Heights and Distances: Overview, Formulas, Uses, Tricks

Heights and Distances is a study about the applications of Trigonometry. It has many real-life applications, such as measuring the object’s height, depth of the object, the distance between two celestial objects etc. The formulas and the ratios of trigonometry are helpful in the study of heights and distances.

In this article, we will discuss heights and distances and provide formulas, and solved problems for a better understanding of the concept. 

Definition of Heights and Distances

Heights and Distances is an application of Trigonometry, and it helps in solving some complex real-life problems such as finding the distance between two or more objects, the height of the mountains, distance between the heavenly objects etc. It is also used to find the angle made by an object at a point without measuring the height and distance of the object.

There are several terms associated with Heights and Distances. Some of them are listed below:

1. Line of Sight
2. Angle of Elevation
3. Angle of Depression

Solving the problems of Heights and Distances require a minimum of any two measurements among the following:

1. The distance of the object from the observer
2. The height of the object
3. Angle of elevation
4. Angle of depression

1. Line of Sight

The imaginary line that is drawn from the observer’s eye to an object is called the line of sight. It gives an idea about where an observer is looking at. In addition, the line of sight tells the type of inclination angle made with horizontal.

In general, the line of sight is shown in the problems with a dotted line.

2. Angle of Elevation

The angle formed by the line of sight with the horizontal when the observer is viewing the object upwards is called the angle of elevation. This angle is formed when the observer raises his head upward over the horizontal line.

In the above figure, the angle formed by the horizontal line with the line of sight when the observer is looking upwards is the angle of elevation.

Here, \(\theta \) is known as “Angle of Elevation”.

3. Angle of Depression

It is the angle of inclination formed by the line of sight with the horizontal when the observer is viewing the object downwards. This angle is formed when the observer lowers his head downwards from the top to the ground.

In the above figure, \(\theta \) is known as “Angle of Depression”.

Trigonometric Ratios

Heights and distance problems are solved by using trigonometric ratios. We know that the trigonometric ratio of the angle gives the relation between any two sides of the triangle. Let us discuss all the trigonometric ratios in detail

Consider a right-angled triangle as shown below:

1. \(\sin \theta \, = \,\,\frac{{{\rm{height}}}}{{{\rm{hypotenuse}}}}\)

2. \(\cos \theta  = \,\frac{{{\rm{distance}}}}{{{\rm{hypotenuse}}}}\)\(\tan \theta  = \frac{{{\rm{height}}}}{{{\rm{distance}}}}\)

3. \(\cot \theta \, = \,\frac{{{\rm{distance}}}}{{{\rm{height}}}}\)

4. \(\sec \theta  = \frac{{{\rm{hypotenuse}}}}{{{\rm{distance}}}}\)

5. \(\cos {\rm{ec}}\theta \, = \,\frac{{{\rm{hypotenuse}}}}{{{\rm{height}}}}\)

Some Specific Ratios of Trigonometry

In the following table, we have mentioned the trigonometric values of some standard angles \(0^\circ ,30^\circ ,45^\circ ,60^\circ \) and \({\text{90}}^\circ \) These values are useful in solving the problems of heights and distances.

The table consists of trigonometric ratios, such as sine, cosine, cotangent, tangent, secant and cosecant. In short, these ratios can be written as \({\text{sin,}}\, {\text{cos,}}\, {\text{cot,}}\, {\text{cosec}}\) and \(\sec \).

How to Find Heights and Distances?

The following Heights and Distances notes will help in solving the questions. To find the heights and distances, we will use trigonometric ratios.

Example:
Finding the height of a tree inclined at an angle.

The height of a tree can be found by using the tangent rule in the above right-angled triangle.
As we know that, \(\tan ({\rm{angle}}) = \frac{{{\rm{height}}}}{{{\rm{distance}}}}\)

\(\therefore \,\,{\rm{Height}}\, = \left\{ {\tan ({\rm{angle}})} \right\} \times {\rm{distance}}\)

For the given figure:

The height of the object is calculated as follows: \(\therefore \,\,{\rm{Height}}\, = \left\{ {\tan ({\rm{angle}})} \right\} \times {\rm{distance}}\)

And, the distance can be calculated as follows: \({\rm{Distance}}\,\,{\rm{ = }}\,\frac{{\left\{ {\tan ({\rm{angle}}} \right\}}}{{{\rm{height}}}}\,\)

Pythagoras theorem is used in heights and distances to find the distance of an object from a point on the ground. It states that the square of the third side (hypotenuse) equals the sum of squares of the other two sides.

