Vertically Opposite Angles

Opposite angles are not the only types of angles that we can find in geometric problems. Even the example seen in the introduction actually includes several other types of angles that we will mention briefly below:

Corresponding angles are a pair of angles that can be found on the same side of a transversal intersecting two parallel lines. These angles are on the same side of the parallel lines to which they belong. Corresponding angles are equal.

If you wish to read a more extensive explanation, you can consult the article on the subject "Corresponding angles".

Alternate angles are a pair of angles that can be found on opposite sides of a transversal intersecting two parallel lines. These angles are on the opposite sides of the parallel lines to which they belong. Alternate angles are equal.

For a more detailed explanation, please refer to the article "Alternate angles".

Collateral angles are a pair of angles that can be found on the same side of a transversal intersecting two parallel lines. These angles are on the opposite side of the parallel lines to which they belong. The sum of the collateral angles equals $180º$.

For a more detailed explanation, please refer to the article "Collateral angles".

In each of the following diagrams, determine if the angles are opposite angles, and if not, specify the type of angle.

Scheme1:

Scheme 2:

Scheme 3:

Solution:

Scheme 1:

In this scheme if we are dealing with opposite angles. The reason is that they meet the criteria, that is to say, two straight lines intersect and the opposite angles are formed at the point of intersection, one in front of the other.

Scheme 2:

In this scheme we are not dealing with opposite angles, but rather with corresponding angles. The reason is that the two angles are on the same side of the transversal that cuts the two parallel lines. Moreover, these angles are on the same side of the parallel lines to which they belong.

Scheme 3:

In this scheme we are not dealing with opposite angles either, but rather collateral angles. The reason is that the two angles are on the same side of the transversal that cuts the two parallel lines. Moreover, these angles are at the opposite side of the parallel lines to which they belong.

So:

Scheme 1: opposite angles

Scheme 2: corresponding angles

Scheme 3: collateral angles

Given the parallelogram $ABCD$ as illustrated in the diagram.

The point $K$ is the point at which the diagonals meet in the parallelogram. $ABCD$.

The angle $AKD$ measures $30º$.

The angle $KBC$ measures $50º$.

Based on the given information calculate the angle $BCK$.

Solution:

First of all, to help us find the answer, we will label the angles as follows:

We will name the angle $AKD K1$ (measure $30º$)

We will name the angle $BKC K2$

We will name the angle $KBC B1$ (measures $50º$)

We will name the angle $BCK C1$ (the angle we are looking for)

First, we will focus on the triangle $BCK$ since the angle $C1$ is in it.

We will build on what we already know. The sum of the interior angles of a triangle is equal to $180º$. The angle $B1$ measures $50º$ according to the given information. That is, if we can find the measure of the angle $K2$ then we can calculate angle $C1$.

As can be understood from the data and the illustration, the point $K$ is the intersection of the diagonals $AC$ and $BD$ in the parallelogram $ABCD$. We can see that, according to its definition, this intersection forms the opposite angles $K1$ and $K2$. The opposite angles are equal, therefore $k1 = k2 = 30º$.

Now we can go back to the triangle $BCK$ and find angle $C1$:

$C1 = 180º − B1 − K2 = 180º − 50º − 30º = 100º$

That is, the angle $C1$which is in fact the angle we are looking for $BCK$ measures $100º$.

Then:

The angle $BCK$ measures $100º$.

Given the isosceles trapezoid $ABCD$ as illustrated in the diagram.

The point $M$ is the point where the diagonals of the trapezoid meet. $ABCD$.

$MA = MB$

The angle $DMC$ measures $120º$.

Based on the given information, calculate the angles of the triangle. $ABM$.

Solution:

First of all, to help us find the answer, we will label the angles as follows:

We will name the angle $DMC \: M1$ (measure $120º$)

We will name the angle $AMB \: M2$ (one of the angles we are looking for)

We will name the angle $MAB \: A1$ (one of the angles we are looking for)

We will name the angle $ABM \: B1$ (one of the angles we are looking for)

We will focus on triangle ABM since we must find its angles.

We will start with angle $M2$.

As we can see from the data and the illustration, the point $M$ is the intersection of the diagonals $AC$ and $BD$ in the trapezoid $ABCD$. According to its definition, this intersection point forms opposite angles $M1$ and $M2$. Opposite angles are equal, therefore, $M1= M2 = 120º$.

Now we turn to another piece of information we have, the side $MA = MB$. This implies that the triangle $ABM$ is an isosceles triangle. We know that in an isosceles triangle the two base angles are equal.

That is to say,

$A1 = B1$

To calculate the angles, remember that the sum of the interior angles of a triangle is $180º$ and that the measure of $M2$ is already known.

Therefore, that gives us:

$180º = A1 + B1 + M2 = 2*A1 + 120º$

$A1 = (180º - 120º) / 2 = 30º$

Then:

The angle $AMB$ (angle $M2$) measures $120º$.

The angle $MAB$ (angle $A1$) measures $30º$.

The angle $ABM$ (angle $B1$) measures $30º$.

What is the value of $X$ given that the lines are parallel?

Solution:

The marked angles are exterior angles to the parallel lines. We can use our knowledge of opposite angles, which share the same properties.

If we take one of the opposite angles and the angle external to the other parallel line, we see that they are actually supplementary angles and therefore are equal.

That is,

$X + 70º = 2X$

$70º = 2X - X$

$70º = X$

Then:

We find that the value of $X = 70º$.

Knowing that $CE$ is parallel to $AD$, find the value of $X$.

Solution:

We have that the angle $∡BCE = X$ as it is opposite to angle $X$. The same is true for the angle $∡BAD = X - 10$.

Then, due to the parallel lines we get: $∡ABC = ∡BAD + ∡BCE$ that is , $3X - 30 = X + (X - 10)$.

Then solving for the value of $X$ we get:

$X = 20º$

Therefore:

$X = 20º$

What are opposite angles?

They are angles that are formed when two straight lines intersect, one in front of the other just at the point of intersection.

What is the main characteristic of opposite angles?

The main characteristic is that opposite angles are equal.

In an illustration of parallel lines cut by a transversal, which pair of angles have the same property of the opposite angles, that is, that they are equal?

The corresponding angles and the alternate angles.

If you are interested in learning more about other angles, you can visit one of the following articles:

• Angle notation
• Sides, vertices, and angles
• Bisector
• Right angles
• Adjacent angles
• Alternate angles
• Corresponding angles
• Sum of the angles of a triangle
• Sum of the angles of a polygon

On Tutorela you will find a variety of articles about mathematics.