frac{2h}{\sqrt{3}}$

Speed of the car ($v$):

$v = \frac{\text{Distance}}{\text{Time}} = \frac{\frac{2h}{\sqrt{3}}}{6} = \frac{h}{3\sqrt{3}}\text{ units/sec}$

Time required to travel remaining distance $BD$:

$\text{Time} = \frac{\text{Distance } BD}{\text{Speed } v} = \frac{\frac{h}{\sqrt{3}}}{\frac{h}{3\sqrt{3}}} = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h} = 3\text{ seconds}$

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

---

Exam Revision & Frequently Asked Questions (FAQs)

FAQ 1: Why is $\tan \theta$ used far more frequently than $\sin \theta$ or $\cos \theta$ in height and distance problems?

In practical real-world scenarios, vertical heights (buildings, trees, towers) and horizontal ground distances are easily accessible and measurable. Since the tangent ratio explicitly links the vertical side (Perpendicular) and horizontal side (Base) without requiring knowledge of the slant hypotenuse, $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$ is the most direct ratio for height-and-distance problems.

FAQ 2: What is the exact geometric relation between angle of elevation and angle of depression?

If an observer at point $A$ looks down at an object at point $B$, the angle of depression is $\phi$. If an observer at point $B$ looks up at point $A$, the angle of elevation is $\theta$.

Because horizontal reference lines through $A$ and $B$ are strictly parallel to each other, the two angles are alternate interior angles, meaning $\theta = \phi$.

FAQ 3: Should final answers be left in square root form ($\sqrt{3}$) or converted to decimals in CBSE Board Exams?

Follow these criteria:

1. If the question explicitly specifies a value (e.g., "Take $\sqrt{3} = 1.732$"), you must substitute the decimal value and calculate the simplified numerical result.
2. If no value for $\sqrt{3}$ is given, leaving your answer in rationalized surd form (e.g., $20\sqrt{3}\text{ m}$) is acceptable and earns full marks. However, ensure that square roots are removed from the denominator (e.g., write $\frac{20\sqrt{3}}{3}\text{ m}$ instead of $\frac{20}{\sqrt{3}}\text{ m}$).

FAQ 4: How do I handle multi-step problems where angles change over time?

For uniform motion problems:

1. Label positions at time $t_1$ and time $t_2$.
2. Express distances $x_1$ and $x_2$ in terms of height $h$ using right-triangle trigonometry.
3. Determine the distance covered $\Delta x = |x_2 - x_1|$.
4. Apply the formula $\text{Speed} = \frac{\Delta x}{\Delta t}$ to derive the rate, and then solve for required unknown times or distances.