For example:

In the right-angled triangle, if \(a\) is the hypotenuse and the other two sides are \(b\) and \(c\) according to the Pythagoras Theorem,

\({c^2}\, = \,{a^2}\, + \,\,{b^2}\)

Heights and Distances Formulas

Some of the useful relations derived from trigonometric ratios are listed below. These will help to solve heights and distances questions easily and quickly.

Formula –\(1\):

In a right triangle \(ABC\),

\(\sin \theta  = \frac{p}{h};\cos \theta  = \frac{b}{h};\tan \theta  = \frac{p}{b}\)

Formula–\(2\):

In a triangle \(ABC\) if \(BD:DC = {\rm{m:}}\,{\rm{n}}\) and \(\angle BAD\, = \alpha ,\angle CAD\, = \,\beta \) and \(\angle ADC\, = \,\theta \) then \((m + n)\cot \theta = m\cot \alpha – n\cot \beta \)

Formula–\(3\): In a triangle \(ABC,\) if \(DE\parallel BA\) then according to basic proportionality theorem,\(\frac{{BD}}{{DC}} = \frac{{AE}}{{EC}}\).

Formula–\(4\):

The angles of elevations of two objects are \(\alpha ,\beta \) respectively, and the distance between two objects is \(d,\) then \(d = h\left({\cot \alpha  – \cot \beta } \right)\).

Heights and Distances – Tricks

There are a few things to be noted while solving the problems of heights and distances. Some of the important points are listed below:

1. In the problems of heights and distances, if the height of the observer is not given, then the observer is denoted by a point.

2. The height of the objects like towers, buildings, celestial bodies, etc., are represented by a line segment.

3.A horizontal line is a line obtained by drawing parallel to the ground.

4. If the observer moves towards the object, then the angle of elevation increases.

5. If the observer moves away from the object, then the angle of depression decreases.

6. Angle of elevation and angle of depression, so formed are equal.

7. An increase in the angle of elevation of the sun results in a decrease in the length of the shadow of an object.

8. A decrease in the angle of elevation of the sun results in an increase in the length of the shadow of an object.

9. Angle of elevation implies that an object is upwards from the position of the observer.

10. The angle of depression implies that an object is downwards from the position of the observer.

Uses of Heights and Distances

Heights and Distances have many real-life applications, such as they are used to find the distance between two or more objects, the height of the mountains etc. Some of the applications are listed below:

1. It is used In finding the slope, distance, height of the objects in the constructions.
2. It is used to calculate the height of the buildings, towers, mountains, celestial bodies.
3. It used to calculate the angle of inclination of the object without knowing the height and distance of the object.
4. It is used in marine and sub-marine systems to find the depth of the water.
5. It is used in navigation systems for the movement of boats, vehicles etc.
6. It is used in satellite systems to find the position and angle of satellites and space air-crafts.
7. It finds its applications in other subjects like physics, chemistry and biology.

Solved Examples on Heights and Distances

Q.1. There is a light-house on one side of the bank of the canal. From a point on the other side of the bank, exactly opposite, the angle of elevation of the top of the light-house is \({60^ \circ }\) From another point, which is \(20\,{\rm{m}}\) away from the point on the line joining the foot of the tower, with an angle of elevation of the top of the light-house is \({30^ \circ }\). Calculate the height of the light-house and also the width of the canal.

