Intercepted arc is that arc which is formed when segments intersect parts of a circle and create arcs.

The intercepted arc in the figure on the left is $\widehat{EF}$. The angle thus formed at the center is called the Central angle or the intercepted angle.

The measure of a central angle in a circle is always equal to the measure of its intercepted arc.
m $\angle$EOF = m$\widehat{EF}$

When two straight lines intersect a circle, that part of the circle between the intersection points is called the intercepted arc. The lines intercept, or 'cut off', the arc of the circle.

Usually, the two lines become the arms of an angle, as shown in the intercepted arcs figure above and such angle is called the intercepted angle. But this is not always the case. For example, in the figure on the left, the two secant lines cut off, or intercept two arcs, $\widehat{CE}$ and $\widehat{BD}$.

In this case the the vertex (at P in the figure on the left) of the angle lies outside the circle. The measure of an angle with its vertex outside the circle is half the difference of the measures of the intercepted arcs.

The formula is, Angle = (measure of difference of intercepted arcs)
The formula is m$\angle$BPD = m($\widehat{BD}$)- m($\widehat{CE}$)
In this case the two lines do not form an angle at the center but become the arms of an angle whose vertex is on the circle. This angle is called the inscribed angle. The sides of this angle (or the arms) become the chords of the circle. They form an intercepted arc BD.
An inscribed angle always has its vertex on the circle. The measure of an inscribed angle equals $\frac{1}{2}$ the measure of the intercepted arc.

The formula is m$\angle$BCD = $\frac{1}{2}$ (m($\widehat{BD}$)
In this case the two lines do not form an angle at the center but part of the lines become the arms of an angle whose vertex is inside the circle. They form intercepted arcs EF and GH.

The measure of an angle with its vertex (at P in the figure on the left) inside the circle is half the sum of the intercepted arcs.
The formula is m$\angle$EPF = m$\angle$GPH = $\frac{1}{2}$(m($\widehat{EF}$) +m($\widehat{GH}$))

## Intercepted Arc Formula

In this diagram, $\theta$ is the central angle (generally measured in radians), r is the radius of the circle, and L is the arc length of the intercepted arc.

Intercepted arc length in radians, L = $\theta$ r.

Intercepted arc length in degrees, L  = $\theta$ $\frac{\pi r}{180}$.

## Intercepted Arc Examples

### Solved Examples

Question 1: In the circle with center O, m$\angle$ COT= 120°. Find the number of degrees in arcs x and y.

Solution:

$\angle$ COT is a central angle. Since the measure of a central angle is equal to the measure of its intercepted arc, we have,

If m$\angle$ COT = 120° then m($\widehat{TC}$) = 120°.

Thus x = 120°.

Y is the major arc in the figure. The sum of measures of x and y together will be 360°. Thus y = 360° - x.
y = 360° - 120°
y = 240°.

Question 2: In the circle with center P, m($\widehat{AC}$)= 106°. Find m$\angle$ ACB.

Solution:

$\widehat{BAC}$ is a semicircle and the measure of angle is 180°.
m($\widehat{CA}$)+ m($\widehat{AB}$)= 180°
Given that m($\widehat{CA}$)= 130°
Therefore we have,  130° + m($\widehat{AB}$)=180°.
m($\widehat{AB}$)= 180 - 130 = 50°.

The measure of an inscribed angle equals half the measure of the intercepted arc. Here $\angle$ ACB is an inscribed angle.

Thus we have, x = $\frac{1}{2}$ m($\widehat{AB}$) = $\frac{1}{2}$(50) = 25°.

Question 3: $\Delta$ABC is inscribed in the circle.
The ratio of m$\widehat{AB}$ : m$\widehat{BC}$ : m$\widehat{AC}$ = 3 : 2 : 5.

(a) Find m($\widehat{AB}$),m($\widehat{BC}$)and m($\widehat{AC}$)

(b) Find each angle of the $\Delta$ ABC.
Solution:

(a) Since the ratio of the three arcs is 3 : 2 : 5, we let m$\widehat{AB}$ = 3x, m$\widehat{BC}$ = 2x and m$\widehat{AC}$ = 5x

Since the measures of the three arcs of the circle must add up to 360°,
We can write the equation as 3x + 2x + 5x = 360°
Grouping like terms, we get,  10x = 360
Divide by 10 and we get,  x = 36
To find the number of degrees in the measures of the three arcs, we plug "x" with 36:
m$\widehat{AB}$ = 3x = 3(36) = 108° and m$\widehat{BC}$ = 2x = 2(36) = 72° and m$\widehat{AC}$ = 5x = 5(36) = 180°.

(b) To find the three angles of the triangle, it can be noticed that each angle of the triangle is an inscribed angle which is equal to half the measure of its intercepted arc.

$\angle$ ABC is an inscribed angle which intercepts $\widehat{AC}$
Thus,m$\angle$ ABC = $\frac{1}{2}$ m($\widehat{AC}$) = $\frac{1}{2}$ (180°) = 90°

$\angle$ BCA is an inscribed angle which intercepts.
Thus,m$\angle$ BCA = $\frac{1}{2}$ m($\widehat{AB}$) = $\frac{1}{2}$ (108°) = 54°

$\angle$ CAB is an inscribed angle which intercepts.
Thus,m$\angle$ CAB = $\frac{1}{2}$ m($\widehat{BC}$) = $\frac{1}{2}$ (72°) = 36°

Question 4: Find the length of a 75°  arc in a circle of radius 12 cm.
Solution:

Intercepted arc length in degrees,L  = $\theta$ $\frac{\pi r}{180}$

Length of the arc = 75 $\frac{\pi \times 12}{180}$

= 75 $\times \pi \times$ ($\frac{1}{15}$)

= 5$\pi$ = 15.71 cm

Length of a 75° arc in a circle of radius 12 cm is 15.71 cm.

Question 5: Calculate the measure of the arc length L in the circle given below.

Solution:

The formula for intercepted arc is given by
Intercepted arc length in radians, L = $\theta$ r.
Here L = 4$\pi$ and r = 5.
Thus, 4$\pi$ = $\theta$ (5)

$\theta$ = $\frac{4 \pi}{5}$.

So, the measure of the arc length L in the given circle is $\frac{4 \pi}{5}$.

Question 6: Calculate the radius of the circle whose intercepted arc is 9cm and its measure is  $\frac{3 \pi}{4}$.
Solution:

The formula for intercepted arc is given by
Intercepted arc length in radians, L = $\theta$ r.

Here L = 9 cm and $\theta$ = $\frac{3 \pi}{4}$.

Thus, 9 = $\frac{3 \pi}{4}$ $\times$ r

r  = $\frac{9 \times 4}{3 \pi}$

r = $\frac{12}{p}$ = $\frac{12}{3.1415}$

r = 3. 82 cm

The radius of the circle whose intercepted arc is 9cm and its measure of $\frac{3 \pi}{4}$ is 3.82cm.