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Home > Maths I > Measure and Shape > Angles involved with circles

Maths I

Angles involved with circles


Angles Involved with Circles - some questions

You should be able to answer all of the following questions by looking at the information and diagrams that you have seen in the previous pages:


In this diagram PR is a diameter and angle PRQ = 25^\circ.

What is the size of angle QPR?

A circle with its diameter marked out vertically. Off the diameter come two lines to the right, angled together to form a right angled triangle reaching to the circumference of the circle. The right angle of the triangle is called Q, the top angle is P, and the bottom angle is R and labelled as being 25 degrees.

PQR is an angle in a semi-circle

so angle PQR = 90^\circ

The three angles in the triangle add up to \;{180^\circ}\;

so angle QPR = 180 - 90 - 25

= 65^\circ

Now try these questions.


In this diagram \;C is the centre of the circle and \;PT is a tangent to the circle.

CT = 5 cm and PT = 12 cm.

What is the length of \;CP?

A circle with it's radius marked out horizontally. From the circumference of the circle where the raduis ends,a line drops vertically. Another line comes from the radius to join up to this line and form a right angled triangle. The radius edge of the triangle is 5 centimetres, the vertical edge is 12 centimetres and the hypotenuse is unmarked. The right angle is t, the bottom angle is p and the angle at the circle's centre is c.

\;PT is a tangent and \;CT is a radius

so angle CTP = 90^\circ

We have a right angled triangle and so we can use Pythagoras:

CP^2 = 12^2 + 5^2

= 144 + 25

\;= 169

CP = \surd{169}

\;= 13cm


In this diagram \;KL is a diameter of the circle and is \;8 cm long. LM = 3 cm.

Calculate the size of \;KM.

A circle with its diameter marked out, labelled 8 centimetres. From both ends of the diameter, lines come out and join together at the circumference to form a right angled triangle. The right angle is called m and the other angles are k and l. The edge between m and l is 3 centimetres.

\;KL is a diameter so we have an angle in a semi-circle and angle KML = 90^\circ.

We have a right angled triangle and so can use Pythagoras.

\;KM is not the hypotenuse so:

KM^2 = 8^2 - 3^2

= 64 - 9

KM^2 = 55

KM = \surd{55}

= 7.4cm (to one decimal place)

Impress your mates - how to remember pi to 8 decimal places:

(Don't worry - you don't need to do this in your exam!)

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