High Impact Tutoring Built By Math Experts

Personalized standards-aligned one-on-one math tutoring for schools and districts

In order to access this I need to be confident with:

Parts of a circle Circle theorems Central angles Segment of a circleHere you will learn about subtended angles of a circle, including, how angles subtend an arc and the relationship between the arc and the angle.

Students first learn about an angle subtended by an arc in geometry where they explore and apply various theorems that are necessary to problem solve as well as standardized testing.

A **subtended angle** of a circle is an angle that is formed by two chords and where the vertex is on the edge of the circle. The angle subtends an arc, meaning that it intercepts the arc.

In circle A, angle FGH is a subtended angle. The angles are made up of chords, the vertices are on the edge of the circle and the angle subtends arc FH.

Letβs look at circle theorems that prove relationships between the angles and arcs they subtend.

Prepare for math tests in your state with these Grade 3 to Grade 8 practice assessments for Common Core and state equivalents. 40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics!

DOWNLOAD FREEPrepare for math tests in your state with these Grade 3 to Grade 8 practice assessments for Common Core and state equivalents. 40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics!

DOWNLOAD FREEInscribed angles are angles that have a vertex on the edge of a circle (or on the circumference of the circle), the sides are made up of chords, and they subtend an arc of a circle. In the figure, angle EDF is an inscribed angle and angle ECF is a central angle. Both angles are subtending minor arc EF.

Inscribed angle theorem: **the inscribed angle is half the measure of the arc it subtends.**

Letβs look at a proof of this theorem.

In the circle below, the center is indicated by the highlighted point. There are two isosceles triangles created by the radii of the circle.

The angles at the center of the circle sum to 360^{\circ} which means that,

s+t+u=360^{\circ}The angles of the triangles sum to 180^{\circ} which means that,

x+x+s=180^{\circ} and y+y+t=180^{\circ}

Using algebra strategies the equations simplify to be:

\begin{aligned} 2x+s&=180 \\\\ s&=180-2x \end{aligned} \begin{aligned} 2y+t&=180\\\\ t&=180-2y \end{aligned}These algebraic equations can be substituted into s+t+u=360^{\circ}

\begin{aligned} (180-2x)+(180-2y)+u&=360 \\\\ 180-2x+180-2y+u&=360 \\\\ 360-2x-2y+u&=360 \\\\ -2x-2y+u&=0 \\\\ u&=2x+2y \\\\ u&=2(x+y) \end{aligned}u represents the central angle which subtends the highlighted arc. The inscribed angle is represented by x+y which is also subtending the same highlighted arc.

The relationship between the central angle (which is also equal to the measure of the subtended arc) and the inscribed angle is that it is two times the measure of it.

u=2(x+y)So, the inscribed angle theorem states that the inscribed angle is half the measure of the arc it subtends.

Inscribed angle LMN and inscribed angle LON both intercept (subtend) minor arc LN.

Letβs say that minor arc LN=40^{\circ}. From the inscribed angle theorem, you can conclude that angle LMN=\cfrac{1}{2} \times 40^{\circ}=20^{\circ} and inscribed angle LON=\cfrac{1}{2} \times 40^{\circ}=20^{\circ}.

Notice how inscribed angle LMN and inscribed angle LON are equal in measure. This will always be the case when the inscribed angles subtend the same arc.

How does this relate to high school math?

**High School Geometry – Circles (HSG-C.A.2):**Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

**High School Geometry – Circles (HSG-C.A.3):**

Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle.

In order to find subtended angles in a circle:

**Use the appropriate circle theorem to find the subtended angle.****Solve for the angle.**

Central angle RTS=121^{\circ}. Find the measure of inscribed angle RWS.

**Use the appropriate circle theorem to find the subtended angle.**

Inscribed angles are half the measure of the subtended arc.

2**Solve for the angle.**

Central angle RTS=121^{\circ} and is subtended by arc RS which is also equal to 121^{\circ}.

