1.  

A circle is drawn with centre O, where A, B, C and D are points on its circumference.

Find in terms of x, the following:

a) angle ADB

b) angle AOD

c) angle ABD

a) Triangles ABD and ABC are drawn in the same direction from the ends of the same chord - hence we can use the "Angles in the same segment" theorem to say:

angle ADB = angle ACB

Hence Angle ADB = 2x.

b) Lines OD and OA are drawn from the centre O to the edge of the circle. As both lines = the radius of the circle, they will be both the same length.

Hence the angle made between line OD and the chord AD is = the angle made between the line OA and the chord AD.

In other words: angle ADO = angle DAO

As we have already calculated angle ADB as 2x, we can see that the angle ADO = 2x + x = 3x.

If ADO = 3x then so does DAO.

We know that the angles in a triangle must add up to 180^o, so we can say that angle AOD = 180 - 3x - 3x

i.e. Angle AOD = 180 - 6x

c) Because two ends of the diameter are drawn into a triangle at point C, we can use the "angle in semicircle" theorem to say that angle ABC = 90^o.

If ABC = 90o, then BAC = 90 - 2x (angles in triangle add up to 180^o).

We know that angle DAO = 3x, so using the "angles in a triangle add up to 180^o" theorem again we can say that:

Angle ABD = 180 - (Angle BAC + Angle DAO) - Angle ADB

Angle ABD = 180 - (90 - 2x + 3x) - 2x

Angle ABD = 90 - 3x

A, B, C and D are points on a circle
AD is parallel to BC
Lines AC and BD intersect at point X
Angle ADX = 36^o

a) Find the angle ACB

b) Find the angle AXB and describe what this tells you about the point X.

a) Triangles ADB and ACB are both drawn from the ends of the chord AB, hence using the "angles in the same segment" theorem we can say that:

Angle DAB = Angle ACB

Hence Angle ACB = 36^o.

b) We are told that line AD is parallel to line BC. So, we can use the angle properties of parallel lines to say:

Angle ADB = Angle DBC

Hence Angle DBC = 36^o.

We calculated in a) that angle ACB = 36^o, so using angle properties of parallel lines again we can say that angle DAC = 36^o also.

Because the angles in a triangle add up to 180^o, we can say that angle DXA = 180 - 36 - 36 = 108^o.

As the triangles DXA and CXB are congruent (have the same angles) the angles CXD and AXB will be the same.

Using "angles around a point add up to 360^o" we can say that

2AXB + 108 + 108 = 360^o.

2AXB = 144

Hence angle AXB = 72^o.

X is the centre of the circle from the theory "angle at centre = twice angle at circumference"

3.

CP and CQ are tangents to a circle which has centre O.

a) Explain why the triangles CQO and CPO are congruent.

b) If the angle PCQ = 72^o, what is the reflex angle POQ ?

a) length OP = length OQ as both are radii of the circle.

Angle CPO = Angle CQO from the theory "angle between tangent and radius = 90^o".

The line CO is common to both triangles.

b) If the angle PCQ = 72^o, the angle PCO = 36^o.

All angles in a triangle add up to 180^o, hence the angle POC = 180 - 90 - 36 = 54^o

Because the two triangles CQO and CPO are the same, the obtuse angle POQ = 2 x 54 = 108^o

If the obtuse angle is 108^o using the property that angles around a point add up to 360^o:

The reflex angle POQ = 360 - 108 = 252^o

4.

ABCD are points on the circumference of the circle.
PC is a tangent to the circle at C
Angle BCP = 20^o and angle ADC = 68^o.

a) Find the angle ABC

b) Find the angle ACB

a) 68 + angle ABC = 180^o (properties of a cyclic quadrilateral)

Angle ABC = 180 - 68 = 112^o

b) Angle BAC = 20^o (from the alternate segment theorem - as it is at the point where the triangle made from the two ends of the chord BC hits the rim of the circle).

We calculated in a) that ABC = 112^o, so because all angles in a triangle add up to 180^o:

ACB = 180 - 112 - 20 = 48^o.

