Arcs and Subtended Angles

1. If the angles subtended by the chords of a circle at a centre are equal, are the chords equal?
2. The chords of congruent circles subtend equal  angles at their centres. Are the chords equal?

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Solved Examples

Example 1

 

 

Find the value of \(x\) (in degrees) in each of the following figures (wherever indicated, \(O\) is the centre of the circle)

I. 

Solution:

From the figure, we can say that \(\angle ACB = \angle ADB = 40^{\circ}\)

Using the angle sum property in \(\Delta ACB\),

\(\begin{align}\angle BAC + \angle ACB + x^{\circ} &=180^{\circ}\\ 30^{\circ} + 40^{\circ} + x^{\circ}&=180^{\circ}\\x^{\circ}=110^{\circ}\end{align}\)

\(\therefore\) The value of \(x\) is 110

II. 

Solution:

\(\angle BAD\) is an angle in a semi-circle, and is therefore, a right angle.

Using the Angle Sum Property in \(\Delta ABD\), we have:

\(\begin{align}\angle ADB&=180^{\circ}-(30^{\circ}+90^{\circ})\\&=60^{\circ}\end{align}\)

Thus, \(\angle ACB=\angle ADB=60^{\circ}\)

\(\therefore\) The value of \(x\) is 60

III.

Solution:

We have \(\angle AOB=2\angle ACB\)

So,

\(\begin{align}\angle AOB &= 2\angle ACB\\ &=2(35^{\circ}+30^{\circ})\\ &=130^{\circ}\end{align}\)

\(\therefore\) The value of \(x\) is 130

IV.

Solution:

We note that,

\(\begin{align} \angle AOC&=360^{\circ}-(150^{\circ}+120^{\circ})\\&=90^{\circ}\end{align}\)

Thus, \(\angle ABC= \begin{align}\frac{1}{2}\end{align} \angle AOC=45^{\circ}\)

\(\therefore\) The value of \(x\) is 45

V.

Solution:

We have, \(\angle BOC=180^{\circ}-40^{\circ}=140^{\circ}\)

\(\implies x^{\circ}= \begin{align}\frac{1}{2}\end{align} \angle BOC=70^{\circ}\)

\(\therefore\) The value of \(x\) is 70

VI.

Solution:

The reflex angle \(AOB\) is equal to \(360^{\circ}-150^{\circ}\) or \(210^{\circ}\)

Thus, \(x^{\circ}= \begin{align}\frac{1}{2}\end{align} \times 210^{\circ}=105^{\circ}\)

\(\therefore\) The value of \(x\) is 105

Example 2

 

 

In the following figure, find \(x^{\circ}\) in terms of \(y^{\circ}\) and \(z^{\circ}\)

\(O\) is the centre of the circle.

Solution:

We label some angles as shown in the figure below.

Using the exterior angle theorem, we have,

\(\angle y = \angle 1 + \angle 2 \;\;\;\; \cdots (1)\)

\(\angle 1 + \angle x = \angle 4 \;\;\;\; \cdots (2)\)

Add Equations \((1)\) and \((2\)

\(\angle y + \angle 1 + \angle x = \angle 1 + \angle 2 + \angle 4\)

\(\implies \angle x + \angle y = \angle 2 + \angle 4\)

Now, since \(z^{\circ} = 2 \angle 2 =2 \angle 4\), we have \(\angle 2 = \angle 4 = \begin{align}\frac{z^{\circ}}{2}\end{align}\) 

So,

\(\begin{align}x^{\circ}+y^{\circ}&=z^{\circ}\\ x^{\circ}&=z^{\circ}-y^{\circ}\end{align}\)

\(\therefore\) \(x^{\circ}=z^{\circ}-y^{\circ}\)

Example 3

 

 

Two circles intersect at \(P\) and \(Q\).

Through \(P\), diameters are drawn for both the circles.

The other end-points of these diameters are \(R\) and \(S\).

Show that \(R\), \(Q\) and \(S\) are collinear.

Solution:

Consider the following figure.

Since angles in semi-circles are \(90^{\circ}\), we have, \(\angle RQP=\angle SQP=90^{\circ}\)

Then,

\(\begin{align}\angle RQS&=\angle RQP+\angle SQP\\&=90^{\circ}+90^{\circ}\\&=180^{\circ}\end{align}\)

Thus, \(RQS\) is a straight line, which means that \(R\), \(Q\) and \(S\) collinear.

\(\therefore\) The points \(R\), \(Q\) and \(S\) are collinear.

Example 4

 

 

\(\Delta ABC\) is isosceles, with vertex \(A\)

Show that the circle through diameter \(AB\) bisects \(BC\) at point \(D\)

Solution:

Observe the following figure.

The circle cuts \(BC\) at \(D\)

Since angles in a semi-circle are of measure \(90^{\circ}\), \(\angle ADB=90^{\circ}\)

This means \(AD\) is the altitude for the isosceles \(\Delta ABC\)

\(\implies\) \(AD\) also bisects \(BC\)

\(\implies\) \(D\) is the mid-point of \(BC\)

Example 5

 

 

From the end-points of a diameter \(AB\) of a circle with centre \(O\), perpendiculars \(AP\) and \(BQ\) are dropped on an external line.

Show that \(AP = CQ\)

Solution:

We note that \(\angle ACB = 90^{\circ}\)

Thus, \(ACQP\) is a rectangle.

Since oppoite sides of a rectangle are equal, \(AP = CQ\)

Hence, proved that \(AP = CQ\)