In this blog, we will learn about the applications of trigonometry. Today we will be studying some ways in which trigonometry is used in the life around us. The knowledge of trigonometry is used in astronomy, to construct maps, determine the position of an island in relation to the longitudes and latitudes, etc.

 

We will see how trigonometry is used for finding the heights and distances of various objects, without actually measuring them.

 

Arun went to see a TV tower. Standing on the ground in front of the Tower he was trying to look at the top of the tower. Let us draw a line from the eye of Arun to the top of the tower. We name it AC. The line AC is called the line of sight.

 

 

Let us draw another line from the eye of Arun to the tower parallel to the ground and name it BC. The angle BAC, so formed by the line of sight with the horizontal, is called the angle of elevation of the top of the tower from the eye of the student.

 

 

Thus, the line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer. The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.

 

We make this angle of elevation many times as when we watch a plane or a bird flying in the sky. When we fly a kite when we talk to someone who is standing at a height from us.

 

Similarly, there is another angle called the angle of depression. This angle is formed when we see from a height downwards. The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level.

 

 

Look at this example. The girl sitting on the balcony is looking down at a flower pot. In this case, the line of sight is below the horizontal level. The angle so formed by the line of sight with the horizontal is called the angle of depression.

 

Let us again look at the figure showing the angle of elevation. If we want to find the height CD of the tower without actually measuring it, what information do we need? We would need to know the following:

 

 

(i) The distance DE at which the student is standing from the foot of the minar.
(ii) the angle of elevation, ?BAC, of the top of the minar.
(iii) the height AE of the student.

With the help of this information and using trigonometry we can find the height of a tower without measuring it. Let us see this through some examples.

 

Example: A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60�. Find the height of the tower.

 

 

Solution: First let us draw a simple diagram to represent the problem. Here AB represents the tower, CB is the distance of the point from the tower and ?ACB is the angle of elevation. We need to determine the height of the tower, i.e, AB. Also, ACB is a triangle, right-angled at B.

 

To solve the problem, we choose the trigonometric ratio tan 60�, as the ratio involves AB and BC. Now, tan 60� = AB/BC i.e., ?3 = AB/15 as value of tan 60� is ? 3 i.e., AB = 15/ ?3. Hence the height of the tower is 15?3 m.

 

Let us take one more example An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m 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� 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?

 

 

Solution: First we will draw a rough sketch. The electrician is required to reach point B on the pole AD.

 

BD = AD � AB = (5 � 1.3)m = 3.7 m.
BD/BC = sin 60�or 3.7/BC =?3/2
Therefore, BC =3.7x 2upon ?3� = 4.28 m
i.e., the length of the ladder should be 4.28 m.
Now, DC/BD = cot 60� =1/?3
i.e., DC = 3.7/?3 = 2.14 m
Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole.

 

Next Example The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30� and 45�, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

 

PC denotes the multistoried building and AB denotes the 8 m tall building. We are interested to determine the height of the multi-storeyed building, i.e., PC, and the distance between the two buildings, i.e., AC. Look at the figure carefully. Observe that PB is a transversal to the parallel lines PQ and BD. Therefore,?QPB and ?PBD are alternate angles, and so are equal.

 

So ?PBD = 30�. Similarly, ?PAC = 45�.
In right  DPAC, we have PC/AC = tan 45� = 1
That is PC = AC
Also, PC = PD + DC,
PD + DC = AC.
Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD = PD?3

 

So, the height of the multi-storeyed building is {4(?3+1) +8}m =4(3+?3)m

 

The distance between the two buildings is also 4(3+?3)m.