4.8. Applications of Trigonometric Functions
Trigonometric functions have many important applications. In this section we discuss three of them and present sample computations involving each one.
Have you ever wondered how one determines the height of a mountain or a tall building? Here is a prime example of the use of trigonometry. The method is known as triangulation. Suppose you are standing a distance d from the base of a building and you wish to determine its height, h. By means of a sextant or a similar device, it is possible to measure the angle of inclination θ to the top of the building from the point where you are standing. This is the angle formed by the base B of the building, the point O where you are standing, and the top of the building P, as shown in the following diargam.
As can be easily seen from the diagram, the formula for computing the height of the building is h = d tan θ as indicated.
Example 1: Suppose you are standing 500 meters away from the base of a tall building. You measure the angle of inclination to the top of the building to be 15 degrees. How tall is the building?
Solution: The height of the building is given by the formula h = d tan θ, where d = 500 m and θ = 15°. Thus, the height of the building is given by
h = (500 m) tan 15° = (500 m)(0.268) = 134 meters.
Sea voyagers and flight navigators have long been using trigonometry to determine their headings. The simplest such application is computing the heading to a nearby destination with known coordinates. For instance, suppose one wants to know what direction to travel along the shortest possible route (a straight line) from A to B, where the distances north and east from A to B are known. Say B lies a distance x0 to the east and y0 to the north of A, as shown in the diagram below. Then the compass heading θ from A to B is given by
- (4.8.1) θ = arctan(x0/ y0)
Example 2: A pilot is flying to a destination 30 km north and 40 km east of his current location. At what heading should he be flying?
Solution: The correct heading θ is given by (4.8.1) with x0 = 40 and y0 = 30. The result is θ = arctan(40/ 30) = 53.1°.
Solar cells convert radiation from the sun to electricity. In order to determine how much power they can generate, we must be able to first determine the amount of solar power which is incident on them. Assuming the sky is clear, the total amount of solar power incident on a solar cell of area A is given by
- (4.8.2) P = AW sin θ
where θ is the angle of inclination of the sun and W is the solar constant, which is equal to 1340 watts per square meter.
Example 1: On a clear day, how much solar power is incident on a solar cell with an area of 10 square meters if the angle of inclination of the sun is 30 degrees? Suppose the solar cell has an efficiency of 20%. How much power will it generate?
Solution: The total amount of solar power incident on the cell is given by Equation (4.8.2) with A = 10, W = 1340, and θ = 30° (all MKS units). Thus we have
P = (10)(1340)(sin 30°) = (10)(1340)(0.5) = 6700 W.
If the cell has an efficiency of 20%, it can generate (0.2)(6700) = 1340 watts of power.