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Mathematics Teaching Self-evaluation Tools

You are viewing a limited version of the NCETM’s self-evaluation tools. Any answers you save during this session will be removed after seven days. Log in or sign up to view, and use, the full version of the tools.

For the following questions, select the statement which most accurately matches your level of confidence (1 is not confident and 4 is very confident) or choose from the alternatives detailed in the question. You do not have to answer all questions. Your answers will be saved so you can exit and come back to your self-evaluation at any time. Click Save and Results to view the next steps for questions you have answered.

1.

How confident are you that you understand

1

2

3

4

a.

the distinction between mass and weight?

Mass is the quantity of matter in an object, measured, for example, in grams and kilograms. Technically this is not the same thing as weight.

Weight is the force of gravity acting on an object and is therefore properly measured in newtons. It varies depending on how far the object is from the centre of the Earth. Colloquially, ‘weight’ is used as a synonym for ‘mass’.

When we weigh an apple and say it weighs 100 grams, we really mean that it weighs the same as a mass of 100 grams. But what we experience when we hold the apple in our hand is its weight, i.e. the force of gravity upon it; we exert a similar force on it to hold it up.

b.

the distinction between volume and capacity?

The volume of an object is the amount of three-dimensional space that it occupies. The capacity of a container is the amount of three-dimensional space inside it, or the maximum amount of liquid volume that it will hold.

Colloquially, volume is often used instead of capacity. For example, we may say a mug has a volume of half-a-pint, meaning its capacity (rather than the volume of clay used to make it).

Liquid volume or capacity and solid volume are conventionally measured in different units, although the concepts are the same. Liquid volume is measured in litres and millilitres (or pints), and solid volume is measured in cubic centimetres, cubic metres, and so on.

In the metric system, units for liquid and solid volume are related in a simple way: 1 ml is the same volume as 1 cm^{3}, and 1 litre is the same volume as 1000 cm^{3}.

c.

the accuracy of measurements?

No measurement on a continuous quantity is exact. Every measurement is limited by the accuracy of the equipment used to make it. It is usual to give any measured value to the nearest whole unit or decimal place (e.g. to the nearest division on a scale).

Measurements given to the nearest whole unit may be inaccurate by up to one half of a unit in either direction. For example:

A length d m is given as 36 m to the nearest metre so 35.5 ≤ d < 36.5.

A volume V cm^{3} is given as 240 cm^{3} to the nearest 10 cm3 so 235 ≤ V < 245.

A mass m kg is given as 2.3 kg to the nearest 0.1 kg so 2.25 ≤ m < 2.35.

d.

compound measures?

Compound measures have two dimensions and require calculation.

Rate is a way of comparing how one quantity changes with another, e.g. a car’s fuel consumption measured in miles per gallon or litres per 100 km.

The two quantities are usually measured in different units, and ‘per’, the abbreviation ‘p’ or a forward slash ‘/’ is used to mean ‘for every’ or ‘in every’.

If a rate is constant, then the two variables are in direct proportion and are connected by a simple formula. For example:

speed =

distance travelled

time taken

density =

mass of object

volume of object

pressure =

force on surface

area of surface

If a rate varies, the same formula can be used to calculate an average rate. For example, if a cyclist travels 36 miles in 3 hours, her average speed is 12 mph.

For example:

A car has a fuel consumption of 7.3 litres per 100 km. The car has 64 litres of fuel when it starts a journey.
How far can it travel without refuelling?

It can travel

64

7.3

× 100 = 877 km without refuelling.

Leroy drives 60 miles in 1 hour 30 minutes.
Give his average speed in miles per hour.

1 hour 30 minutes is 1.5 hours so the average speed is

The bearing of a point A from an observer O is the angle between the line OA and the north line through O, measured in a clockwise direction. Bearings are always given using three figures. For example, in the diagram:

C is 12 km from B on a bearing of 065°, and D is 10 km on a bearing of 150° from C.
Construct a scale drawing, using a scale of 1 cm to 2 km, and give the distance and bearing of B from D.

The angle x is 284° and the distance from D to B (when the scale is 1 cm to 2 km) is 8.2 cm. So B is 16.4 km from D on a bearing of 284°.

3.

How confident are you that you know how to find:

1

2

3

4

a.

the area and perimeter of sectors and segments of a circle?

Find the area and arc length for the shaded sector shown.

Area of sector

Arc length

The perimeter of the sector can then be found using the radius:
10 + 10 + 6.98 ≈ 27.0 cm.

Find the area of the shaded segment.

Area of segment = area of corresponding sector – area of subtended triangle

Area of sector =

Area of triangle

So

Area of segment

The chord length forming the boundary of the segment with the arc can be found using the isosceles triangle: chord = 2 × 9 sin 60 = 9 √3

Perimeter = .

4.

How confident are you that you know how to use:

1

2

3

4

a.

the properties of simple solids to solve problems involving surface area and volume?

Here are some key volume formulae:

Volume of a prism = cross-sectional area × length

Volume of a cylinder = πr^{2}h
where r is the radius of the cross-section and h the height (or length)

Volume of a pyramid =

1

3

(base area) × height

Volume of a cone =

1

3

πr^{2}h
where r is the radius of the base and h the perpendicular height (or length)

Volume of a sphere =

4

3

πr^{3}
where r is the radius

A frustum is a truncated cone. Find the volume of this frustum.

