Inscribed Angle of a Circle

An inscribed angle is formed when two secant lines intersect on a circle. In other words, the inscribed angle in a circle is formed when one of the end points of two chords in a circle meet in a point.

                Inscribed angle 

In the above picture, L ABC is the of the circle.


Inscribed Angle of a Circle


      Inscribe angle                                                               

From the picture above,


  •  Angle ABC is the inscribed angle.

  • CB and BA are the chords.

  •  Arc CA is the intercepted arc.

  • Formula: Angle CAB = Arc AC/2.


Inscribed Angle of a Circle


  • Two or more inscribed angles intercepting a same arc will be equal.


      BAC angle = BDC angle


  • An inscribed angle is the measure of half the central angle intercepting the same arc.


     CABangle = (1 / 2) BOC angle

     CDB angle = (1 / 2) BOC angle

Another Formula for inscribed angle:

If we know the length of the minor arc, radius, the inscribed angle is found by:

Angle= `(90 L)/(pi* R)`


L is the length of the shortest arc BA

R is defined as radius.


I am planning to write more post on Inscribed Angle Theorem, Consecutive Interior Angles Keep checking my blog.


Example Problems


From the circle diagram below, the chord CA has a length of 12 cm and center at O. The circle has a radius of 14 cm. Find the measure of the inscribed angle ABC.

                   Inscribed Angle Examples


1. First we calculate the central angle AOC. The triangle AOC is a isosceles triangle. Distance of OC = Distance of OA = radius = 14 cm. Here we use cosine law to find value of cos (angle AOC).

AC2 = OC2 + OA2 - 2 OC OA cos (angle COA)

2. Substitute the value of the angles AC, OC and AO in cos (angle AOC) as follows

cos(angle COA) = `[ 14^2 + 14^2- 122 ] / [28 * 14 ]`

                        = `62 / 98`

3. The measure of the angle COA is given by.

Angle COA = `arccos (62 / 98)`

According to the theorem described before, the size of angle CBA will be equal to half the size of angle COA.

Angle CBA = `(1/ 2) arccos (62 / 98)`

                 = 25.38 degrees.


Practice Problems


Problem 1:

 Given that the central angle of circle is 50 degrees. Find the angle of inscribed circle.

Answer = 100 degree

Problem 2:

 Given that the central angle of circle is 120 degrees. Find the angle of inscribed circle.

Answer = 240 degree.