Chapter 9 Some Applications of Trigonometry (Concepts)

Introduction to Heights and Distances

One of the significant practical applications of trigonometry is in finding the heights and distances of objects or locations that are difficult or impossible to measure directly. For example, we can determine the height of a tall building, a mountain, or the distance across a river using trigonometric principles without physically climbing the building or crossing the river. These applications primarily rely on forming right-angled triangles and using the trigonometric ratios defined in the previous chapter.

Key Terms in Heights and Distances

To solve problems involving heights and distances using trigonometry, we need to understand a few specific terms related to observation and angles:

Line of Sight:

The line of sight is an imaginary straight line that connects the eye of an observer to the point on the object that the observer is looking at. It represents the direction in which the observer is looking.

Angle of Elevation:

When an observer looks at an object that is located above the observer's horizontal level, the angle of elevation is the angle formed between the line of sight and the horizontal line (drawn from the observer's eye parallel to the ground). This angle is always measured upwards from the horizontal.

In the figure, if the observer's eye is at A, and the object is at B, with AC being the horizontal line from A, then $\angle \text{BAC}$ is the angle of elevation.

Angle of Depression:

When an observer looks at an object that is located below the observer's horizontal level, the angle of depression is the angle formed between the line of sight and the horizontal line (drawn from the observer's eye parallel to the ground). This angle is always measured downwards from the horizontal.

In the figure, if the observer's eye is at A (at a certain height above B), and the object is at B, with AC being the horizontal line from A, then $\angle \text{BAC}$ is the angle of depression.

Important Relationship: The angle of elevation of an object from an observer's eye is equal to the angle of depression of the observer's eye from the object, assuming the horizontal lines are parallel (which they are in a plane). In the diagram for the angle of depression, if there is a point D on the ground directly below A, and a horizontal line through B, then $\angle \text{BAC} = \angle \text{ABD}$ (alternate interior angles formed by parallel horizontal lines and the line of sight as a transversal).

Using Trigonometric Ratios to Solve Problems

Problems involving heights and distances can typically be solved by creating a simplified diagram that represents the given situation as a right-angled triangle. The unknown height or distance will be one side of this triangle, and the angle of elevation or depression will be one of the acute angles. Known lengths will be other sides.

We use the trigonometric ratios (sine, cosine, or tangent, and their reciprocals) that relate the given angle, the known side(s), and the unknown side. Once the correct trigonometric ratio is identified and the equation is set up, we can solve for the unknown quantity using the values of trigonometric ratios for specific angles (Section I4).

• If the relationship between the side opposite to the angle and the hypotenuse is relevant, use the sine ratio: $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$.
• If the relationship between the side adjacent to the angle and the hypotenuse is relevant, use the cosine ratio: $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$.
• If the relationship between the side opposite to the angle and the side adjacent to the angle is relevant, use the tangent ratio: $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$.

Answer:

To Find:

The height of the tower.

Given:

Distance from the foot of the tower to the observation point = 15 m.

Angle of elevation of the top of the tower from the observation point = $60^\circ$.

Solution:

Let the tower be represented by the vertical line segment BC, where B is the foot of the tower and C is the top. Let A be the point on the ground 15 m away from the foot of the tower B. So, the distance AB = 15 m.

The angle of elevation of the top of the tower C from point A is $\angle \text{BAC} = 60^\circ$.

Since the tower stands vertically on the ground, $\triangle \text{ABC}$ is a right-angled triangle with the right angle at B ($\angle \text{ABC} = 90^\circ$).

In the right triangle $\triangle \text{ABC}$, we are given the angle $\angle \text{A} = 60^\circ$ and the side adjacent to this angle, AB = 15 m. We need to find the height of the tower, which is the side opposite to angle A, BC. The trigonometric ratio that relates the opposite side and the adjacent side to an angle is the tangent ($\tan$).

$\tan A = \frac{\text{Side opposite to } A}{\text{Side adjacent to } A} = \frac{\text{BC}}{\text{AB}}$

Substitute the given values $\angle \text{A} = 60^\circ$ and AB = 15 m:

$\tan 60^\circ = \frac{\text{BC}}{15}$

... (1)

From the table of trigonometric ratios for specific angles (Section I4), we know that $\tan 60^\circ = \sqrt{3}$.

Substitute this value into equation (1):

$\sqrt{3} = \frac{\text{BC}}{15}$

Solve for BC by multiplying both sides by 15:

$\text{BC} = 15 \times \sqrt{3}$ metres

... (2)

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

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

$\text{BC} \approx 15 \times 1.732 = 25.98$ metres

Answer: The height of the tower is $15\sqrt{3}$ metres (approximately 25.98 m).

Answer:

To Find:

The height of the chimney.

Given:

Height of the observer = 1.5 m.

Distance of the observer from the chimney = 28.5 m.

Angle of elevation of the top of the chimney from the observer's eyes = $45^\circ$.

Solution:

Let the observer be represented by the line segment AB, where A is the position of her eyes and B is the point on the ground below her. So, AB = 1.5 m. Let the chimney be represented by the line segment EF, where E is the foot of the chimney and F is the top. The distance between the observer and the chimney on the ground is BE = 28.5 m.

Draw a horizontal line from the observer's eyes A, parallel to the ground BE. Let this line be AC, meeting the chimney EF at C. So, AC is horizontal, and BE is horizontal. This makes ABEC a rectangle, so AB = EC = 1.5 m and AC = BE = 28.5 m.

The angle of elevation of the top of the chimney F from the observer's eyes A is $\angle \text{FAC} = 45^\circ$.

Consider the right-angled triangle $\triangle \text{ACF}$, right-angled at C ($\angle \text{ACF} = 90^\circ$ because AC is horizontal and CF is vertical part of chimney). In $\triangle \text{ACF}$, we are given the angle $\angle \text{FAC} = 45^\circ$ and the side adjacent to this angle, AC = 28.5 m. We need to find the side opposite to this angle, CF.

Using the tangent ratio in $\triangle \text{ACF}$:

$\tan (\angle \text{FAC}) = \frac{\text{Side opposite to } \angle \text{FAC}}{\text{Side adjacent to } \angle \text{FAC}} = \frac{\text{CF}}{\text{AC}}$

Substitute the given values $\angle \text{FAC} = 45^\circ$ and AC = 28.5 m:

$\tan 45^\circ = \frac{\text{CF}}{28.5}$

... (1)

From the table of trigonometric ratios for specific angles, we know that $\tan 45^\circ = 1$.

Substitute this value into equation (1):

$1 = \frac{\text{CF}}{28.5}$

Solve for CF by multiplying both sides by 28.5:

$\text{CF} = 28.5 \times 1 = 28.5$ metres

... (2)

The total height of the chimney is EF. From the diagram, EF = EC + CF.

Since ABEC is a rectangle, EC = AB (height of the observer) = 1.5 m.

Total height of chimney (EF) = EC + CF

Substitute the values EC = 1.5 m and CF = 28.5 m (from equation 2):

$\text{EF} = 1.5 + 28.5$

$\text{EF} = 30.0$ metres

... (3)

Answer: The height of the chimney is 30 metres.