Properties of parallel lines. The angles 1-3, 2-4, 6-8 and 5-7 are vertically opposite angles. Vertically opposite angles are equal . The angles 2-6, 1-5, 7-3 and 8-4 are corresponding angles. Corresponding angles are equal . The angles 3-4 and 4-6 are alternate angles.
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The angles 1-3, 2-4, 6-8 and 5-7 are vertically opposite angles.
Vertically opposite angles are equal.
The angles 2-6, 1-5, 7-3 and 8-4 are corresponding angles.
Corresponding angles are equal.
The angles 3-4 and 4-6 are alternate angles.
Alternate angles are equal.
The angles 3-6 and 4-5 are cointerior angles.
Cointerior angles are supplementary (180∘).
1. a ° , b° and c° are the magnitudes of the interior angles of the triangle ABC.
d° is the magnitude of an exterior angle at C. A triangle is said to be a right-angled triangle if it has one angle of magnitude 90°. 2. The sum of the magnitudes of the interior angles of a triangle is equal to
a ◦ + b◦ + c◦ = 180° . 3. The magnitude of an exterior angle is equal to the sum of the magnitudes of the two opposite interior angles. b° + a ° = d °.
4. A triangle is said to be equilateral if all its sides are of the same length: AB = BC = CA.
5. The angles of an equilateral triangle are all of magnitude 60◦ .
6.The bisector of each of the angles of an equilateral triangle meets the opposite side at right angles and passes through the midpoint of that side.
7. A triangle is said to be isosceles if it has two sides of equal length. If a triangle is isosceles, the angles opposite each of the equal sides are equal.
8. The sum of the magnitudes of the exterior angles of a triangle is equal to 360°: e° +d°+f°=360°
Example 1: Find the values of the pronumerals.
Solution: 20° +22°+x°=180° (sum angles△=180°)
∴42° +x°=180° or x =138° 138°+y°=180° (sum angles=180°) ∴y =42°
Solution:100°+2x°=180° (sum angles△=180°)
∴2x° =80° or x =40°
A regular polygon has all sides of equal length and all angles of equal magnitude. A polygon with n sides can be divided into n triangles.
The angle sum of the interior angles of an n-sided polygon is given by the
formula: S = [180(n − 2)]◦ = (180n − 360)◦
The magnitude of each of the interior angles of an n-sided polygon is given by: x = (180n − 360)◦
n interior angle
The sum of the exterior angles of a regular polygon is 360◦ .
Example1: The diagram opposite shows a regular octagon.
a Show that x = 45.
b Find the size of angle y.
ax = 360°÷ 8 = 45°
b BOC is isosceles
then OBC and OCB are equal
x° + 2y°= 180°
45° + 2y° = 180°
2y°= 180° - 45°
y° = 67.5°
Example:Find the sum of the interior angles of an 8-sided polygon(octagon).
Use the formula x°= (180n-360)°
x°= 180×8-360 = 1080 °
Example 1: Find the value, correct to two decimal places, of the unknown length for the triangle below.
Solution: x2 = 5.32 + 6.12
√x2 = √(5.32 + 6.12)
x = 8.08cm
Example 2: Find the value, correct to two decimal places, of the unknown length for the triangle below.
Solution: 8.62 = y2 + 5.62
y2 = 8.62 - 5.62
y = √( 8.62 - 5.62)
y = 6.53cm
Shapes are similar when they have the same shape but not the same size.
Two triangles are similar if one of the following conditions holds:
1.The corresponding angles in the triangles are equal
A = A′, B = B′, C = C′
2. The ratio of the corresponding sides is equal
A′B′ = B′C′ = A′C′ =k
AB BC AC
k is the scale factor.
If AB = 2, BC = 3, AC =4 and A′B′ = 6, B′C′ = 9, A′C′= 12 then 6 = 9 = 12 = 3= k
2 3 4
3.Two pairs of corresponding sides have the same ratio and the included angles are equal
Example 1: Find the value of length of side AC in △ABC,correct to two decimal places.
Solution: Prove that the triangles are similar.
B= B′and 5 = 4 3 = 4
6.25 5 3.75 5
k = 4
(Two pairs of corresponding sides of equal ratio and the included angles equal).
Use k to find x
x = 5
x = 3.013 x 5 = 2.4103 ≈2.41cm
Find the value of length of side AB in △ABC.
Solution: Prove that the triangles ACB and AYX are similar.
A is a common angle
C = Y ( corresponding angles)
B = X ( corresponding angles)
Use the ratio to find x
x = 3
x + 6 3+ 2.5
x = 3 cross multiply
5.5x = 3(x + 6)
5.5x = 3x + 18
2.5x = 18
x = 18
x = 7.2cm
A prism is a solid which has a constant cross-section. Examples are cubes, cylinders, rectangular prisms and triangular prisms
Cross section is the side of the prism that does not change.
volume of prism =area of cross-section × height (or length)
V = A × h
Example 1:Find the volume of the cylinder which has radius 3 cm and height 4cm, correct to two decimal places.