Imagine the section of cone cut off. It has a base radius of 10 cm. This is half the base radius of the ‘complete’ cone so the complete cone had a height of 60 cm and the truncated part had a height of 30 cm.

Volume of frustum

= volume of complete cone – volume of truncated section
=

Surface area of a cylinder (including the ends) = 2πrh + 2πr^{2} = 2πr(h + r) where r is the radius of the cross-section and h the height (or length)

Surface area of a cone (including the base) = πrs + πr^{2} = πr(s + r) where r is the radius of the base, h the perpendicular height (or length) and s is the slant height of the cone ()

Surface area of a sphere = 4πr^{2} where r is the radius

5.

How confident are you that you understand:

1

2

3

4

a.

The trigonometric ratios sine, cosine and tangent and know how to use them to solve relevant problems?

The main trigonometric functions are cosine, sine and tangent. Other functions are reciprocals of these.

The diagram shows a circle centre O with radius r. The point P on the circumference has coordinates (x, y). The angle θ is measured anticlockwise from the x-axis. The trigonometric functions are defined in terms of r and the coordinates x and y.

The functions may be introduced as functions of angles in a right-angled triangle. Cosine, sine and tangent are defined as the ratios of sides in the triangle.

where a is the side opposite the angle A, b is the side adjacent to the angle A and c is the hypotenuse.

Example: In the diagram below which sides can be given by:

(a) 8 tan 30

(b) 8 cos 60

(c) 8 sin 60

(d) 8 cos 30

(e) 8 sin 30

b.

the equivalences of the trigonometric functions beyond the range –180° to 180°?

The graphs of the trigonometric functions illustrate their cyclic nature.

In particular:

sin x = sin (180° – x)cos x = –cos (180° – x)

sin 30° = sin 150°

cos 60° = cos 120°

sin (–x) = –sin(x)

cos (–x) = cos x

For example:

sin –30°

= –sin 30°

cos –60°

= cos 60°

Also, adding or subtracting any multiple of 360° to an angle will leave the value of both sine and cosine unchanged.

Whereas the sine and cosine functions are applied to right-angled triangles, the sine rule and cosine rule show relationships between sides and angles that are true for all triangles.

Using the label conventions indicated in this diagram:

the sine rule states:

a

sin A

=

b

sin B

=

c

sin C

or

sin A

a

=

sin B

b

=

sin C

c

You can use the sine rule when you know two sides and one of the angles which is not between the two sides, or if you know a side and two angles (since you can find the third angle from the angle sum in a triangle)

The cosine rule states:

a^{2} = b^{2} + c^{2} – (2bc cos A)

or cos A =

b^{2} + c^{2} –a^{2}

2bc

You have to use the cosine rule if you only know two sides and the included angle, or if you know all three sides.

If you know all three sides, you should find the angle opposite the largest side first since it is the only one which might possibly be obtuse.

You only ever need to use the cosine rule once – you must then have enough information to allow you to use the simpler sine rule

For each rule, the first formulation is more useful for finding missing side lengths while the second is more useful for finding missing angles. If you have a right angle in the triangle then it is better to use standard trigonometry and Pythagoras than the sine or cosine rule.

AB = 8.7 m, BC = 7.5 m, Angle DAB = 620, Angle DBA = 650, Angle DBC = 420. Calculate the length CD.

In triangle ABD, you have two angles and can therefore find the third angle ADB is 53°, and then use the sine rule to find length BD. In triangle DBC you then have 2 sides and the included angle and the cosine rule will give the required length CD.

Because AB and CD are parallel, CAB = 320 and so ABC = 950. The sine rule in ABC will give the length BC and then the cosine rule in ACD will give the required length AD.

AC

sin 95

=

65

sin 53

⇒ AC = 81.079

AD^{2} = AC^{2} + 93^{2} – 2 × AC × 93 cos 32 = 2433.65

AD = 49.332 = 49.3 cm

d.

why caution is sometimes needed in interpreting results from the sine rule?

In some situations the sine rule can give ambiguous results. In the following diagram, a triangle is being constructed with an angle of 30°. One side adjacent to the known angle has lengths 7.2 cm and the side opposite the angle is 4 cm. The diagram shows that there are two possible triangles with these properties.

This ambiguity is reflected when we use the sine rule to find angle C.

sin C

c

=

sin A

a

So sin C = 7.2 ×

sin 30

4

= 0.9

So C = sin^{-1} 0.9

Angle C must be between 0° and 180°. There are two such angles with sine of 0.9: 64° and (180 – 64)° = 116° (both to the nearest degree).

So angle C_{1} = 116° and angle C_{2} = 64°

6.

How confident are you that you understand and can use

1

2

3

4

a.

the formula for the area of a triangle: A =

1

2

ab sin C ?

The triangle area formulaA =

1

2

ab sin C is true for any triangle. It can be used to find the area of a triangle when two side lengths and the angle between them are known. By rearranging, it can also be used to find a missing angle or side if the area and the other corresponding dimensions are known.

Find the area of this triangle.

Using the formula A =

1

2

ab sin C with a = 3.2, b = 7.1 and C = 46 gives:

A =

1

2

× 3.2 × 7.1 × sin 46
= 8.2 (to 1 d.p.)

The area of the triangle is 8.2 m^{2} (to 1 d.p.).