5.

O is the centre of the circle
A,B,C and D are points on the circumference of the circle
PQ is the tangent to the circle at point A

a) Calculate the angle between the tangent and the line AB.

b) Calculate the angle ABC.

a) The angle between the tangent PQ and the line AB = 35^o (from the alternate segment theorem).

b) Angle ADC = 54^o (angle at centre is twice the angle at the circumference).

Angle ADC = Angle ABC (from: "opposite angles in a cyclic quadrilateral add up to 180^o")

Hence Angle ABC + 54 = 180^o

Angle ABC = 180 - 54 = 126^o

6.

Line PQ is the tangent to the circle at point A
The line AB is extended to point X
The angle XBC = 105^o
The angle PAB = 3x
The angle BAC = 4x

a) Determine the angle QAC in terms of x.

b) Calculate the value of x.

a) Sometimes a simple solution can be staring you in the face like this one:

PQ is a straight line so the angle QAC = 180 - 4x - 3x.

Angle QAC = 180 - 7x.

b) Line ABX is a straight line so,
the angle ABC = 180 - 105 = 75^o

Using the alternate segment theorem, we can see that the angle ABC is the same as the angle QAC = 180 - 7x

Hence 180 - 7x = 75

Add 7x to both sides:
180 = 75 + 7x

Subtract 75 from both sides:
105 = 7x

Divide both sides by 7:
x = 15^o

7.

AB and BC are parallel lines between the two straight lines ABX and DCX.

a) Work out the following angles:
i) ACD
ii) BCA
iii) CDA

b) Show that the triangle EBA is isosceles.

a)
i) ACD = Angle between tangent and line AD (alternate segment theorem).

ACD = 52^o.

ii) BCA = Angle between tangent and line AB (alternate segment theorem again).

BCA = 47^o.

iii) As DA and CB are parallel, we can use the properties of angles with parallel lines to say that angle DAC = angle BCA = 47^o.

Knowing that the angles in a triangle add up to 180^o, we can say that CDA = 180 - angle ACD - angle DAC

CDA = 180 - 54 - 47 = 81^o.

b) We have just calculated that angle CDA = 81^o.

If the angle DAB = CDA and the angle DXA is different then the triangle is isosceles.

DAB = 180 - angle DCB

DAB = 180 - (52 + 47) = 81^o

If angles DAB and CDA = 81^o then angle DXA = 180 - 81 - 81 (sum of angles in a triangle)

= 18^o, hence the triangle is isosceles.

8.

ABC and ADE are straight lines. The diameter of the larger circle is CE. PQ is the tangent to the larger circle at E.

Express in terms of x, the following angles:

a) ABD

b) DBE

c) BAD

Explain why the statement BE x AC = AE x CD is true.

a) Angle CDE = 90o (angle in a semi-circle). Hence as all the angles in a triangle add up to 180o:

the angle DEC = 90 - x

angle DEC + angle DBC = 180o (opposite angles add up to 180o in a cyclic quadrilateral)

i.e. 90 - x  +  angle DBC = 180
angle DBC = 180 - 90 + x = 90 + x

As AC is a straight line:

angle ABD = 180 - angle DBC
= 180 - (90 + x)
= 90 - x

b) The angle DBE is the same as the angle between the tangent and the chord DE (alternate segment theorem).

Angle PEC = 90^o (property of tangent)

As angle AEC = 90 - x, then

Angle PED = 90 - (90 - x) = x

Hence, Angle DBE = x

c) From the sum of the angles in a triangle = 180^o:

BAD + ACE + AEC
BAD + 3x + 90 - x = 180

Angle BAD = 90 - 2x

d) CD and BE are both altitudes (line vertically up the centre of a triangle from its base to its tip).

So the area of the triangle = ½ AC x BE
and the area of the triangle = ½ CD x AE

i.e. ½ AC x BE = ½ CD x AE

divide both sides by ½ :

AC x BE = CD x AE