Trigonometric ratios examples, exercises and applications

The trigonometric ratios are the quotients or ratios that can be made with the value of the sides of a right triangle. These sides are: two legs that form 90º with each other and the hypotenuse, which forms the acute angle θ with one of the legs.

You can form 6 quotients. Their names and respective abbreviations are:

• bosom (sen)
• cosine (cos)
• tangent (tg or tan)
• cotangent (ctg or cotan)
• secant (sec) and
• cosecant (cosec)

All of them referred to angle θ, as shown in the following figure:

The basic trigonometric ratios of the angle θ are sin θ, cos θ, and tan θ, while the remaining ratios can be expressed in terms of these three. From the above table it can be seen that:

• sec θ = 1 / cos θ
• cosec θ = 1 / sin θ
• cot θ = 1 / tg θ

The size of the sides of the triangle does not influence the value of the ratios, since two triangles whose angles measure the same are similar triangles and the respective ratios between the sides have the same value.

Article index

• 1 Example
• 2 Solved exercises
  • 2.1 - Exercise 1
  • 2.2 - Exercise 2
• 3 Applications of trigonometric ratios
• 4 References

Example

For example, let's calculate the trigonometric ratios of the angle θ in the following triangles:

For the small triangle we have the three basic ratios of the angle θ:

sin θ = 3/5

cos θ = 4/5

tg θ = ¾

And now let's calculate the three basic ratios of θ with the large triangle:

sin θ = 30/50 = 3/5

cos θ = 40/50 = 4/5

tg θ = 30/40 = ¾

An important detail to take into account is the following: both sin θ and cos θ are less than 1, since the legs always measure less than the hypotenuse. Indeed:

sin θ = 3/5 = 0.6

cos θ = 4/5 = 0.8

Solved exercises

In the following exercises you are asked to solve the right triangle, which means finding the length of its three sides and the measure of its internal angles, one of which always measures 90º.

The Pythagorean theorem applies to right triangles and is very useful when two of the sides are known and the missing side needs to be determined. The theorem goes like this:

Hypotenuse^two = opposite leg^two + adjacent leg^two

We can check the Pythagorean theorem with the small triangle in Figure 2, whose legs are 3 and 4. The order in which the legs are taken does not matter. Applying the theorem we have:

Hypotenuse^two = 3^two + 4^two = 9 + 16 = 25

Therefore the hypotenuse is:

Hypotenuse = √25 = 5

- Exercise 1

Calculate the trigonometric ratios of the angles shown in the following triangles:

Solution to

This triangle is the same as in figure 3, but we are asked for the trigonometric ratios of the other acute angle, denoted α. The statement does not offer the value of the hypotenuse, however, by applying the Pythagorean theorem we know that it is worth 5.

The ratios can be calculated directly from the definition, taking care when selecting the leg that is the opposite of the angle α to calculate sin α. Let's see:

• sin α = 4/5
• cos α = 3/5
• tg α = 4/3
• cot α = ¾
• sec α = 1 / (3/5) = 5/3
• cosec α = 1 / (4/5) = 5/4

And as we can see, the values ​​of the trigonometric ratios have been swapped. Indeed, α and θ are complementary angles, which means that they add up to 90º. In this case it is true that sin α = cos θ and so on for the other reasons.

Solution b

Let's calculate the hypotenuse of the triangle using the Pythagorean theorem:

Hypotenuse^two = 20^two + twenty-one^two = 841

√841 = 29

Then the 6 trigonometric ratios of the angle β are:

• sin β = 20/29
• cos β = 21/29
• tg β = 20/21
• cot β = 21/20
• sec β = 1 / (21/29) = 29/21
• cosec β = 1 / (20/29) = 20/29

- Exercise 2

a) Find the value of x in the figure.

b) Calculate the perimeter of the 3 triangles shown.

Solution to

In the figure we can identify several triangles, in particular the right triangle on the left, which has a leg equal to 85 and the acute angle 60º.

With the information from this triangle we can calculate side b. It is not the measure requested by the statement, but knowing its value is a prior step.

To determine this, the appropriate ratio is tg 60º = 85 / b, since b is the leg adjacent to 60º and 85 is the opposite of said angle. Therefore:

b = 85 / tg 60º = 85 / √3

Once b is known, we will use the large and outer right triangle, which has a common side with the previous triangle: the one that measures 85. This is the leg opposite the 30º angle..

Thence:

Leg adjacent to 30º = (85 / √3) + x

Now we can propose the following:

85 / [(85 / √3) + x] = tg 30º

What is in brackets happens to multiply the tg 30º:

85 = [(85 / √3) + x]. tg 30º

Applying the distributive property of multiplication:

85 = tg 30 °. (85 / √3) + x. tg 30º

Therefore:

x.tg 30º = 85 - tg 30º. (85 / √3) = 85 [1 - tg 30º. (1 / √3)] = 85. (2/3) = 170/3

Substituting the value tg 30º = √3 / 3:

x = (170/3) ÷ (√3 / 3) = 98.15

Solution b

Perimeter of the small triangle

Let h₁ the hypotenuse of this triangle, which can be calculated by the Pythagorean theorem or by means of a trigonometric ratio, for example cos 60º:

cos 60 º = 85 / √3 / h₁→ h₁ = (85 / √3) ÷ cos 60º = 98.1

To find P, the perimeter of this triangle, we simply add the 3 sides:

P = 85 + (85 / √3) + 98.1 = 232.2

Perimeter of the outer triangle

Let h_two to the hypotenuse of the outer triangle:

sin 30º = 85 ÷ h_two  

h_two = 85 ÷ sin 30º = 170

For this triangle the perimeter is:

P = 85 + [(85 / √3) + 98.15] + 170 = 402.22

Perimeter of non-right triangle

We already know all its sides of this triangle:

P = x + h₁ + h_two = 98.15 + 98.15 + 170 = 366.3

Applications of trigonometric ratios

Trigonometric ratios have numerous practical applications, for example heights can be calculated.

Suppose a water tower is 325 feet from a building. An observer at a window notices that the elevation angle of the upper end of the tower is 39º, while the angle of depression with which the base of the tower is viewed is 25º. He wonders:

a) What is the height of the tower?

b) How high is the window?

Solution to

From the leg opposite 39º of the upper triangle we obtain a part of the answer:

h₁/ 325 = tg 39º → h₁ = 325. tg 39º feet = 263.2 feet

In a similar way we obtain the remainder of the height of the tower, called h_two starting from the lower triangle:

h_two/ 325 = tg 25º → h_two = 325. tg 25º feet = 151.6 feet

The total height of the tower is h₁ + h_two = 263.2 + 151.6 feet = 414.7 feet.

Solution b

The window is precisely at a height h_two ground:

h_two = 151.6 feet.

References

1. Carena, M. 2019. Pre-University Mathematics Manual. National University of the Litoral.
2. Hoffman, J. Selection of Mathematics Topics. Volume 3.
3. Jiménez, R. 2008. Algebra. Prentice hall.
4. Stewart, J. 2006. Precalculus: Mathematics for Calculus. 5th. Edition. Cengage Learning.
5. Zill, D. 1984. Algebra and Trigonometry. Mcgraw hill.