Solution: The cross section is the circular part of the cylinder
cross section = πr2
Volume of cylinder = πr2h =
π(3)24 = 113. 10 cm3
The formulas for determining the volumes of some ‘standard’ prisms are given here.
Pyramidsare shapes where the outer surfaces are triangular and converge at a point.The base of the pyramid can be square,triangular...
Volume of pyramid = 1 × base area × perpendicular height
For the square pyramid shown: V = 1 x2 h
Example: Find the volume of this hexagonal pyramid with a base area of 40cm2
and a height of 20 cm.Give the answer correct to one decimal place.
V = 1 ×A×h = 1 ×40×20 =266.7cm3
with each edge 10 cm and a height of 27 cm.
V = 1 ×A×h = 1 ×10×10 x 27 = 900 cm3
Volume of cone
Cones are shapes where the outer surfaces are rounded and converge at a point.
The base of the cone is circular.
The formula for ﬁnding the volume of a cone can be stated as:
Volume of cone = 1 × base area × height
Volume of cone = 1 × π r2 × height
The formula for the volume of a sphere is:
V = 4 π r3 where r is the radius of the sphere.
Example: Find the volume of the sphere with radius 4cm.
Volume of sphere = 4 π r3 = 4 × π × 43 = 268.08cm3
Using the shapes above, new shapes can be made.The volumes of these can be found by summing the volumes of the component solids.
Example: A hemisphere is placed on top of a cylinder to form a capsule. The radius of both the hemisphere and the cylinder is 5 mm. The height of the cylinder is also 5 mm. What is the volume of the composite solid in cubic millimeters, cor rect to two decimal places?
Volume of the composite = volume of cylinder+volume of hemisphere
= π r 2 h + 1 (4 π r3)
= π 52 × 5 + 1 (4 π 53)
= 654.50 mm3
The surface area of a solid can be found by calculating and totalling the area of each of its
surfaces. The net of the cylinder in the diagram demonstrates how this can be done.
Here are some more formulas for the surface areas of some solids.
Example:Find the surface of the right square pyramid shown if the square base has each edge 10 cm in length and the isosceles triangles each have height 15 cm.
TSA = 4 × 1 10 × 15 + 10 × 10 = 300 + 100 = 400cm2
Lengths, Areas,Volumes and Similarity
Definition 1: When two shapes are similar the ratio of their sides is k.
where k = scale factor
Ratio of lengths = length 1 =3 = k
length 2 4
Ratio of areas = area 1=π× 32=(3)2 = 9 =k2
area 2 π× 42 (4)2 16
Definition 3: When two shapes are similar the ratio of their volumes is k3.
Ratio of volumes = of volume 1=4/3π× 33 =(3)3 = 27 =k3
of volume 24/3π× 43 (4)3 64
The two triangles shown are similar. The base of the smaller triangle has a length of 10 cm. Its area is 40 cm2.
The base of the larger triangle has a length of 25 cm. Determine its area.
Ratio of lengths = length small = 10 = 2 = k
length big 255
k2 = 4 1
Ratio of areas = k2 = 40 2
1=2 and 4 = 40
x = 25 ×40 = 250 cm2
Example 2: The two cuboids shown are similar solids. The height of the larger cuboid is 6 cm. Its volume is 120 cm3. The height of the smaller cuboid is 1.5 cm. Determine its volume.
Ratio of lengths = length small = 1.5 = 1 = k
length big 64
k3 = 1 1
Ratio of volume= k3 = x 2
1=2 and 1 = x
x = 120 = 1.875 cm3
Example 1: Find the value of x correct to two decimal places.
Solution : sin 29.6°= x
x = 80sin 29.6°
x = 39.52 cm
Example 2: Find the length of the hypotenuse correct to two decimal places.
Solution : cos 15°= 10
AB = 10
x = 10.35 cm
tan x° = 11
x° = tan-1 11
When the triangles are not right-angled then the pythagoras theorem and the trigonometric
ratios are not appropriate for finding unknown lenghts and angles. The appropriate rules to use
are the sine rule and the cosine rule.
The sine rule
We use the sine rule when we are given:
1. Two sides and the non- included angle
2. One side and two angles
Considering that the lower- case letters represent the sides of the triangle and the upper- case represent angles, then the following formula states the sine rule.
a = b = c
sinA sinB sinC
We use the cosine rule when is given:
1. Two sides and the included angle of the triangle.
2. All sides.
Solution: Two angles and a side -sine rule
b = c
sin B sinC
10 = c
sin 70° sin 31°
c = 10 sin 31° = 5.48cm
Example 2: For triangleABC, ﬁnd the length ofABin centimetres,
correct to two decimal places.
