Lesson Explainer: Angles of Tangency | Nagwa Lesson Explainer: Angles of Tangency | Nagwa

Lesson Explainer: Angles of Tangency Mathematics

In this explainer, we will learn how to identify the angle of tangency in a circle and find its measure using the measure of its subtended arc, inscribed angle, or central angle subtended by the same arc.

Let us begin by recalling that a tangent line to a circle is a line that intersects the circle at just one point, as shown below.

We note that the line segment from the point of intersection 𝐴 to the center of the circle 𝑀 is a radius of the circle. Furthermore, this radius is perpendicular (i.e., at 90 degrees) to the tangent line.

In this explainer, we want to discuss angles of tangency. Consider a tangent to a circle that meets with a chord of the circle (i.e., a line segment on the inside of the circle) at the point 𝐴.

The angle 𝜃 between the chord and the tangent is known as an angle of tangency. Calculating this angle can be done with the help of several theorems and observations that we will discuss over the course of this explainer.

It will also be important to keep in mind some of the properties of triangles. Isosceles and equilateral triangles are triangles that have two or three equal sides, respectively, as shown below.

It is important to note that because the radii of a circle all have the same length, two radii can form the sides of an isosceles triangle as shown below.

This can be useful to know, since it tells us that the measures of the angles at 𝐴 and 𝐵 are equal.

Additionally, let us review the inscribed angle theorem (also called the central angle theorem), which is crucial for upcoming calculations that deal with angles within a circle.

Recall that two points on a circle, 𝐴 and 𝐵, divide a circle into two arcs: a major arc and a minor arc (when the arcs are the same length, they divide the circle into two semicircular arcs). We can also form an inscribed angle with any point 𝐶 on the major arc as shown.

Note that sometimes, the major arc is referred to as the subtended arc and the minor arc the intercepted arc. Then, we have the following theorem.

Theorem: Inscribed Angle Theorem

Let 𝐴 and 𝐵 be two points on a circle, 𝑀 be the center of the circle, and 𝐶 be any point on the major arc. Then, the measure of the central angle𝐴𝑀𝐵 is double the measure of the inscribed angle𝐴𝐶𝐵, as shown.

Another way to phrase this theorem is that the measure of the central angle subtended by two points on a circle is twice the measure of the inscribed angle subtended by those points.

Having recalled this theorem, let us learn a new theorem that deals with angles of tangency in a circle.

Theorem: Alternate Segment Theorem

Let 𝐴 and 𝐵 be two points on a circle and 𝐶 be the point where a tangent (passing through 𝐸, 𝐶, and 𝐷) intersects the circle. Then, the angles of tangency 𝐴𝐶𝐷 and 𝐵𝐶𝐸 are equal to the angles in the alternate segments 𝐴𝐵𝐶 and 𝐵𝐴𝐶 respectively. This is shown below.

Let us prove this theorem. To start with, recall that a tangent of the circle at point 𝐶 forms a right angle with the radius 𝑀𝐶 at the point of intersection.

Now, we begin by considering one of the angles of tangency 𝐴𝐶𝐷 that we label as 𝜃 below.

Because the tangent and the radius form a right angle, we know that 𝑚𝐴𝐶𝐷+𝑚𝐴𝐶𝑀=90. Therefore, 𝑚𝐴𝐶𝑀=90𝜃. We also know that the inner triangle 𝐴𝐶𝑀 is an isosceles triangle because two of its sides are radii and are therefore equal. So, 𝑚𝐶𝐴𝑀=90𝜃 as well.

Since a triangle always has all of its angles adding up to 180, we have (90𝜃)+(90𝜃)+𝑚𝐴𝑀𝐶=180.

Rearranging, we have 𝑚𝐴𝑀𝐶=180(90𝜃)(90𝜃)=2𝜃.

This gives us that the measure of the central angle between 𝐴 and 𝐶 is 2𝜃. Finally, using the inscribed angle theorem, we can conclude that the measure of 𝐴𝐵𝐶 is half the central angle 2𝜃, since it is the inscribed angle of 𝐴 and 𝐶. Thus, 𝑚𝐴𝐵𝐶=𝜃.

We note that the same method applies to the other angle of tangency 𝐵𝐶𝐸, since we can just repeat the process with the chord 𝐵𝐶. Thus, we have proved the alternate segment theorem.

Let us see an example where we can put this theorem directly to use.

Example 1: Finding the Measure of an Angle of Tangency given the Measure of the Inscribed Angle Subtended by the Same Arc

Given that 𝐵𝐶 is a tangent to the circle, find 𝑚𝐴𝐵𝐶.

Answer

Let us first of all mark the angle we want to find on the diagram with 𝜃.

For any question where we need to find the angle of tangency, we need to ask ourselves whether we can use the alternate segment theorem to help us.