Ans: Let \(AB\) be the height of the light-house.
Given the distance between two points, \(C\),\(D\) is \(20\,{\rm{m}}\)
In a right triangle \(ABC\),
\(\tan {60^ \circ }\, = \,\frac{{AB}}{{BC}}\)
\( \Rightarrow \sqrt 3  = \frac{{AB}}{{BC}}\)
\( \Rightarrow AB = \sqrt 3 \, \times \,BC\) …(1)

In the right triangle \(ADB\),
\(\tan {30^ \circ }\, = \,\frac{{AD}}{{BD}} = \frac{{AB}}{{BC + CD}}\)
\( \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{AB}}{{BC + 20}}\)
\( \Rightarrow \frac{{BC + 20}}{{\sqrt 3 }} = AB\) …(2)

From \((1),(2)\)
\( \Rightarrow \frac{{BC + 20}}{{\sqrt 3 }} = \sqrt 3  \times BC\)
\( \Rightarrow BC + 20 = 3\,BC\)
\( \Rightarrow 2\,BC\, = \,20\)
\( \Rightarrow BC = \frac{{20}}{2} = 10\)

So, the width of the canal is \(10\,{\rm{m}}\).
From \((1)\), \(AB = 10\sqrt 3 \).
Hence, the height of the light-house is \(10\sqrt 3 \,{\rm{m}}\).

Q.2. A \({\rm{1}}{\rm{.2}}\,{\rm{m}}\) tall boy saw a parachute moving with the wind in a horizontal line of a height \({\rm{88}}{\rm{.2}}\,{\rm{m}}\) above with angle of elevation of the parachute being \({60^ \circ }\) After some time, the parachute moves away from the boy, which reduces the angle of elevation to \({30^ \circ }\) Find the distance travelled by the parachute during that interval.

Ans: Let \(BC\) be the height of the parachute from the horizontal line.
\({\rm{BC}}{\rm{  = }}\,{\rm{DE}}\,{\rm{ = }}\,{\rm{88}}{\rm{.2}}\,{\rm{m}}\)
In the right triangle \(ABC\)
\({\rm{tan3}}{{\rm{0}}^ \circ } = \frac{{BC}}{{AB}}\)
\( \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{88.2}}{{AB}}\)
\( \Rightarrow AB = 88.2\sqrt 3 \,{\rm{m}}\) …(1)

In triangle \(ADE\),
\({\rm{tan6}}{{\rm{0}}^ \circ } = \frac{{DE}}{{AE}}\)
\( \Rightarrow \sqrt 3 = \frac{{88.2}}{{AE}}\)
\( \Rightarrow AE = \frac{{88.2}}{{\sqrt 3 }} \) …(2)
From \((1)\) and \((2)\),
\(AB = AE\, + BE\)
\( \Rightarrow 88.2\sqrt 3  = \frac{{88.2}}{{\sqrt 3 }} = BE\)
\( \Rightarrow 88.2\sqrt 3  – \frac{{88.2}}{{\sqrt 3 }} = BE\)
\( \Rightarrow BE = \frac{{88.2 \times 3 – 88.2}}{{\sqrt 3 }}\)
\( \Rightarrow BE = \frac{{88.2 \times 2}}{{\sqrt 3 }}\)
\( \Rightarrow BE = \frac{{176.4}}{{\sqrt 3 }}\)
\( \Rightarrow BE = \frac{{176.4}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}\)
\( \Rightarrow BE = \frac{{176.4\sqrt 3 }}{3}\)
\( \Rightarrow BE = 58.8\sqrt 3 \,{\rm{m}}\)
Hence, the distance travelled by the parashot during the time period is \(58.8\sqrt 3 \,{\rm{m}}\)

Q.3. A tree breaks due to a 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 \({\rm{8}}\,{\rm{m}}\) Find the height of the tree.

Ans: Here,\(AC\) is the height of the tree, and \(BD\) is the broken part of the tree.
In triangle \(BCD\)
\(\tan {30^ \circ } = \frac{{BC}}{{CD}}\)
\( \Rightarrow \frac{1}{{\sqrt 3 }}\, = \frac{{BC}}{8}\)
\( \Rightarrow BC = \frac{8}{{\sqrt 3 }}\)

Here, \(BD = AB\)
\(\cos {30^ \circ } = \frac{{CD}}{{BD}}\)
\( \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{8}{{BD}}\)
\( \Rightarrow BD = \frac{{16}}{{\sqrt 3 }}\)
The height of the tree is \(AC\, = BD + BC\)
\( = \frac{{16}}{{\sqrt 3 }} + \frac{8}{{\sqrt 3 }}\)
\( = \frac{{24}}{{\sqrt 3 }}\)
\( = 8\sqrt 3 \,{\rm{m}}\)
The total height of the tree given is \(8\sqrt 3 \,{\rm{m}}\)

Q.4. The angle of elevation of the top of the tower from the top of the building of height \(30\,\,{\rm{m}}\) is \({30^ \circ }\) The distance between the tower and building is given by \(40\,\,{\rm{m}}\) Find the height of the tower.

