Lesson Explainer: Inscribed Angles in a Circle | Nagwa Lesson Explainer: Inscribed Angles in a Circle | Nagwa

Lesson Explainer: Inscribed Angles in a Circle Mathematics • Third Year of Preparatory School

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In this explainer, we will learn how to identify theorems of finding the measure of an inscribed angle with respect to its subtended arc or central angle subtended by the same arc and the measures of inscribed angles in a semicircle.

Let us first define an inscribed angle.

Definition: Inscribed Angle

An inscribed angle is the interior angle between two chords intersecting on a circle’s circumference.

We are now going to prove an important relationship between the measure of an inscribed angle and the measure of the central angle subtended by the same arc. Note that the measure of the central angle subtended by the same arc is, per definition, the same as the measure of the subtended arc of the inscribed angle.

Let us first consider the case where the center of the circle, 𝑀, belongs to one of the sides of the inscribed angle.

Inscribed angle ∠𝐴𝐢𝐡 and central angle βˆ π΄π‘€π΅ are subtended by the same arc, 𝐴𝐡. It is worth noting that the arc 𝐴𝐡 that is said to subtend angle ∠𝐴𝐢𝐡 is the arc that does not contain point 𝐢 (it is the arc in red in the diagram).

As two sides of the triangle 𝐴𝐢𝑀 are two radii of the circle, it is an isosceles triangle. This means that, in triangle 𝐴𝐢𝑀, π‘šβˆ π΄πΆπ΅=π‘šβˆ π΄.

Hence, as the angles in a triangle add to 180∘, we have

π‘šβˆ πΆπ‘€π΄=180βˆ’2π‘šβˆ π΄πΆπ΅.∘(1)

Angles βˆ πΆπ‘€π΄ and βˆ π΄π‘€π΅ are on a straight line; hence, we have

π‘šβˆ π΄π‘€π΅+π‘šβˆ πΆπ‘€π΄=180π‘šβˆ πΆπ‘€π΄=180βˆ’π‘šβˆ π΄π‘€π΅.∘∘(2)

Equations (1) and (2) lead to π‘šβˆ π΄π‘€π΅=2π‘šβˆ π΄πΆπ΅; that is, π‘šβˆ π΄πΆπ΅=12π‘šβˆ π΄π‘€π΅.

We now consider another situation for an inscribed angle and a central angle subtended by the same arc, namely, when the circle center, 𝑀, is a point inside the inscribed angle.

We can use our previous result for an inscribed angle that has a side containing the circle center by splitting the inscribed angle ∠𝐴𝐢𝐡 into two inscribed angles, ∠𝐴𝐢𝐷 and ∠𝐷𝐢𝐡, that have a side containing the circle center (as 𝐢𝐷 is a diameter of the circle).

We have π‘šβˆ π΄π‘€π·=2π‘šβˆ π΄πΆπ· and π‘šβˆ π·π‘€π΅=2π‘šβˆ π·πΆπ΅.

Since π‘šβˆ π΄π‘€π΅=π‘šβˆ π΄π‘€π·+π‘šβˆ π·π‘€π΅ and π‘šβˆ π΄πΆπ΅=π‘šβˆ π΄πΆπ·+π‘šβˆ π·πΆπ΅, we find that π‘šβˆ π΄π‘€π΅=2π‘šβˆ π΄πΆπ·+2π‘šβˆ π·πΆπ΅π‘šβˆ π΄π‘€π΅=2(π‘šβˆ π΄πΆπ·+π‘šβˆ π·πΆπ΅)π‘šβˆ π΄π‘€π΅=2π‘šβˆ π΄πΆπ΅.

So, π‘šβˆ π΄πΆπ΅=12π‘šβˆ π΄π‘€π΅.

Finally, let us consider the third situation, that is, when the circle center, 𝑀, is outside the inscribed angle.

As for the previous situation, we consider the two inscribed angles that have a side containing the circle center, ∠𝐴𝐢𝐷 and ∠𝐡𝐢𝐷, where 𝐢𝐷 is a diameter of the circle.

