If an observer is observing an object, then line of sight is an imaginary line drawn between observer's eye and the point at which observer is looking at. Let us suppose that a person or observer "O" is looking at the object, then line segment joining the point O with the point A is called line of sight. The angle between the line of sight and ground is called angle of elevation. The figure shown below demonstrates the line of sight as well as angle of elevation:
Line of Sight
If the object is below the level of the observer, then the angle between the ground and the line of sight is called the angle of depression. The figure shown below demonstrates line of sight as well as angle of depression.
Lines of Sight :


Propagation of line of sight follows the path of electromagnetic waves. Electromagnetic waves includes light waves which travel in a straight line. These waves can reflect, refract and deflect. These can be absorbed by the obstacles. But in mathematics, all these phenomena are ignored. We assume the propagation of line of sight as a straight line only.
There is no particular formula for finding length of line of sight. Line of sight formula varies on the basis of the given information. The figure shown below shows the line of sight:
Line of Sight Formula
Since triangle so formed is a right triangle, we can use trigonometric ratio formulas to find the length of line of sight. If angle θ is given, then we may find line of sight distance by using Sine or Cosine formulas, which can be explained as given below:
Line of Sight Formulas

Lines of Sight Formula

If both the height and distance of the base are given and θ is unknown, then we can use Pythagorean theorem to find the length of the line of sight.
Line of sight distance can be calculated in many ways. The method of finding line of sight distance depends upon what is given in the question. Following are few examples which demonstrate how to find line of sight distance:

Solved Examples

Question 1: A flag of height 9 meter is subtending an angle of 60$^o$ from a point located at some distance from the base of the tower. What is the length of the line of sight at flag top?
Solution:
The following figure is obtained from the given information:
Line of Sight Distance
Triangle ABC is a right-angled triangle. Let us use Sine formula:
$\sin 60^{\circ}$ = $\frac{AB}{AC}$

$\frac{\sqrt{3}}{2}$ = $\frac{9}{AC}$

$AC$ = $\frac{18}{\sqrt{3}}$

$AC$ = $\frac{18}{\sqrt{3}}$ $\times$ $\frac{\sqrt{3}}{\sqrt{3}}$

AC = 6$\sqrt{3}$ = 10.392 m

Question 2: A tower is subtending an angle of 45$^o$ from a point which is 10 meter away from the bottom of the tower. What would be the line of sight distance from this point to the top of the tower?
Solution:
The figure described in the question is drawn below:
Line of Sight Distances
Triangle ABC is a right-angled triangle. Let us use Cosine formula:
$\cos 45^{\circ}$ = $\frac{BC}{AC}$

$\frac{1}{\sqrt{2}}$ = $\frac{10}{AC}$

AC = 10$\sqrt{2}$ = 14.14 m

A horizontal line which goes parallel to the plane at which the observer is situated and passes through the object, is called horizon line of sight. The angle between this line and line of sight is known as angle of depression. In this case, the angle of depression is equal to the angle of elevation. Following figure shows a horizon line:
Horizon Line of Sight
Given below are few problems based on line of sight:

Solved Examples

Question 1: A 2-meter-long flag is making an angle of 30$^o$ on an observer's eye. Determine the length of the line of sight.
Solution:
The image explained in the question looks as shown below:
Line of Sight Problems
In triangle ABC:
$\sin 30^{\circ}$ = $\frac{2}{AC}$

$\frac{1}{2}$ = $\frac{2}{AC}$

AC = 4 m

Question 2: An observer observes an angle of 45$^o$ at some distance from a 50 meter tall building. This building has a tower at its top which is making an angle of 15$^o$ on observer's eye as shown in the following figure:
Line of Sight Problem
Find the length of the line of sight of the top of the tower.
Solution:
Let us suppose that the distance between base of building and observer is "x".
In triangle ORQ:
$\tan 45^{\circ}$ = $\frac{50}{x}$

$1$ = $\frac{50}{x}$
x = 50 m

Total angle subtended by tower and building is 60$^o$.
In tringle ORP:
$\cos 60^{\circ}$ = $\frac{x}{OP}$

$\frac{1}{2}$ = $\frac{50}{OP}$

OP = 100 m