COMPLEMENTARY ANGLES: WHICH ONES AND HOW THEY ARE CALCULATED, EXAMPLES, EXERCISES - MATHS - 2024

Two or more angles are complementary angles if the sum of their measures corresponds to that of a right angle. As is known, the measure of a right angle in degrees is 90º, and in radians it is π / 2.

For example, the two angles adjacent to the hypotenuse of a right triangle are complementary to each other, since the sum of their measures is 90º. The following figure is very illustrative in this regard:

Figure 1. On the left, several angles with a common vertex. To the right, an angle of 60º that complements the angle α (alpha). Source: F. Zapata.

A total of four angles are shown in figure 1. α and β are complementary since they are adjacent and their sum completes a right angle. Similarly β is complementary to γ, from which it follows that γ and α are of equal measure.

Now, since the sum of α and δ is equal to 90 degrees, it can be stated that α and δ are complementary. Furthermore, since β and δ have the same complementary α, it can be said that β and δ have the same measure.

Examples of complementary angles

The following examples ask to find the unknown angles, marked with question marks in figure 2.

Figure 2. Various examples of complementary angles. Source: F. Zapata.

- Examples A, B and C

The following examples are in order of complexity.

Example A

In the figure above we have that the adjacent angles α and 40º add up to a right angle. That is, α + 40º = 90º, therefore α = 90º- 40º = 50º.

Example B

Since β is complementary to the angle of 35º, then β = 90º - 35º = 55º.

Example C

From figure 2C we have that the sum of γ + 15º + 15º = 90º. In other words, γ is complementary to the angle 30º = 15º + 15º. So that:

γ = 90º- 30º = 60º

- Examples D, E and F

In these examples there are more angles involved. To find the unknowns, the reader must apply the concept of complementary angle as many times as necessary.

Example D

Since X is complementary to 72º, it follows that X = 90º - 72º = 18º. Furthermore Y is complementary to X, so Y = 90º - 18º = 72º.

Finally Z is complementary with Y. From all the above it follows that:

Z = 90º - 72º = 18º

Example E

The angles δ and 2δ are complementary, therefore δ + 2δ = 90º.

That is, 3δ = 90º, which implies that δ = 90º / 3 = 30º.

Example F

If we call the angle between que and 10º U, then U is supplementary to them, because it is observed that their sum completes a right angle. From which it follows that U = 80º. Since U is complementary to ω, then ω = 10º.

Exercises

Three exercises are proposed below. In all of them the value of angles A and B in degrees must be found, so that the relationships shown in figure 3 are fulfilled.

Figure 3. Illustrations for complementary angle exercises. Source: F. Zapata.

- Exercise 1

Determine the values ​​of angles A and B from part I) of Figure 3.

Solution

From the figure shown it can be seen that A and B are complementary, therefore A + B = 90º. We substitute the expression for A and B as a function of x given in part I):

(x / 2 + 7) + (2x + 15) = 90

The terms are then grouped appropriately and a simple linear equation is obtained:

(5x / 2) + 22 = 90

Subtracting 22 in both members we have:

5x / 2 = 90 -22 = 68

And finally the value of x is cleared:

x = 2 * 68/5 = 136/5

Now the angle A is found by substituting the value of X:

A = (136/5) / 2 +7 = 103/5 = 20.6 º.

While angle B is:

B = 2 * 136/5 + 15 = 347 / 5th = 69.4º.

- Exercise 2

Find the values ​​of the angles A and B of image II, figure 3.

Solution

Again, since A and B are complementary angles, it follows that: A + B = 90º. Substituting the expression for A and B as a function of x given in part II) of figure 3, we have:

(2x - 10) + (4x +40) = 90

Like terms are grouped together to obtain the equation:

6 x + 30 = 90

Dividing both members by 6 you get:

x + 5 = 15

From which it follows that x = 10º.

Thus:

A = 2 * 10 - 10 = 10º

B = 4 * 10 + 40 = 80º.

- Exercise 3

Determine the values ​​of angles A and B from part III) of Figure 3.

Solution

Again the figure is carefully analyzed to find the complementary angles. In this case we have that A + B = 90 degrees. Substituting the expression for A and B as a function of x given in the figure, we have:

(-x +45) + (4x -15) = 90

3 x + 30 = 90

Dividing both members by 3 results in the following:

x + 10 = 30

From which it follows that x = 20º.

In other words, the angle A = -20 +45 = 25º. And for its part: B = 4 * 20 -15 = 65º.

Perpendicular side angles

Two angles are said to have perpendicular sides if each side has a corresponding perpendicular on the other. The following figure clarifies the concept:

Figure 4. Angles of perpendicular sides. Source: F. Zapata.

In figure 4 the angles α and θ are observed, for example. Now notice that each angle has its corresponding perpendicular at the other angle.

It is also seen that α and θ have the same complementary angle z, therefore the observer immediately concludes that α and θ have the same measure. It seems then that if two angles have sides perpendicular to each other, they are equal, but let's look at another case.

Now consider the angles α and ω. These two angles also have corresponding perpendicular sides, however they cannot be said to be of equal measure, since one is acute and the other is obtuse.

Note that ω + θ = 180º. Furthermore θ = α. If you substitute this expression for z in the first equation you get:

δ + α = 180º, where δ and α are mutually perpendicular angles of sides.

General rule for angles of perpendicular sides

1. Baldor, JA 1973. Plane and space geometry. Central American Cultural.
2. Mathematical laws and formulas. Angle measurement systems. Recovered from: ingemecanica.com.
3. Wentworth, G. Plane Geometry. Recovered from: gutenberg.org.
4. Wikipedia. Complementary angles. Recovered from: es.wikipedia.com
5. Wikipedia. Conveyor. Recovered from: es.wikipedia.com
6. Zapata F. Goniómetro: history, parts, operation. Recovered from: lifeder.com