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This topic is simply about remembering a set of rules, so even if you struggle with math, this topic is where you can “Pick up” marks quite easily during the CXC exam. Here are a few tips to solve the majority of these problems. Always observe the diagrams carefully and be on the look-out for :

1. Parallel lines - alternate angles, (Z angles)2. Isosceles triangle ( base angles are equal)3. Tangents ( see below)4. Cyclic quadrilateral ( opp. Angs are supp)5. Angles at the center = twice ang. at circum.6. Angles in a semicircle = 900

7. angles in the same segment are equal8. angle between radius and tangent = 900

Once you understand what to do with the information above, then you should see the solution in less than 1 minute, and be able to complete the problem in less than 5 minutes.

Examples:• Parallel lines - alternate angles, (Z angles)

Alternate angles are equal ( angle x at the elbows of the “Z” are the same)************************************************************

Angle between radius and tangent = 900

Problems generally fall into a few basic groups: viz;

• Problems with No tangent• Problems with One tangent• Problems with Two tangents

Problems with One TangentIf the circle has one tangent, then look for any opportunity to use the theorem which says that the angle between tangent and chord is equal to angle in the alternate segment.

Notice that angle between Tangent FG and Chord HI is 20o , so the angle in the

Alternate segment x = 20. Similarly y = 35

Problems with Two Tangents

You will need to remember the following:

• <SCT + <SUT = 180 (supplementary) • Where <SCT = 2y , and <SUT = 2x • <STU = <UCT = y• Quadrilateral SCTU is cyclic ( can u say why?)

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Introduction:- Circle Theorem

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In all other cases, tangent or no tangent, look out for the following

• Angles at center (100) = twice angle at circumference (y=50)

• Isosceles triangles, are formed whenever one point of a triangle is at the center of the circle and the other two points touch the circumference ( Triangle OJR). This means that the two base angles (x) are equal.

• Angles in the same segment (JR) are equal <RKJ = <RAJ = x ( if the chord is not given on the diagram, use your pencil and create it in dotted lines as shown.

• Angle in a semicircle <PQN = 90o This topic is simply about remembering a set of rules, so even if you struggle with math, this topic is where you can “Pick up” marks quite easily during the CXC exam. Here are a few tips to solve the majority of these problems.

Construction using using Ruler and Compass onlyTips: on Constructing angles:

All the angles that you will be asked to construct using compass only, can be completed once you know three (3) basic things.

1. How to construct a 900 angle

2. How to construct a 600

3. How to bisect an angle

All other angles are derived from the above:

Terms you need to know:

• Perpendicular – at right angles

• Bisect – divide into two equal parts

• Perpendicular bisector – a line which is at right angles to another line and which passes through its mid point.

Example:

Explain how you would construct the following angles using ruler and compass only.

450 ; 1350 ; 300 1650 ; 1200 , 157.50

Solution:

1) 450 is found by first constructing a 900 and then bisecting ( divide equally) the 900 .

2) 1350 = 900 + 450

3) 300 = ½ 600 (construct 600 and then bisect)

4) 1650 = 1350 + 300

5) 1200 = 1800 - 600

6) 22.5 = 1/2(45)

7) 157.5 = 180 – 22.5

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