Solution: Two sides and the included angle - cosine rule
AB2 = 102 + 52- 2×5×10 cos67°
AB = √(100 + 25 - 100 cos67°)
AB = 9.27cm
Solution: Three sides- cosine rule
cosB = 62 + 122 - 152
B = cos-1(62 + 122 - 152)
B =cos-1(-45/144) = 108.21°
Area of the triangle
It is known that the area of a triangle is given by theformula
Area = 1 × base length ×height
A = 1bh
When the triangle is not a right angled and the 2 sides and the included angle is given thenthe following formula can be used
A = 1 × side a × side b × sin (angle in between a and b)
Example 1: Find the area of triangleABC. Give your answercorrect to two decimal places.
The triangle is not right angled and we are given 2 sides and the included angle.
We can use the formula A =1/2 ac sinB
A= 1/2 × 6.5 × 7.2 × sin 140° = 15.04°
When the 3 sides of the triangle are given, then Heron’s Formula can be used to determine the area of the triangle:
Example 1: Find the area of the triangle with sides 6 cm, 4 cm and 4 cm. Give your answer correct to twodecimal places.
s = (6+4+4) = 7
A = √7(7 - 6)(7 - 4)(7 - 4) = √ 7 × 1 × 3 × 3 = √63 = 7.94cm2
Theangle of elevationis the anglebetween the horizontal and adirection above the horizontal.
Theangle of depressionis the anglebetween the horizontal and a direction below the horizontal.
Example 1: The pilot of a helicopter ﬂying at 400 m observes a small boat at an angle of depression of1.2◦. Draw a diagram and calculate the horizontal distance of the boat to the helicopter, correctto the nearest 10 metres.
Solution: The horizontal distance of the boat to the helicopter is AB. The angle of depression H = B ( alternate).
tan(1.2◦) = 400
AB = 400
AB = 19 095.80 m
Example 2: The light on a cliff-top lighthouse, known to be 75 m above sea level, is observed from a boatat an angle of elevation of 7.1◦. Draw a diagram and calculate the distance of the boat from thelighthouse, to the nearest metre.
Solution: tan (7.1◦) = 75
AB = 75
AB =602 m
Example 3: From a pointA, a man observes that the angle of elevation of the summit of a hill is 10◦. Hethen walks towards the hill for 500 m along ﬂat ground. The summit of the hill is now at anangle of elevation of 14◦. Draw a diagram and ﬁnd the height of the hill above the level ofA, tothe nearest metre.
Solution: We need to find HC
To find HC we need to find HB
3 angles and a side are given- sine rule
HB = 500
HB = 500 sin(10◦) = 1244.67 m
sin(14◦) = HC
HC = 1244.67 sin(14◦) = 301.11m
Bearings are used to indicate direction.
Thethree-ﬁgure bearing the True Bearing is the direction measured clockwise fromnorth and starts from 0° to 360°.
For the Compass Bearing we seperate the plane in 4 sections of 90°.
A True Bearing is 30°.
Compass Bearing = N 30°E
B True Bearing = 150°
Compass Bearing = E 30°B
C True Bearing = 210°
Compass Bearing = S 30°W
D True Bearing = 330°
Compass Bearing = W60°N
Example: The road from townAruns due west for 14 km to townB. A television mast is located duesouth ofBat a distance of 23 km. Draw a diagram and calculate the distance of the mast fromthe centre of townA, to the nearest kilometre. Find the bearing of the mast from the centre ofthe town.
Solution: AT2 = 142 + 232
AT = √( 142 + 232 ) = 26.93 km
For Bearing find θ
tan θ = 23
θ = tan-1(23/14) = 58.67°
Bearing = 270°- 58.67°= 211.33°
The bearing of the mast from A is 211.33°T
Example: A yacht starts from a point A and sails on a bearing of 038◦ for3000m. It then alters its course to a bearing of 318◦, and aftersailing for 3300 m it reaches a point B.
aFind the distance AB,correct to the nearest metre.
bFind the bearing of B from A, correct to the nearest degree.
Solution: a 2 sides and the included angle are given- cosine rule
AB = √(33002 +30002 - 2 × 3300 × 3000 cos 100°)
AB = 4830 m
b To find the bearing of B from A we need to find angle A
We may use sine and cosine rule
3300 = 4830
A°= sin-1 (3300sin100° ) = 42.28°
The bearing of B from A = 360◦ −(42.29◦- 38◦) = 355.71◦
The bearing of B from A is 356◦T , to thenearest degree.