Since π‘šβˆ π΄π‘€π΅=π‘šβˆ π΄π‘€π·βˆ’π‘šβˆ π΅π‘€π· and π‘šβˆ π΄πΆπ΅=π‘šβˆ π΄πΆπ·βˆ’π‘šβˆ π΅πΆπ·, we find that π‘šβˆ π΄π‘€π΅=2π‘šβˆ π΄πΆπ·βˆ’2π‘šβˆ π΅πΆπ·π‘šβˆ π΄π‘€π΅=2(π‘šβˆ π΄πΆπ·βˆ’π‘šβˆ π΅πΆπ·)π‘šβˆ π΄π‘€π΅=2π‘šβˆ π΄πΆπ΅.

So, π‘šβˆ π΄πΆπ΅=12π‘šβˆ π΄π‘€π΅.

We found the same result in the three possible positions for the circle center, 𝑀, with respect to the inscribed angle: (i) on one side of the inscribed angle, (ii) inside the inscribed angle, and (iii) outside the inscribed angle.

Remember that the measure of a central angle subtended by an arc is the same as the measure of this arc.

Theorem: Inscribed Angle Theorem

The measure of an inscribed angle subtended by an arc is half the measure of this arc, that is, half the measure of the central angle subtended by this arc.

Let us now see with our first example how to use this theorem to find the measure of an inscribed angle.

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

Find π‘šβˆ π΄πΆπ·.

Answer

Let us call 𝑀 the center of the circle. It is the intersection point of 𝐢𝐷 and 𝐴𝐡.

Angle ∠𝐴𝐢𝐷 is an inscribed angle because points 𝐴, 𝐢, and 𝐷 are on the circle. βˆ πΆπ‘€π΅ and βˆ π΄π‘€π· are vertically opposite angles; therefore, they have the same measure, 72∘. βˆ π΄π‘€π· is the central angle subtended by the same arc as ∠𝐴𝐢𝐷. The inscribed angle theorem states that the measure of an inscribed angle subtended by an arc is half the measure of the central angle subtended by this arc.

Hence, we have π‘šβˆ π΄πΆπ·=12π‘šβˆ π΄π‘€π·π‘šβˆ π΄πΆπ·=12Γ—72π‘šβˆ π΄πΆπ·=36.∘

Let us look at an example involving solving linear equations.

Example 2: Finding the Measure of an Inscribed Angle given Its Arc’s Measure by Solving Two Linear Equations

From the figure, what is π‘₯?

Answer

In the circle of center 𝑀, ∠𝐴𝐢𝐡 is an inscribed angle because points 𝐴, 𝐢, and 𝐡 are on the circle. The central angle subtended by the same arc (major arc 𝐴𝐡) has a measure of (2π‘₯+8)∘. The inscribed angle theorem states that the measure of an inscribed angle subtended by an arc is half the measure of the central angle subtended by this arc.

Hence, we have π‘šβˆ π΄πΆπ΅=12π‘šβˆ π΄π‘€π΅101=12(2π‘₯+8)101=π‘₯+4101βˆ’4=π‘₯+4βˆ’4π‘₯=97.∘∘

Let us look now at an example involving the measure of an arc and solving a linear equation.

Example 3: Solving Equations Using the Measure of an Inscribed Angle given Its Arc’s Measure

Given that π‘šβˆ π΅π΄πΆ=(π‘₯+15)∘, find π‘₯.

Answer

Angle ∠𝐡𝐴𝐢 is an inscribed angle subtended by the arc 𝐡𝐢 of measure 118∘.

The inscribed angle theorem states that the measure of an inscribed angle subtended by an arc is half the measure of this arc. Therefore, we have π‘šβˆ π΅π΄πΆ=12π‘šπ΅πΆπ‘šβˆ π΅π΄πΆ=12Γ—118=59.∘∘

In addition, we are told that π‘šβˆ π΅π΄πΆ=(π‘₯+15)∘; hence, π‘₯+15=59π‘₯=59βˆ’15π‘₯=44.

In the next example, we are going to solve a multistep problem where we are given the measure of an arc.