Ans: Here,\(AB\, = \,30\,\,{\rm{m}}\) is the height of the building.
\(CD\) is the height of the tower.

In triangle, \(ACE\)
\(\tan {30^ \circ } = \frac{{CE}}{{AE}}\)
\( \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{40}}\)
\( \Rightarrow h = \frac{{40}}{{\sqrt 3 }}\,\,{\rm{m}}\)

The height of the tower is \(\frac{{40}}{{\sqrt 3 }} + 30 = \frac{{40 + 30\sqrt 3 }}{{\sqrt 3 }} = \frac{{\sqrt 3 (40 + 30\sqrt 3 )}}{{\sqrt 3  \times \sqrt 3 }} = \frac{{90 + 40\sqrt 3 }}{3}\,{\rm{m}}\)
Fence, the height of the building is \(\frac{{90 + 40\sqrt 3 }}{3}\,{\rm{m}}\)

Q.5. The angle of elevation of the building, which is \(30\,\,{\rm{m}}\) away from the foot of the building, is \({30^ \circ }\). Find the height of the building.

Ans: Here,\(BC\) is the height of the building.
In a triangle \(ABC\),
\(\tan {30^ \circ } = \frac{{BC}}{{AB}}\)
\( \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{BC}}{{30}}\)
\( \Rightarrow BC = \frac{{30}}{{\sqrt 3 }}\)
\( \Rightarrow BC = 10\sqrt 3 \)
Therefore, the height of the building is \(10\sqrt 3 \,{\rm{m}}\)

Summary

Heights and Distances is an application of Trigonometry. It helps us in solving complex real-life issues like calculating the distance between two or more objects, the height of mountains, etc. Heights and Distances are helpful in finding the distance, height, slope, etc., of the objects in the constructions. It is helpful in calculating the height of the towers, buildings, celestial bodies, etc. Furthermore, it is also helpful to calculate the angle of inclination of the object without the height and distance of a particular object.

SOLVE QUESTIONS ON HEIGHTS & DISTANCES

FAQs on Heights and Distances

Q.1. How do you find heights and distance problems in trigonometry?
Ans: By using the trigonometric ratios and the trigonometric ratios of some specific angles such as\(30^\circ ,45^\circ ,60^\circ \) and \(90^\circ \).
Example:
\(\tan ({\rm{angle}}) = \frac{{{\rm{opposite}}\,{\rm{side}}}}{{{\rm{adjacent}}\,{\rm{side}}}}\)

Q.2. How do you find height and distance?
Ans: For the given figure:

The height of the object is calculated as follows:  \({\rm{Height}}\,\, = \,\left\{ {\tan ({\rm{angle}})} \right\} \times {\rm{distance}}\)
And, the distance can be calculated as follows: \({\rm{Distance}} = \frac{{\left\{ {\tan ({\rm{angle}})} \right\}}}{{{\rm{height}}}}\)

Q.3. Which type of triangle is used in the problems of heights and distances?
Ans: The most commonly used triangle to solve the problems of heights and distances is the right-angled triangle.

Q.4. What is the angle of elevation? Explain with an example
Ans: The angle of inclination is formed by the line of sight with the horizontal when the observer is viewing the object upwards.
Example: The angle formed when an observer is looking at a bird on the tree from the ground.

Q.5. What is the relation between the angle of elevation and the angle of depression?
Ans: The angle of elevation and angle of depression of the triangle so formed are equal.

Q.6. What are the applications of heights and distances?
Ans: Some of the applications of heights and distances are:
1. In finding the slope, distance, height of the objects at the construction sites.
2. It is used in the height of the buildings, towers, mountains, celestial bodies etc.

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