# Unit 3 Revision Notes - Higher

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Unit 3 Revision Notes - Higher
Angles: Parallel Lines a b Corresponding angles ‘a’ and ‘b’ are equal Parallel lines are shown by having arrows drawn on them c d Alternate angles ‘c’ and ‘d’ are equal Notes: 1) Use the correct words: Alternate and Corresponding in the exam 2) Useful for Bearings questions. The Bearing of A from B is 120 degrees what is the Bearing of B from A? N Bearing of B from A = 300 degrees N 120 120 N B B A A 180

why each angle is the same!
3) Used when showing that two triangles are similar/congruent – remember to explain why each angle is the same! A B C D E Show that ABC is similar to CED Angle A = Angle E (alternate) Angle B = Angle D (alternate) Angles at C are the same (vertically opposite) Angles: Circle Thoerems Chord my not always be visible! The angle subtended by a chord at the circumference is half that at the centre. The angles subtended (made) by a common chord at the circumference are equal. The angle made by a diameter at the circumference is 90 degrees.

The angle between a tangent and chord is equal to the angle subtended (made) by a chord in the alternate segment. Note when doing circle theorems questions watch out for isosceles triangles – in particular triangles with two sides as a radius of the circle. Also remember that a tangent to a circle will always make a 90 degree angle with the radius at the point where they meet.

THESE FORMULA ONLY WORK FOR RIGHT ANGLED TRIGONOMETRY!!
Eg. What are the missing angles? Give reasons. A = 120 (angle at centre = twice that at circumference) B = 60 (angles at circumference are constant) C = 120 (opposite angles in a cyclic quad add to 180) Right Angled Trigonometry THESE FORMULA ONLY WORK FOR RIGHT ANGLED TRIGONOMETRY!! IF YOU SEE A RIGHT ANGLED TRIANGLE (INDEED A RIGHT ANGLE!) ANYWHERE IN THE EXAM THINK PYTHAGORAS AND (SOHCAHTOA) Eg1 Find the angle X We know 8 = adjacent 12 = hypotenuse Cos X = 8/12 X = Cos-1 (8/12) X = 48.2 degrees (1 d.p.) 12 cm X 8 cm Remember to use shift button on calculator to get the angle using cos inverse)

Both of these are Useful for finding an angle
Eg2 Find the side x We know 12 = opposite 70 = angle Tan 70 = 12/x x = 12/tan(70) x = 4.4 cm 12 cm 70o This is what I calla ‘hard side’ question. You need to get the rearrangement correct from the first to the second line. x Eg3 Find the side x We know 6 = opposite 70 = angle Sin 70 = x/6 6 × Sin 70 x = 5.6 cm 6 cm x cm 70o This is what I call an ‘easy side’ question. You need to get the rearrangement correct from the first to the second line – but it is easier to do than in eg 2 above. Trigonometry for general Triangles These rules work for any triangle. If you see a non right angled triangle you have to use them. Useful rearrangements but you need to learn them if you want to use them On the formula sheet in the exam: Both of these are Useful for finding an angle

Use the Cosine rule when:
You want to find a third side and you have 2 sides and an angle in between You want to find an angle and you have 3 sides 50O a 10 12 A 8 10 12 Use the Sine rule when: The Cosine rule doesn’t work! This will be when you have a side and an opposite angle as a pair. 50O 10 12 B

Eg 3. A yacht sets off from A and sails 5km on a bearing of 067 degrees to a point B so that it can clear the headland before it turns onto a bearing of 146 degrees. It then stays on that course for 8km until it reaches point C. Find the distance AC. The trick here is to find the angle at B. Clearly this is a cosine rule question. Use a diagram and corresponding angles to do this. N N N 2 sides but need angle in between so redraw and use angle facts with parallel lines to find it! 146 – 67 = 79 67 146 67 B 67 B 5 A A 8 C C 101 b2 = c2 + a2 - 2×c×a×Cos B b2 = ×5×8×Cos 101 b2 = ×Cos 101 b2 = b2 = So b = 10.2 km Be careful with the double negative!!!!! (Cos of a value above 90 is always negative)

Internal and External Angles of Regular Polygons
Internal angles are those on the inside of the polygon For a triangle the internal angles sum to 180o For a quadrilateral they sum to 360o For a pentagon 540o. (Each extra side adds on 180o) Regular shapes are shapes where the internal angles are all the same and where the side lengths are all the same. This is a Regular Pentagon. External angles always add up to 360 degrees and are the amount you turn at each corner. The external angle of a regular polygon with ‘n’ sides is 360 ÷ n The internal angle of a regular polygon = 180 – external angle Answer: External angle in a regular octagon = 360 ÷ 8 = 45o Internal angle = 1800 – 45 = 135o At Angle here = 360 – 135 – 135 = 90o

Graph Transformations
Vertical stretch: Times whole expression y = 2f(x) Move up: Add on the end y = f(x) + 1 Original Graph y = f(x) This is drawn as dotted line in each example Move sideways: Add to x (but works ‘wrong’ way) y = f(x + 2) Stretch sideways:– Times the x (but works ‘wrong’ way) y = f(3x)