Example 4: Finding the Measure of an Inscribed Angle Using Its Arc’s Measure

Find π‘šβˆ π·πΆπ΅.

Answer

Angle ∠𝐴𝐷𝐡 is an inscribed angle subtended by the arc 𝐴𝐡. The inscribed angle theorem states that the measure of an inscribed angle subtended by an arc is half the measure of this arc. Therefore, we have π‘šβˆ π΄π·π΅=12π‘šπ΄π΅52=12Γ—π‘šπ΄π΅π‘šπ΄π΅=2Γ—52=104.∘∘∘

Angle ∠𝐷𝐢𝐡 is an inscribed angle subtended by the arc 𝐷𝐡=𝐴𝐡+𝐴𝐷. Hence, π‘šπ·π΅=π‘šπ΄π΅+π‘šπ΄π·π‘šπ·π΅=104+60=164.∘

The inscribed angle theorem states that the measure of an inscribed angle subtended by an arc is half the measure of this arc. Therefore, we have π‘šβˆ π·πΆπ΅=12π‘šπ·π΅π‘šβˆ π·πΆπ΅=12Γ—164=82.∘∘

Let us look at a corollary of the inscribed angle theorem, namely, when the inscribed angle is drawn in a semicircle (which means that the inscribed angle is subtended by an arc of measure 180∘) or, in other words, when the central angle is a straight angle (the central angle βˆ π΄π‘€π΅ has a measure of 180∘).

Applying the inscribed angle theorem gives us π‘šβˆ π΄πΆπ΅=12Γ—180=90.∘∘

Corollary: Inscribed Angle in a Semicircle

An inscribed angle drawn in a semicircle is a right angle.

Let us now solve a system of linear equations to find the measure of an inscribed angle in a semicircle.

Example 5: Finding the Measure of an Inscribed Angle in a Semicircle

Given that π‘šβˆ πΆπ΄π΅=31∘, find 𝑦 and π‘₯.

Answer

The inscribed angle ∠𝐡𝐢𝐴 is drawn in a semicircle since 𝐴𝐡 is a diameter of the circle. An inscribed angle drawn in a semicircle is a right angle. Therefore, we have π‘šβˆ π΅πΆπ΄=𝑦=90.∘∘

In addition, the sum of the angles in a triangle is 180∘, which gives π‘₯+𝑦+31=180.

Substituting the value we have found for 𝑦 into this equation gives π‘₯+90+31=180π‘₯+121=180π‘₯=180βˆ’121=59.

We have found that 𝑦=90π‘₯=59.and

In our last example, we solve a problem involving an inscribed angle drawn in a semicircle and solving an equation.

Example 6: Solving Equations Using the Measure of an Inscribed Angle in a Semicircle

Given that π‘šβˆ π΄π΅πΆ=(6π‘₯+15)∘ and π‘šβˆ πΆπ΄π΅=(11π‘₯βˆ’10)∘, find the value of π‘₯.

Answer

The inscribed angle ∠𝐡𝐢𝐴 is drawn in a semicircle since 𝐴𝐡 is a diameter of the circle. An inscribed angle drawn in a semicircle is a right angle. Therefore, we have π‘šβˆ π΅πΆπ΄=90.∘

As the angles in a triangle add to 180∘, we find, considering triangle 𝐴𝐡𝐢, that π‘šβˆ πΆπ΄π΅+π‘šβˆ π΄π΅πΆ+90=180π‘šβˆ πΆπ΄π΅+π‘šβˆ π΄π΅πΆ=9011π‘₯βˆ’10+6π‘₯+15=9017π‘₯+5=9017π‘₯=90βˆ’5=85π‘₯=85Γ·17π‘₯=5.∘∘∘

Let us summarize the key points of this explainer.

Key Points

  • An inscribed angle is an angle whose vertex lies on the circle and whose sides contain two chords of the circle.
  • The inscribed angle theorem states that the measure of an inscribed angle subtended by an arc is half the measure of this arc, that is, half the measure of the central angle subtended by this arc.
  • A corollary to the inscribed angle theorem is that an inscribed angle drawn in a semicircle is a right angle.

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