Now, we know that 𝐵𝐶 is a tangent to the circle at 𝐵, that 𝐴, 𝐵, and 𝐷 are three points on the circle, and that the angle we want to find is the angle of tangency 𝐴𝐵𝐶. Therefore, we can use the alternate segment theorem to find this angle. That is, 𝑚𝐴𝐵𝐶=𝑚𝐴𝐷𝐵𝑚𝐴𝐵𝐶=78.

Since 𝐴𝐷𝐵 is the angle in the alternate segment to 𝐴𝐵𝐶, we directly use the theorem to find that 𝑚𝐴𝐵𝐶=78.

We have seen how the alternate segment theorem can be used directly to find inscribed angles given the angle of tangency and vice versa, but we can also use other aspects of the theorem to help solve different problems. For instance, let us consider how angles of tangency relate to central angles.

Corollary: Angles of Tangency and Central Angles

Let 𝐴 be a point on a circle of center 𝑀 and 𝐵 be the point where a tangent (passing through 𝐵 and 𝐶) intersects the circle. Then, the angle of tangency 𝐴𝐵𝐶 is half of the central angle 𝐴𝑀𝐵. This is shown below.

Another way to phrase this is that the angle of tangency is half the central angle subtended by the same arc (i.e., the arc 𝐴𝐵).

Note that as this corollary is something we demonstrated during the proof of the alternate segment theorem, we therefore do not need to prove it again. Let us consider an example where we can use this theorem directly to find the angle of tangency.

Example 2: Finding the Measure of an Angle of Tangency given the Measure of the Central Angle Subtended by the Same Arc

Find 𝑚𝐵𝐴𝐶.

Answer

Let us begin by marking the angle we have been asked to find on the diagram:

We can see that we have been asked to find the measure of an angle of tangency 𝐵𝐴𝐶. Note that the angle 𝐴𝑀𝐵 on the diagram is a right angle because it is marked with a square, thus telling us its measure is 90. We also note that 𝐴𝑀𝐵 is the central angle subtended by the same arc (i.e., 𝐴𝐵) as the angle of tangency.

Recall that the measure of the angle of tangency 𝑚𝐵𝐴𝐶 is equal to half the measure of the central angle subtended by the same arc, which is 𝑚𝐴𝑀𝐵. Thus, we have 𝑚𝐵𝐴𝐶=12𝑚𝐴𝑀𝐵=12(90)=45.

So far, we have seen how to calculate an angle of tangency using a central angle and using an inscribed angle. We can also calculate angles of tangency using the measure of an arc.

Recall that the measure of an arc is the angle that the arc makes at the center of the circle. For instance, consider the diagram below.

Here, we can see that the measure of the major arc of the circle created by the two points 𝐴 and 𝐵 (denoted by 𝜃 on the outside of the circle) is equivalent to the measure of the angle between the radii formed by 𝐴 and 𝐵 in the inside of the circle.

Recall that we already have a corollary that relates the central angle of a circle to the angle of tangency. By using the equivalency between the central angle and the arc measure, we can extend this corollary to apply to the arc of a circle. This equivalency can be seen in the following diagram.

In other words, since both the measure of the central angle and the arc measure are 2𝜃, they are both twice the angle of tangency. Thus, we have the following corollary.

Corollary: Angles of Tangency and Arc Measures

Let 𝐴 be a point on a circle and 𝐵 be the point where a tangent (passing through 𝐵 and 𝐶) intersects the circle. Then, the angle of tangency 𝐴𝐵𝐶 is half of the measure of the arc 𝐴𝐵 formed on the same side. This is shown below.

It is important to realize which arc 𝐴𝐵 refers to, since there are two possibilities: the major arc and the minor arc (also known as the subtended and intercepted arcs). In the diagram above, the minor arc was used, since it is on the same side as the angle of tangency. However, the opposite would be true if we considered an obtuse angle, as shown below.

Here, 𝐴𝐵 is now a major arc of the circle. It is also important to note that if we are given the opposite arc to the one we need, we can make use of the fact that the measures of the two arcs of a circle add up to 360. So, it is always possible to find the major arc if we are given the minor arc, or vice versa.

Let us consider an example where we can use this theorem to calculate an angle of tangency using the measure of an arc.

Example 3: Finding the Measure of an Angle of Tangency Using Arc Measure

Given that 𝐵𝐶 is a tangent to the circle below, find 𝑚𝐴𝐵𝐶.

Answer

Let us begin by marking the angle we need to find on the diagram.

Recall that the angle of tangency 𝐴𝐵𝐶 is half of the measure of the arc 𝐴𝐵 formed on the same side. In this example, we can see that the arc measure we have been given (i.e., 190) is not on the same side. However, we can find the correct arc measure by using the fact that the measures of the two arcs of a circle have to sum to 360. Thus, the correct arc measure is equal to 360190=170.

