Applications of Trigonometry: Heights and Distances

Basics of Heights and Distances (Angle of Elevation, Angle of Depression, Line of Sight)

Heights and Distances is a significant application of trigonometry. It deals with finding the height of objects or the distance between objects indirectly, using the angles of elevation or depression and the known length of one side in a right-angled triangle. This is particularly useful when direct measurement is difficult or impossible, such as determining the height of a mountain, a tall building, or the distance across a river.

The core principles applied are the properties of right-angled triangles and the definitions of trigonometric ratios (sine, cosine, tangent, etc.) relating the sides and angles.

Line of Sight

When an observer looks at an object, the imaginary straight line drawn from the eye of the observer to the point on the object being viewed is called the line of sight.

Angle of Elevation

The angle of elevation is defined when the object being viewed is located above the horizontal level of the observer's eye.

It is the angle formed between the horizontal line (at the level of the observer's eye) and the line of sight, when the observer is looking upwards at the object.

In the diagram, let O be the position of the observer's eye and P be the object being viewed. Let OH be the horizontal line passing through the observer's eye O. The line OP is the line of sight. The angle formed between the horizontal line OH and the line of sight OP, which is $ \angle HOP $, is the angle of elevation of the object P from the point O.

Think of it as the angle you have to elevate (raise) your line of sight from the horizontal to look up at something.

Example Scenario:

If you are standing on the ground and looking up at the top of a tree, the angle of elevation is the angle between your horizontal line of sight (straight ahead) and your line of sight pointing towards the top of the tree.

Angle of Depression

The angle of depression is defined when the object being viewed is located below the horizontal level of the observer's eye.

It is the angle formed between the horizontal line (at the level of the observer's eye) and the line of sight, when the observer is looking downwards at the object.

In the diagram, let O be the position of the observer's eye and Q be the object being viewed below the horizontal level. Let OH be the horizontal line passing through O. The line OQ is the line of sight. The angle formed between the horizontal line OH and the line of sight OQ, which is $ \angle HOQ $, is the angle of depression of the object Q from the point O.

Think of it as the angle you have to depress (lower) your line of sight from the horizontal to look down at something.

Example Scenario:

If you are standing on top of a building and looking down at a car parked on the road below, the angle of depression is the angle between your horizontal line of sight (straight ahead) and your line of sight pointing towards the car.

Relationship between Angle of Elevation and Depression

When an observer at point A looks at an object at point B above their level, the angle of elevation is formed at A. If another observer at point B looks at the first observer at point A (which is below their level), the angle of depression is formed at B. These two angles are always equal, provided the horizontal lines are parallel (which they are in this context).

In the diagram, let O be the position of the first observer and P be the position of the object (or second observer). Let OH be the horizontal line at O, and PK be the horizontal line at P. Since OH and PK are both horizontal, they are parallel ($ OH \parallel PK $). The line of sight OP acts as a transversal cutting these parallel lines.

The angle of elevation of P from O is $ \angle HOP $ (marked as $ \alpha $).

The angle of depression of O from P is $ \angle OPK $ (marked as $ \beta $).

Since $ OH \parallel PK $ and OP is a transversal, the alternate interior angles are equal. Therefore, $ \angle HOP = \angle OPK $.

Conclusion: The angle of elevation of an object from an observer is equal to the angle of depression of the observer from the object.

Trigonometric Ratios Used in Heights and Distances

Problems in heights and distances almost always involve setting up one or more right-angled triangles. The unknown height or distance becomes a side of this triangle. The angles of elevation or depression are used as the acute angles in the triangle.

Once the right triangle is identified and the known side and angle are marked, we use the appropriate trigonometric ratio to relate the known information to the unknown side. The primary trigonometric ratios relating the acute angle $\theta$ and the sides (Opposite/Perpendicular, Adjacent/Base, Hypotenuse) in a right triangle are:

• Sine (sin): $ \sin \theta = \frac{\text{Opposite Side}}{\text{Hypotenuse}} $

• Cosine (cos): $ \cos \theta = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} $

• Tangent (tan): $ \tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}} $

The choice of which ratio to use depends on which sides are involved with respect to the known angle:

• If you know or need the Opposite side and know or need the Adjacent side, use the Tangent ratio.

• If you know or need the Opposite side and know or need the Hypotenuse, use the Sine ratio.

• If you know or need the Adjacent side and know or need the Hypotenuse, use the Cosine ratio.

In typical height and distance problems, we often deal with vertical heights (Opposite side) and horizontal distances (Adjacent side). Thus, the tangent ratio ($ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} $) is very frequently used.

Solving these problems requires knowing the trigonometric values for common angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$).