Example : Two points,AandB, are on opposite sides of a lake so that the distancebetween them cannot be measured directly. A third point,C, is chosenat a distance of 100 m fromAand with anglesBACandBCAof 65◦and 55◦, respectively. Calculate the distance betweenAandB,correct to two decimal places.
Solution: 2 angles and a side- sine rule
AB = 100
AB = 100sin55°
AB = 94.59m
Problems in three dimensions are solved by picking out triangles from a main ﬁgure and
ﬁnding lengths and angles through these triangles.
Example 1:ABCDEFGHis a cuboid. Find:
cthe magnitude of angleHBD
ethe magnitude of angleHBA
a.Byselecting the appropriate triangle from the diagram and by using the Pythagoras Theorem:
DB2 = 82 + 102
DB = √(82 + 102) = √164 =12.81 cm
b. Using the information from a
HB2 = 72 + √1642
HB = √(49 + 164)
HB = √213 = 14.59 cm
θ = tan-1 7
θ = 28.66◦
d. Use the triangle HAD
HA2 = 82 + 72
HA = √(82 + 72 ) = √113 cm
e. 3 sides-cosine rule
cosB = 102 +√2132 -√1132
Example 2: The diagram shows a pyramid with a square base.The base hassides 6 cm long and the edges VA,VB,VC,VD are each10 cm long.
aFind the length of DB.
bFind the length of BE.
cFind the length of VE.
dFind the magnitude of angle VBE.
Give all answers correct to two decimal places.
a DB2 = 62 + 62
DB = √( 62 + 62 )
DB = 8.49cm
b BE = 8.49/2 = 4.24 cm
c VE = √( 102 - 4.242) = 9.06 cm
d sin (VBE) = 9.06
VBE = sin-1(9.06/10) = 64.90°
Example 3: A communications mast is erected at the cornerAof a rectangularcourtyardABCDwhose sides measure 60 m and 45 m. If theangle of elevation of the top of the mast fromCis 12◦, ﬁnd:
athe height of the mast
bthe angle of elevation of the top of the mastfromB(whereAB = 45 m)
Give answers cor rect to two decimal places.
aWe need to find AC first.
AC = √(452 + 602) = 75m
Then we can use AC to find the height of the mast.
tan(12°) = HA
HA = 75tan (12°) = 15.94m
b tanθ = 15.94
θ = tan-1( 15.94/45) = 19.51◦
The diagram above is called contour map.
The lines are called contour linesand they represent different hights above sea level.
The map gives the horizontal scaled distance between the lines and not the actual distance.
The real/cross-sectional representation of the contour map above is as follows:
To ﬁnd the distance between B and C, ﬁrst determine from the diagram the horizontal distance B′C′. Suppose this distance is 80 m.Then triangle BCH in theﬁrst diagram can be used to ﬁnd the distance between B and C and the average slope betweenB and C.
A cross-sectional proﬁle can be drawn from a contour map for a given cross-section AB.
This is illustrated below. The horizontal distance that is represented on the cross-sectional profile is the real distance.The distance AB on the contour map is 6cm then the distance on the cross-sectional profile will be 600cm which is 6m.
CONVERSION OF UNITS
When we are converting from smaller units to larger we divide
When we are converting from larger to smaller units we multiply
Example 1: Convert 10cm to m
Answer: There are 100 cm in a m
10cm /100 = 0.1m
a 1 cm (mm) = 1 × 10 = 10 mm
b 1 m (cm) = 1 × 100 = 100 cm
c 1 km (m) = 1 × 1000 = 1000km
d 1 mm(cm) = 1/10 = 0.1 cm
e 1 cm ( m) = 1/100 = 0.01 m
f 1m (km) = 1/1000 = 0.001 km
Conversion of units of area
1 cm2 (mm2) = 1 × 102 = 100 mm2
1 m2 (cm2) = 1 × 1002 = 10000 cm2 = 104 cm2
1 km2 (m2) = 1 × 10002 = 1000000km2 = 106 km2
1 mm2(cm2) = 1/102 = 0.01 cm2 = 10-2 cm2
1 cm2 (m2) = 1/1002 = 0.0001 m2 = 10-4 m2
1m2 (km2) = 1/10002 = 0.000001 km2= 10-6
aConvert 156000 m2 to km2
Answer: 156000 m2 / 10002 = 156000/1000000 = 0.156 km2
b Convert 20000mm2 to cm2
Answer: 20000mm2 /102 = 20000/100 = 200 cm2
1000mm3 (cm3) = 1000/103 = 1000/1000 = 1 cm3
1000000cm3(m3) = 1000000/1003 = 1000000/1000000 = 1 m3
1 litre = 1000cm3
1000 litres = 1m3