Let us mark this on the diagram.

Now, we can use the fact that the angle of tangency is half the measure of the arc on the same side to get 𝐴𝐵𝐶=12170=85.

One other type of question involving angles of tangency has a point outside of the circle that goes through two different tangent lines. Let us consider an example of this.

Example 4: Finding the Measures of Two Inscribed Angles given the Measures of Angles of Tangency Subtended by the Same Arcs to Find Other Unknown Angles

Given that 𝑚𝐸𝐶𝐷=54 and 𝑚𝐹𝐵𝐷=78, find 𝑥 and 𝑦.

Answer

Let us begin by putting the information we have been given, 𝑚𝐸𝐶𝐷=54 and 𝑚𝐹𝐵𝐷=78, into the diagram.

To start with, let us try to find 𝑥. As this is a question involving angles of tangency, we ask ourselves whether the alternate segment theorem can be used to help us. We can see that as there are two angles of tangency and three points on the circle, we can use the theorem twice in the inner triangle to find missing angles. This is shown below.

Now, we notice that we have two of the three angles of the inner triangle. Since the angles of a triangle add up to 180, we have 𝑥+78+54=180𝑥=48.

Having found 𝑥, we now want to find 𝑦, which means finding the angles of the triangle that 𝑦 is in. To find the other angles, we can make use of the fact that the angles of a straight line have to add up to 180. Thus, using the exact calculation as before, we find that the remaining angles must be 48 as well, giving us the following.

Recalling that two tangent segments meeting at a point have the same length and thus form an isosceles triangle when connected by a chord, we can confirm that these two angles being equal is consistent with this rule. Finally, we can calculate 𝑦 using the sum of angles in a triangle. 48+48+𝑦=180𝑦=84.

So, all in all, we have 𝑥=48 and 𝑦=84.

Up until now, we have used the alternate segment theorem and variations of it to find the angle that a tangent makes with a chord in a circle. Just as it is possible to use the theorem to find angles in this way, we can also use the converse to prove that a given ray or segment is a tangent to the circle if the corresponding angles match up. Formally, we have the following corollary.

Corollary: Converse of Alternate Segment Theorem

If a ray or line segment meets with a chord of a circle on the outside of the circle and the angle it makes with the chord is equal in measure to the angle in an alternate segment of the circle, then that ray or line segment must be a tangent to the circle.

If the angles are not equal, then that ray or line segment is not tangent to the circle.

Let us illustrate what this means specifically. Suppose we have a ray 𝐶𝐷 and we are given the angle that it makes with the chord 𝐴𝐶. Then, we have two possibilities, as shown below.

In the first case, the measure of 𝐴𝐶𝐷 is equal to that of the measure of the angle in the alternate segment 𝐴𝐵𝐶; consequently, 𝐶𝐷 must be a tangent. In the second case, the angles are not equal, since 𝑚𝐴𝐶𝐷>𝑚𝐴𝐵𝐶; hence, 𝐶𝐷 cannot be a tangent.

One particular situation we may have to use this corollary is with circumcircles. Recall that if we are given a triangle, there exists exactly one circle that passes through all the vertices of the triangle. We call this circle a circumcircle.

Sometimes, we may be asked questions regarding tangents to the circumcircle of a triangle. For instance, we may have to verify that a ray is a tangent to a circumcircle by measuring its angle with a chord and confirming that it satisfies the properties of an angle of tangency. Let us explore this idea in the following example.

Example 5: Finding the Angle of Tangency given That a Line Is Tangent to the Circumcircle of a Triangle

In the given figure, if 𝐶𝐷𝐷𝐴, which of the following is a tangent to the circle that passes through the vertices of the triangle 𝐴𝐵𝐸?

  1. 𝐸𝐷
  2. 𝐸𝐶
  3. 𝐴𝐷
  4. 𝐵𝑌
  5. 𝐵𝐶

Answer

To best understand this question, let us annotate on the diagram the information we are being asked to find. Specifically, we need to consider the circle that passes through the vertices of the triangle 𝐴𝐵𝐸 (i.e., a circumcircle). Let us highlight this triangle.

Although we have not yet drawn a circumcircle around 𝐴𝐵𝐸, we can immediately see that 𝐸𝐷 and 𝐸𝐶 cannot be tangential to the circle, since they are extensions of the sides of the triangle (in other words, they are secants). It remains to be seen whether 𝐴𝐷, 𝐵𝑌, or 𝐵𝐶 is tangent to the circle.

In order to verify these other options, we need to investigate the surrounding angles and find whether it makes sense for them to be tangents. In particular, we can use the converse of the alternate segment theorem to prove in each case whether they are a tangent or not.

