Applications of trigonometric functions

The tools of trigonometry are used wherever a problem is seen to involve lengths and angles.

This can be when a distance or an angle is required in ordinary space. It can also be when a problem involves more abstract vector quantities like force, velocity, acceleration and so on, that can be represented by diagrams with arrows set at various angles.

We can represent any vector quantity by an arrow whose length and direction correspond to the magnitude and direction of the vector.

Distance and direction in ordinary space

Example 1

A building can be seen on the far side of a flooded river, near the edge of the water at point $D$D. We wish to know how wide the flood is, adjacent to the building.

From a point $A$A on the bank opposite the building, a surveyor walks to point $C$C along a line perpendicular to a line directly across the flood until there is an angle of $45^\circ$45° between the line along the bank and the line from the surveyor's position to the building. At this point, there is an angle of $15^\circ$15° between the line along the bank and a line from the surveyor's position to the point $B$B where the line $AD$AD intersects the edge of the water.

Since triangle $CAD$CAD is isosceles, the distance $AD$AD must be equal to the distance $AC$AC. That is, $250\ m$250 m.

To obtain the distance $BD$BD across the flood, we need distance $AB$AB which we assume cannot be directly measured for some reason.

We can apply the sine ratio in triangle $CAB$CAB. 

$\sin15^\circ=\frac{BD}{250}$sin15°=BD250​

Therefore, $BD=250\sin15^\circ=64.7\ m$BD=250sin15°=64.7 m , and so, $BD=250-64.7=185.3\ m$BD=250−64.7=185.3 m.

 

Vector addition

Example 2

 

The floodwaters are flowing at the rate of $2\ m/s$2 m/s. To travel directly across the river by a small boat that can move through the water at $4\ m/s$4 m/s, the boat will have to aim slightly upstream by an angle $\theta$θ from the line directly across.

What is the angle $\theta$θ, and what will be the boat's speed in the line directly across the water?

Arrows representing the velocity vectors are drawn as in the diagram. (If you don't know much about vectors yet, that's ok, it's not necessary to solve this problem).   We have $\sin\theta=\frac{2}{4}=\frac{1}{2}$sinθ=24​=12​. Therefore, $\theta=30^\circ$θ=30°. 

Also, $\tan\theta=\frac{2}{v}$tanθ=2v​. So, $v=\frac{2}{\tan30^\circ}=2\sqrt{3}\approx3.5\ m/s$v=2tan30°​=2√3≈3.5 m/s.

 

It will take approximately $\frac{185.3}{3.5}\approx53$185.33.5​≈53 seconds to make the crossing.

Example 3

 

Example 4