Finding angles in Circles

In this chapter, we are going to look at some different rules we can use to find angles within circles. You can use all the rules you already know, so let's start with a recap.

 

Inscribed Angles

Have a look a the diagram given here:

The angle formed by chord $AC$AC and chord $BC$BC at the common point $C$C is called the inscribed angle. This angle as the vertex at point $C$C, we would write it as $\angle ACB$∠ACB or $\angle BCA$∠BCA

Note also the language of major arc and minor arc.  

Have a play with this mathlet and see what you can find.  

• Did you notice when the angle was constant?
• Did you notice a connection when the angle is on the major or minor arc? 

What you may have just discovered is that if you keep points A and B fixed, and just rotate C around the circle then the ANGLE is always the same.  This has a special name: 

Angles Subtended by Same Arc Theorem

When there are two fixed endpoints, the angle $\left(a\right)$(a) is always the same, no matter where it is on the circumference.

 

Angle at the Centre Theorem

Let's now add some more line segments to our diagram.  Particularly the ones from $A$A and $B$B to the centre $O$O. Now we have two angles of interest, the inscribed angle as mentioned before and another angle we call the Central Angle.  

Have a play with this mathlet and see if you can find a connection between the two.  

 

An inscribed angle $\left(\angle ACB\right)$(∠ACB) is half of the central angle $\left(\angle AOB\right)$(∠AOB) and in geometric reasoning, we often say the angle at the centre is twice the angle at its circumference.

Proof:

We have drawn a green line from the origin to Point $C$C on the circumference, to form two isosceles triangles (since the radii makes a pair of sides in each triangle equal).

Let $\angle CAO$∠CAO be $x$x and $\angle CBO$∠CBO be $y$y.

$\angle ACB=\left(x+y\right)$∠ACB=(x+y)

$\angle CAO=\angle ACO$∠CAO=∠ACO$=$=$x$x (equal base angles in isosceles triangle)

$\angle CBO=\angle BCO$∠CBO=∠BCO$=$=$y$y (similarly)

$\angle AOC=\left(180-2x\right)$∠AOC=(180−2x) (angle sum of a triangle is $180^\circ$180°)

$\angle BOC=\left(180-2y\right)$∠BOC=(180−2y) (similarly)

$\angle AOB=360-\left(180-2x+180-2y\right)$∠AOB=360−(180−2x+180−2y) (angles at a point add up to $360^\circ$360°)

          $=$=$2x+2y$2x+2y

          $=$=$2\left(x+y\right)$2(x+y)

$\therefore$∴ the central angle is double the inscribed angle.

 

Angle in a Semicircle

A particularly interesting event is when points $A$A and $B$B create a diameter, thus creating a semicircle. The angle at point $C$C is a right angle. (Use the mathlet above to check)

We are also able to deduce that if the inscribed angle is a right angle, then the chord $AB$AB is a diameter. 

The angle inscribed in a semicircle is always a right angle ($90^\circ$90°).

Proof:

$\triangle AOC$△AOC and$\triangle BOC$△BOC are isosceles (radii are equal lengths in a circle)

Hence, $\angle OCB=\angle OBC$∠OCB=∠OBC$=$=$\alpha$α (equal base angles in an isosceles triangles)

Similarly, $\angle OAC=\angle OCA$∠OAC=∠OCA$=$=$\beta$β 

 

In $\triangle ABC$△ABC:

$\alpha+\beta+\alpha+\beta$α+β+α+β	$=$=	$180^\circ$180°	(angle sum of a triangle is $180^\circ$180°)
$2\alpha+2\beta$2α+2β	$=$=	$180^\circ$180°
$\alpha+\beta$α+β	$=$=	$90^\circ$90°
Hence, $\angle ACB$∠ACB	$=$=	$90^\circ$90°

 

Cyclic Quadrilaterals

A cyclic quadrilateral is a four-sided shape that has all its vertices touching the circle's circumference.

The opposite angles in a cyclic quadrilateral add up to $180^\circ$180°.

Proof:

Let $\angle ADC$∠ADC be $a$a, $\angle ABC$∠ABC be $b$b and the centre of the circle be $O$O.

$\angle AOC=2a$∠AOC=2a (a central angle is double the inscribed angle) 

$\angle AOC$∠AOC (reflex)$=$=$2b$2b (similarly)

$2a+2b=360^\circ$2a+2b=360° (angles at a point add up to $360^\circ$360°)

$\therefore$∴$a+b=180^\circ$a+b=180°

 

Worked Examples

Question 1

Question 2