Let us begin by considering 𝐴𝐷. Recall that in the problem description, it has been given to us that 𝐶𝐷𝐷𝐴. In particular, this means that the corresponding arcs are not equal in length. Let us consider the arcs 𝐷𝐴 and 𝐶𝐷. We note that the angle 𝐴𝐵𝐸 is an inscribed angle of 𝐷𝐴 and 𝐷𝐴𝐸 is an inscribed angle of 𝐶𝐷. We highlight these arcs and the corresponding inscribed angles below.

Recall that the measure of an inscribed angle subtended by an arc is half the measure of the arc. Since arcs 𝐷𝐴 and 𝐶𝐷 have different lengths, this means 𝑚𝐴𝐵𝐸 and 𝑚𝐷𝐴𝐸 are not equal.

Now, let us recall the converse of the alternate segment theorem: if the angle between 𝐴𝐷 and 𝐴𝐸 (which is a chord of the circumcircle) is not equal to the angle in the alternate segment (i.e., 𝐴𝐵𝐸), then 𝐴𝐷 cannot be a tangent to this circle. Thus, 𝐴𝐷 is not a tangent.

Next, let us look at 𝐵𝑌. Consider the triangle 𝐸𝐵𝐶. We can see that 𝐴𝐸𝐵 is an exterior angle to this triangle, which means that it is equal to the sum of the remote interior angles (by the exterior angle theorem). In other words,

𝑚𝐴𝐸𝐵=𝑚𝐴𝐶𝐵+𝑚𝐶𝐵𝐸.(1)

We highlight this below.

Now, since 𝐵𝑌 is a tangent to the larger circle, we can use the alternate segment theorem on it. In particular, if we consider the triangle 𝐴𝐵𝐶, we can see that

𝑚𝐴𝐶𝐵=𝑚𝐴𝐵𝑌.(2)

We indicate this on the diagram too.

However, now we can use the converse of the alternate segment theorem in the triangle 𝐴𝐵𝐸: substituting equation (2) into (1), we have 𝑚𝐴𝐸𝐵=𝑚𝐴𝐵𝑌+𝑚𝐶𝐵𝐸.

Since 𝑚𝐶𝐵𝐸>0, this means that 𝑚𝐴𝐸𝐵>𝑚𝐴𝐵𝑌.

So, the two angles cannot be equal, which means 𝐵𝑌 cannot be tangent to the circle passing through triangle 𝐴𝐸𝐵.

Finally, let us consider 𝐵𝐶. To begin with, let us use the information given in the question. We can see that 𝑋𝐶 and 𝐵𝐸 are parallel lines, since they have been marked with double arrows. This means that the alternate angles 𝑋𝐶𝐵 and 𝐶𝐵𝐸 must be equal. We highlight this below.

Now, we can see that 𝑋𝐶 is a tangent to the larger circle and that there are multiple triangles inside this circle, so we can use the alternate segment theorem here. In particular, consider the triangle 𝐴𝐵𝐶. We can see that the angle of tangency 𝐵𝐶𝑋 is equal to the angle in the alternate segment, 𝐵𝐴𝐸, as shown.

Now, returning to the triangle 𝐴𝐵𝐸, we sketch the circumcircle going round it and highlight the angles 𝐶𝐵𝐸 and 𝐵𝐴𝐸:

Let us use the converse of the alternate segment theorem once more. Specifically, since the angles 𝐶𝐵𝐸 and 𝐵𝐴𝐸 have been shown to be equal, it must therefore be the case that 𝐵𝐶 is a tangent to the circle.

Thus, the answer is E: 𝐵𝐶.

Let us finish off by recapping the alternate segment theorem and the places we can use it.

Key Points

  • Let 𝐴 and 𝐵 be two points on a circle and 𝐶 be the point where a tangent (passing through 𝐸, 𝐶, and 𝐷) intersects the circle. Then, the angles of tangency 𝐴𝐶𝐷 and 𝐵𝐶𝐸 are equal to the angles in the alternate segments 𝐴𝐵𝐶 and 𝐵𝐴𝐶 respectively.
  • As an extension of the above, for a circle of center 𝑀, the angle of tangency 𝐴𝐵𝐶 is half of the central angle 𝐴𝑀𝐵.
  • Additionally, the angle of tangency 𝐴𝐵𝐶 is half of the measure of the arc 𝐴𝐵 formed on the same side.
  • Conversely, if a ray or line segment meets with a chord of a circle on the outside of the circle and the angle it makes with the chord is equal in measure to the angle in an alternate segment of the circle, then that ray or line segment must be a tangent to the circle.
    On the other hand, if the angles are not equal, then that ray or line segment is not tangent to the circle.

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