In this case, angle RWS= \cfrac{1}{2}\times{121}^{\circ}=60.5^{\circ}

Angle RWS=60.5^{\circ}

In the diagram, \angle{BAC}=70^{\circ}, \angle{ACB}=40^{\circ}. Find \angle{ADC}.

**Use the appropriate circle theorem to find the subtended angle.**

Use the theorem that the inscribed angle is half the measure of the subtended arc.

Also, use the theorem that inscribed angles that subtend the same arc are equal.

**Solve for the angle.**

Since inscribed angle ACB=40^{\circ}, the subtended arc is 2\times{40}^{\circ}=80^{\circ} because the angle is half the measure of the arc.

Multiply the angle measurement by 2 to find the arc measurement.

Arc AB=80^{\circ}

Inscribed angle BAC=70^{\circ}, so the subtended arc, arc BC=2\times{70}=140^{\circ}

Arc BC=140^{\circ}

The angles of a triangle sum to 180^{\circ}. In triangle ABC, two of the angles are known, one is 70^{\circ} and the other is 40^{\circ}.

To find the third angle, subtract those values from 180^{\circ} .

180-(40+70)=70^{\circ}

The third angle of triangle ABC, angle ABC=70^{\circ}. Angle ABC subtends arc AC.

So, arc AC=2\times{70}^{\circ}=140^{\circ}.

Angle ADC subtends the same arc as angle AB.

Therefore, the measure of angle ADC is the same, angle ADC=70^{\circ}.

The diagram shows a circle with center D and \angle{CAD}=42^{\circ}.

Find \angle{ABC}.

**Use the appropriate circle theorem to find the subtended angle.**

Use the theorems that state that central angles are equal in measure to the subtended arcs and that inscribed angles are half the measure of the subtended arcs.

**Solve for the angle.**

In circle D, AD and CD are radii.

So, triangle ADC is an isosceles triangle. If angle CAD=42^{\circ} then angle ACD is also 42^{\circ} because they are the base angles of the isosceles triangle.

Angle ADC is the vertex angle of the isosceles triangle and also a central angle of circle D.

To find the measure, you can subtract the sum of 42^{\circ} and 42^{\circ} from 180^{\circ}.

180-(42+42)=96^{\circ}

Angle ADC=96^{\circ} , so the arc it intercepts also equals 96^{\circ}.

Arc AC=96^{\circ}

Inscribed angle ABC subtends arc AC.

Inscribed angles are half the measure of the arcs they intercept, so angle ABC=\cfrac{1}{2}\times{96}=48

Angle ABC=48^{\circ}

The diagram shows a circle with center C and \angle{ABC}=27^{\circ}.

Find \angle{FAD}.

**Use the appropriate circle theorem to find the subtended angle.**

The inscribed angle is half the measure of the subtended arc.

**Solve for the angle.**

Since angle ABD=27^{\circ} then the subtended arc is double that measure, which is arc AD=2\times{27}=54^{\circ}

Chord AD is the diameter of the circle because C is the center of the circle. That means that arc AF is a semi-circle which equals 180^{\circ}.

Therefore, arc FD can be found by subtracting 54^{\circ} from 180^{\circ}.

Arc FD=180-54=126^{\circ}

Arc FD=126^{\circ}

Inscribed angle FAD subtends arc FD , so it is half the measure of that arc.

Angle FAD=\cfrac{1}{2}\times{126}=63^{\circ}

In the diagram, \angle{CBD}=43^{\circ} and \angle{CDB}=22^{\circ}. Find \angle{BAD}.

**Use the appropriate circle theorem to find the subtended angle.**

Use that an inscribed quadrilateral has opposite angles that are supplementary.

Also use that inscribed angles are half the measure of the subtended arcs.

**Solve for the angle.**

Angles in a triangle sum to 180^{\circ}.

To find the missing angle of triangle BCD , sum 22^{\circ} and 43^{\circ} and subtract it from 180^{\circ}.

180-(43+22)=115^{\circ}

Angle BCD=115^{\circ}

Since Quadrilateral ABCD is inscribed in the circle, the opposite angles have to be supplementary; angle BAD=180-115=65^{\circ}

Angle BAD=65^{\circ}

In the diagram, F is the center, \angle{ABE}=72^{\circ} and \angle{AGB}=80^{\circ}. Find \angle{BED}.