Note for Competitive Exams

Solving Problems involving Heights and Distances using Trigonometric Ratios (Single Right Triangle)

Many basic problems in Heights and Distances can be solved by applying trigonometric ratios to a single right-angled triangle. These problems typically involve an observer looking at a single object, where the relative positions form a right triangle. The height of a vertical object, the distance to an object, or the length of a line of sight (hypotenuse) can be found if one side and one acute angle are known.

Steps to Solve Problems with a Single Right Triangle:

1. Understand the Problem: Read the problem statement carefully to identify all given information (known lengths, known angles of elevation or depression) and determine what quantity needs to be calculated.

2. Draw a Clear Diagram: This is the most crucial step. Draw a diagram representing the situation. Use lines to represent vertical objects (towers, buildings, trees), horizontal distances (ground level), the observer's position, the object's position, the line of sight, and the horizontal line through the observer's eye. Label the points clearly (e.g., A, B, C). Indicate the right angle(s). Mark the known lengths and angles and assign a variable to the unknown quantity you need to find.

3. Identify the Right Triangle: From your diagram, pinpoint the right-angled triangle that contains the known angle, the known side, and the unknown side.

4. Relate Sides to the Angle: In the identified right triangle, relative to the given acute angle (angle of elevation or depression), identify which sides are the Opposite, Adjacent, and Hypotenuse.

5. Choose the Correct Trigonometric Ratio: Select the trigonometric ratio (sine, cosine, or tangent) that relates the known side and the unknown side you need to find with respect to the known angle.

   • Use $ \tan \theta $ if the Opposite and Adjacent sides are involved.
   • Use $ \sin \theta $ if the Opposite side and Hypotenuse are involved.
   • Use $ \cos \theta $ if the Adjacent side and Hypotenuse are involved.

6. Set up the Equation: Write down the equation using the chosen trigonometric ratio, the known angle, the known side length, and the variable for the unknown side length.

7. Solve the Equation: Solve the equation for the unknown variable. This may involve using the values of trigonometric ratios for specific angles or using a calculator if the angles are not standard.

8. Write the Final Answer: State the answer clearly with the appropriate units.

Examples

Example 1: Finding Height using Angle of Elevation

Answer:

Given:

• Distance from the observation point to the foot of the tower = 15 m.
• Angle of elevation of the top of the tower = $60^\circ$.

To Find:

• The height of the tower.

Diagram:

Let AB represent the tower, with height $h$ meters. B is the foot of the tower, and A is the top. Let C be the point on the ground where the observer is located. The distance from C to the foot of the tower is given as BC = 15 m.

The angle of elevation from the point C to the top of the tower A is $ \angle ACB $. This is given as $60^\circ$.

Since the tower is vertical to the ground, the triangle ABC is a right-angled triangle, with the right angle at B ($ \angle ABC = 90^\circ $).

Solution:

In the right-angled triangle ABC, with respect to the angle $ \angle ACB = 60^\circ $:

• The side opposite to $60^\circ$ is AB, which is the height $h$ we need to find (Opposite side).
• The side adjacent to $60^\circ$ is BC, which is the given distance 15 m (Adjacent side).

We need to relate the Opposite side and the Adjacent side, so we use the tangent ratio:

$\tan(\angle ACB) = \frac{\text{Opposite}}{\text{Adjacent}}$

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

Substitute the known values:

$\tan 60^\circ = \frac{h}{15}$

From the table of specific angles, we know that $ \tan 60^\circ = \sqrt{3} $.

$\sqrt{3} = \frac{h}{15}$

Solve for $h$ by multiplying both sides by 15:

$h = 15 \sqrt{3}$

The height of the tower is $15 \sqrt{3}$ meters.

If a numerical approximation is required, using $ \sqrt{3} \approx 1.732 $:

$ h \approx 15 \times 1.732 = 25.98 $ meters.

Example 2: Finding Distance using Angle of Depression

Answer:

Given:

• Height of the cliff = 75 m.
• Angle of depression of a boat from the top of the cliff = $30^\circ$.

To Find:

• The distance of the boat from the foot of the cliff.

Diagram:

Let AB be the cliff, with height 75 m (A is the top, B is the foot). Let C be the position of the boat on the sea level. Let AX be the horizontal line through the top of the cliff A.

The angle of depression from A to the boat C is $ \angle XAC = 30^\circ $. We need to find the horizontal distance BC. Let BC = $d$ meters.

Since AX is a horizontal line and BC represents the horizontal distance on the ground/sea, $ AX \parallel BC $. The line of sight AC is a transversal. Therefore, the angle of depression $ \angle XAC $ is equal to the angle of elevation $ \angle ACB $ (alternate interior angles).

$\angle ACB = \angle XAC = 30^\circ$

[Alternate Interior Angles]

The triangle ABC is a right-angled triangle at B ($ \angle ABC = 90^\circ $).