**Use the appropriate circle theorem to find the subtended angle.**

Use the theorem that states that inscribed angles are half the measure of the subtended arc and that inscribed angles that subtend the same arc are equal.

**Solve for the angle.**

Angle ADE is an inscribed angle and subtends the same arc as angle ABE. So, angle ABE is equal to angle ADE.

Angle ABE=72^{\circ} so arc AE=2\times{72}=144^{\circ} and angle ADE=72^{\circ}

From the diagram, \angle{EGB}=\angle{AGB} because they are vertical angles. Therefore, both angles have a measure of 28^{\circ}.

Triangle DEG has two known angles, angle ADE=72^{\circ} and angle DGE=80^{\circ}, so to find the third angle, sum the measures and subtract from 180^{\circ}.

180-(72+80)=28^{\circ}

Angle DEG which is the same as angle BED has a measure of 28^{\circ} .

- Provide opportunity for students to discover the theorems and make conjectures using compasses and protractors.

- Instead of worksheets, have students use online platforms for game playing to practice skills.

- Incorporate digital platforms for students to explore and investigate the theorems.

- Incorporate digital platforms that have tutorials to assist struggling students.

- Integrate vocabulary into geometry lessons (word lists) due to the high frequency of new words. For example, incorporating βgeometry word of the dayβ or having students use the words in example sentences to help students remember and apply the vocabulary words.

Also, link the words to other languages to compare and contrast how the word is derived and used in American English versus Spanish or German or Italian or French or even British English.

**Forgetting previously learned relationships**

When working with circles, often you will need to recall triangle relationships and other angle relationships. Forgetting those previously learned skills can result in getting incorrect solutions.

Make sure you know the other angle facts including:- Angles on a straight line total 180^{\circ}
- Angles at a point total 360^{\circ}
- Angles in a triangle total 180^{\circ}
- Angles in a quadrilateral total 360^{\circ}
- Vertical angles are equal

**Forgetting when to half or double a measurement**By remembering the inscribed angle theorem incorrectly, the student could double the angle at the center, or half the angle at the circumference.

Top tip: look at the angles. The angle at the center is always larger than the angle at the circumference (on the edge).

**Forgetting triangle types**

Special types of triangles inside circles can help with angle relationships. For example, in the diagram below, the isosceles triangle is made up of the radii of the circle, which are equal. Isosceles triangles also have equal angles, so you can also conclude that angle A and angle B are equal.

**Recognizing when a chord is not the diameter**

In order to assume that a chord is the diameter of the circle, you have to know that it contains the center of the circle as a point. For example, in the figure below, you cannot assume that the chord is the diameter because there is not any given information stating that it contains the center of the circle.

- Chords of a circle
- Circle chord theorems
- Tangent of a circle

1. In circle B below, find the measure of angle LMK.

90^{\circ}

120^{\circ}

45^{\circ}

180^{\circ}

Angle LMK is an inscribed angle that intercepts arc KL.

Central angle LBK is a right angle and also intercepts arc KL which means that arc KL is 90^{\circ}. Inscribed angles are half the measure of the arcs they intersect so

Angle LMK=\cfrac{1}{2}\times{90}=45^{\circ}

2. Find the measure of angle BED in the diagram below.

55^{\circ}

6^{\circ}

116^{\circ}

58^{\circ}

Triangle BED is inscribed in the circle.

The angles of a triangle sum to 180^{\circ}, angle DBE is given to be 67^{\circ}, and angle BDE is inscribed and intercepts arc BE which is 116^{\circ}.

The inscribed angle is half the measure of the arc it intercepts so angle BDE=\cfrac{1}{2}\times{116}=58^{\circ}.

Since angle BDE=58^{\circ}, to find the third angle of the triangle add 58 and 67 and subtract it from 180.

180-(67+58)=55

Angle BED = 55^{\circ}

3. In the diagram, \angle{BAD}=83^{\circ}, \angle{ABD}=41^{\circ}. Find \angle{ACB}.

83^{\circ}

41^{\circ}

56^{\circ}

82^{\circ}

To find angle ACB add 83 and 41 and subtract it from 180 because all three angles of a triangle sum to 180^{\circ}, angle ADB = 180 – (83+41) = 56^{\circ}

Inscribed angle ADB and inscribed angle ADC intercept the same arc, which is arc AB. Two inscribed angles that intercept the same arc are equal to each other.

Since angle ADB=56^{\circ}, angle ADC is also equal to 56^{\circ}.

4. In the diagram, D is the center, \angle{ABC}=38^{\circ}. Find \angle{ACD}.

38^{\circ}

52^{\circ}

76^{\circ}

104^{\circ}

Inscribed angle ABC subtends arc AC. So, the measure of arc AC is 2\times{38}=76, which means that arc AC=76^{\circ}.

Central angle ADC subtends arc AC and is equal in measure to that arc. Therefore, you can conclude that angle ADC=76^{\circ}.

Triangle ADC is an isosceles triangle with equal sides being radii, AD and CD and equal angles being angle ACD and angle CAD.

All the angles in a triangle sum to 180^{\circ}. The vertex angle of the isosceles triangle is also the central angle of the circle, which is 76^{\circ}.

To find the measure of the base angles of the isosceles triangle subtract 76 from 180 and divide the difference by 2.

180-76=104 and 104\div2=52

So, each base angle of the isosceles triangle is equal to 52^{\circ}.

Angle ACD=52^{\circ}

5. In the diagram, \angle{ACB}=53^{\circ}. Find \angle{BAE}.

26.5^{\circ}

37^{\circ}

53^{\circ}

90^{\circ}

In circle D, BE is the diameter of the circle (longest line segment in a circle) with endpoints on the circumference of the circle.

The diameter is also one of the sides of the inscribed triangle ABE. The endpoints of the diameter also indicate the endpoints of the semicircle BE.

Angle BAE is an inscribed angle that subtends the same arc that the diameter intercepts. That means that angle BAE subtends a semicircle.

The semicircle is equal to 180^{\circ} which means angle BAE is equal to \cfrac{1}{2}\times{180}=90^{\circ}

Angle BAE=90^{\circ} so, it is a right angle.

6. In the diagram, \angle{BCD}=88^{\circ} , \angle{ADE}=49^{\circ} , \angle{AED}=78^{\circ}. Find \angle{BAE}.

39^{\circ}

53^{\circ}

78^{\circ}

44^{\circ}

Quadrilateral ABCD is inscribed in the circle so it is a cyclic quadrilateral.

The opposite angles must be supplementary in an inscribed quadrilateral Angle BCD is 88^{\circ} so to find angle BAD subtract 88 from 180,

180-88=92^{\circ}

Angle BAD=92^{\circ}

All the angles in a triangle sum to 180^{\circ}. In triangle AED, you are given two of those angles. If you subtract them from 180, you find the measure of angle DAE.

180-(78+49)=53^{\circ}

Angle DAE=53^{\circ}

Using the angle addition postulate, from the diagram you know that angle DAE + \text{angle BAE} = \text{angle BAD}, substituting in the known values you can solve for AB.

\text{angle }DAE+\text{angle }BAE=\text{angle }BAD

53+\text {angle }BAE=92

Angle BAE=39^{\circ}

Yes, in the American form of English, the term subtended arc is also known as an intercepted arc.

Yes, in botany subtend means to be located beneath or close to another structure such as a leaf , bract or branch. You can use a thesaurus to find the other meanings for subtend. You can also explore how the word was derived from Latin subtendere.

- Types of data
- Averages and range
- Representing data

At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.

Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.

Find out how we can help your students achieve success with our math tutoring programs.

x
####
[FREE] Common Core Practice Tests (3rd to 8th Grade)

Download free

Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents.

Get your 6 multiple choice practice tests with detailed answers to support test prep, created by US math teachers for US math teachers!