Solution:

In the right-angled triangle ABC, with respect to the angle $ \angle ACB = 30^\circ $:

• The side opposite to $30^\circ$ is AB, which is the height of the cliff 75 m (Opposite side).
• The side adjacent to $30^\circ$ is BC, which is the distance $d$ we need to find (Adjacent side).

We need to relate the Opposite side and the Adjacent side, so we use the tangent ratio:

$\tan(\angle ACB) = \frac{\text{Opposite}}{\text{Adjacent}}$

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

Substitute the known values:

$\tan 30^\circ = \frac{75}{d}$

From the table of specific angles, we know that $ \tan 30^\circ = \frac{1}{\sqrt{3}} $.

$\frac{1}{\sqrt{3}} = \frac{75}{d}$

Solve for $d$ by cross-multiplying:

$1 \times d = 75 \times \sqrt{3}$

$d = 75 \sqrt{3}$

The distance of the boat from the foot of the cliff is $75 \sqrt{3}$ meters.

If a numerical approximation is required, using $ \sqrt{3} \approx 1.732 $:

$ d \approx 75 \times 1.732 = 129.9 $ meters.

Note for Competitive Exams

Solving Problems involving Heights and Distances (Multiple Right Triangles)

While many basic problems in Heights and Distances can be solved using a single right-angled triangle, more complex and realistic scenarios often involve situations that require analysing two or more right-angled triangles. These triangles typically share a common side (like the height of a tower or a horizontal distance) or have dimensions that are directly related.

Solving such problems involves identifying these triangles, setting up trigonometric equations for each, and then solving the resulting system of equations simultaneously to find the required unknown quantity.

Common Scenarios Leading to Multiple Triangles

You will often encounter multiple triangles in problems where:

• An observer moves towards or away from an object, taking observations (angles of elevation/depression) from two different positions.

• Observations of the same object are made from two different points (e.g., from the tops of two buildings or from two points on the ground).

• An object consists of two parts at different heights, and observations are made to the top and bottom of the upper part (e.g., a flagstaff on top of a building, a statue on a pedestal).

• Observations of two different objects are made from the same point.

General Strategy for Solving Problems with Multiple Triangles

1. Draw a Clear and Accurate Diagram: Represent the ground level, vertical objects, observer positions, object positions, lines of sight, and horizontal lines. Label all points (A, B, C, etc.), known lengths, and known angles (angles of elevation/depression). Mark the right angles. Use variables (like $h$ for height, $x$ for distance) for unknown quantities. Ensure angles of depression are correctly shown from the horizontal downwards and then use the alternate interior angle property to relate them to angles inside the triangle.

2. Identify All Relevant Right Triangles: Carefully look at the diagram and identify the right-angled triangles that are formed by the given information. These triangles will contain the known angles and the sides related to the quantities you need to find.

3. Identify Common or Related Sides: Notice which sides are shared by the triangles or how the sides in one triangle relate to the sides in another (e.g., one distance is the sum or difference of other distances).

4. Set Up Trigonometric Equations: For each relevant right triangle, choose the appropriate trigonometric ratio (most often tangent, as it relates opposite and adjacent sides) and write down an equation involving the known angle, a known side (if any), and the side(s) related to the unknowns.

5. Solve the System of Equations: You will typically have a system of two or more equations with two or more unknowns. Use algebraic methods (like substitution or elimination) to solve this system for the required unknown quantity.

6. Calculate and State the Answer: Perform the final calculations and state the answer clearly with appropriate units.

Examples

Example 1: Observer Moving Towards a Tower

Answer:

Given:

• Angle of elevation from point A = $30^\circ$.
• Angle of elevation from point B = $60^\circ$.
• Distance AB = 20 m (distance moved towards the tower).

To Find:

• The height of the tower.

Diagram:

Let CD represent the tower, with height $h$ meters. D is the foot of the tower, and C is the top. Let A and B be the two points on the ground in line with D such that B is between A and D. Point A is 20 m away from B, so AB = 20 m.

The angle of elevation from A to C is $ \angle CAD = 30^\circ $. The angle of elevation from B to C is $ \angle CBD = 60^\circ $.

Since the tower is vertical to the ground, the triangles CBD and CAD are right-angled triangles at D ($ \angle CDB = \angle CDA = 90^\circ $).

Let the distance BD be $x$ meters. Then the distance AD = AB + BD = $20 + x$ meters.

Solution:

We have two right-angled triangles involving the height $h$ and the distances $x$ and $20+x$. We will use the tangent ratio as it relates the opposite side (height $h$) and the adjacent sides (distances from the foot of the tower).

In right $\triangle CBD$:

With respect to the angle $ \angle CBD = 60^\circ $:

• Opposite side = CD = $h$.
• Adjacent side = BD = $x$.

$\tan 60^\circ = \frac{CD}{BD}$

Substitute the values:

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

From this, we can express $h$ in terms of $x$ (or vice versa). Let's express $h$ in terms of $x$:

$h = x\sqrt{3}$

... (i)

In right $\triangle CAD$:

With respect to the angle $ \angle CAD = 30^\circ $:

• Opposite side = CD = $h$.
• Adjacent side = AD = $20 + x$.

$\tan 30^\circ = \frac{CD}{AD}$

Substitute the values:

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

Cross-multiply:

$1 \times (20+x) = h \times \sqrt{3}$

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

... (ii)

Now we solve the system of two equations (i) and (ii) with two unknowns ($h$ and $x$). We can substitute the expression for $h$ from equation (i) into equation (ii):

$20+x = (x\sqrt{3})\sqrt{3} $

[Substituting $h = x\sqrt{3}$]

$20+x = x(\sqrt{3})^2 = 3x$

Rearrange the equation to solve for $x$:

$20 = 3x - x$

$20 = 2x$

$x = \frac{20}{2} = 10$

So, the distance BD = 10 m.

Now we can find the height $h$ by substituting the value of $x$ back into equation (i):

$h = x\sqrt{3} = 10\sqrt{3}$

[Substituting $x=10$ into (i)]

The height of the tower is $10 \sqrt{3}$ meters.

If a numerical approximation is required, using $ \sqrt{3} \approx 1.732 $:

$ h \approx 10 \times 1.732 = 17.32 $ meters.

Example 2: Object with Two Parts (Statue on Pedestal)

Answer:

Given:

• Height of the statue = 1.6 m.
• Angle of elevation to the top of the statue = $60^\circ$.
• Angle of elevation to the top of the pedestal = $45^\circ$.
• Observations are made from the same point on the ground.

To Find:

• The height of the pedestal.

Diagram:

Let BC represent the pedestal, with height $h$ meters. Let CD represent the statue standing on the pedestal, with height 1.6 m. The total height from the ground to the top of the statue is BD = BC + CD = $h + 1.6$ meters.

Let A be the point of observation on the ground. Let the horizontal distance from the observation point to the foot of the pedestal/statue (point B) be $x$ meters. So, AB = $x$.

The angle of elevation from A to the top of the pedestal (point C) is $ \angle BAC = 45^\circ $.

The angle of elevation from A to the top of the statue (point D) is $ \angle BAD = 60^\circ $.

Since the pedestal is vertical to the ground, the triangles ABC and ABD are right-angled triangles at B ($ \angle ABC = \angle ABD = 90^\circ $).

Solution:

We have two right-angled triangles: $\triangle ABC$ and $\triangle ABD$. Both share the common side AB ($x$). We will use the tangent ratio.

In right $\triangle ABC$ (for the pedestal height):

With respect to the angle $ \angle BAC = 45^\circ $:

• Opposite side = BC = $h$.
• Adjacent side = AB = $x$.

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

Substitute the values:

$1 = \frac{h}{x}$

[Since $\tan 45^\circ = 1$]

From this equation, we get:

$h = x$

... (i)

The height of the pedestal is equal to the distance of the observer from the foot of the pedestal.

In right $\triangle ABD$ (for the total height):

With respect to the angle $ \angle BAD = 60^\circ $:

• Opposite side = BD = $h + 1.6$.
• Adjacent side = AB = $x$.

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

Substitute the values:

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

[Since $\tan 60^\circ = \sqrt{3}$]

Cross-multiply:

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

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

... (ii)

Now we solve the system of two equations (i) and (ii) for $h$. We can substitute $ x = h $ from equation (i) into equation (ii):

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

[Substituting $x=h$]

Rearrange the equation to isolate $h$ terms:

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

Factor out $h$ from the left side:

$h(\sqrt{3} - 1) = 1.6$

Solve for $h$ by dividing both sides by $ (\sqrt{3} - 1) $:

$h = \frac{1.6}{\sqrt{3} - 1}$

To simplify and rationalise the denominator, multiply the numerator and denominator by the conjugate of the denominator, $ (\sqrt{3} + 1) $:

$h = \frac{1.6}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$

Use the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ in the denominator:

$h = \frac{1.6 (\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{1.6 (\sqrt{3} + 1)}{3 - 1} = \frac{1.6 (\sqrt{3} + 1)}{2}$

Divide the numerator by 2:

$h = 0.8 (\sqrt{3} + 1)$

The height of the pedestal is $0.8(\sqrt{3} + 1)$ meters.

If a numerical approximation is required, using $ \sqrt{3} \approx 1.732 $:

$ h \approx 0.8 \times (1.732 + 1) = 0.8 \times 2.732 = 2.1856 $ meters.

Note for Competitive Exams