Look at the graph of y = f(x) opposite. On the same axes plot
y = f(x)-3s Translate down 3 b) y= 2f(x) Stretch by factor 2 vertically – all the points are twice as Far away from the X-axis as they Were to begin with y = f(x)-3

½ horizontally – all the points are half as far away from the
Look at the graph of y = f(x) opposite. On the same axes plot the graph of: y = f(2x) Stretch by factor ½ horizontally – all the points are half as far away from the y-axis as they were to begin with b) y = f(x - 4) Translate +4 horizontally y = f(2x) y = f(x - 4) Note: The graphs y = f(x) and y = -f(x) are related by reflecting the shape in the x axis Similarly y = -2f(x) is a reflection and a stretch of scale factor 2 vertically

Using Graphs to solve equations
An important principle here is that if you want to solve any equation: Eg 2x + 1 = 3x – 5 Then you can do this by plotting the graphs of what is on each side and then seeing where they meet. In this case plot the graphs y = 2x + 1 (y = what is on the left) and y = 3x – 5 (y = what is on the right) Where the graphs meet the x values will give the same y vale ie the equation will be sovled. There may be several values for x. Eg. Use the graph of y = cos Ө shown below to find all the solutions to cos Ө = in the range 0o ≤ Ө ≤ 720o Draw the line y = -0.7 Find one solution using cos inverse: cos Ө = - 0.7 Then Ө = cos-1 (- 0.7) = 134.4 Then use the fact that the graph is symmetrical about 180 to find 3 other solutions: or or 585.6 134 225.6 134 225.6 and (Remember the graph repeats every 360 degrees, so adding or taking 360 degrees to any solution will get another)

This is the graph of y = x2 – 5x Use it to solve x2 – 5x = 0
Easy just see where y = x2 – 5x meets the graph y = 0 (ie the x axis) Clearly 0 and 5 b) Again just see where y = x2 – 5x meets the graph y = -4 Clearly 1and 4 by drawing the line y = -4 on the graph This is the hard one! Start with the equation you want to solve x2 – 6x + 5 = 0 Add and subtract things to both sides to rearrange the equation to make one side look like the equation of the graph you have been given (in this case x2 – 5x). x2 – 6x + 5 = Add x and take 5 from both sides give us: x2 – 5x = x - 5 Now plot the graph of y = x – 5 and see where the lines meet. In this case x = 1 & 5.

Graphs you need to know:
General quadratic y = ax2 +bx + c (usually have two solutions which are the x values where the graph crosses the x axis) General cubic y = ax3 +bx2 + cx + d (usually have three solutions which Inverse Proportion graph y = k/x Straight line graph y = mx + c straight line graph The direct proportion graph y = kx looks like this but always Go through the origin y = sin(x) y = cos(x) and y = tan(x) you should be familiar with but you don’t need to be able to sketch them for GCSE maths

Exponential graphs like
y = 2x, y = 3x , y = 4x etc.. These all look like the shape on the right, which is the Graph of y = 2x when x = 1 y = 21 = 2 when x = 2 y = 22 = 4 when x = 3 y = 23 = 8 etc Transformations Transformations are ways to change a shape. There are 4 main transformations: Enlargements (shape’s size is changed so that is still in proportion) Translation (shape’s position is changed) Rotation (shape’s orientation is changed) Reflection (shape’s orientation is changed) You can get other transformations by combining these together. You can also define new transformations (like a stretch which enlarges a shape in 1 direction only – these are not on the syllabus apart from as part of graph transformations)

A reflection is obtained by ‘folding’ the space over a mirror line.
Reflections A reflection is obtained by ‘folding’ the space over a mirror line. The shortest distance of any point on the original object to the mirror line should be equal to the shortest distance of its image to the mirror line. To describe a reflection you must have: An Object A line of reflection described as an equation (Remember vertical lines are x = lines, horizontal lines are y = lines, the line y = x goes through (0,0), (1,1), (2,2) etc When reflecting in a diagonal line it is easier to turn the line of reflection so that it is horizontal or vertical) OBJECT IMAGE

Rotations To describe a rotation you must have: An Object An Centre of Rotation (an axis) described as a coordinate An Angle of Rotation A Direction to rotate in (usually clockwise/anticlockwise) (Note: Tracing the object onto scrap paper and rotating the paper using a pencil nib as an axis is a good way to avoid making mistakes. Ask for tracing paper if you need it!) A rotation is obtained by turning a space around a fixed point (we call this the centre of rotation). The distance of every point from the centre of rotation will remain constant as the space is turned.

To describe a translation you must have: An Object A Vector
Translations A translation is obtained by moving a shape in space. Each corner of the object is moved by an amount described in a vector. The image should be congruent to the object and of the same orientation. To describe a translation you must have: An Object A Vector The top line of the vector describes the horizontal movement (+ve right, -ve left) The bottom line describes the vertical movement (+ve up, -ve down) In this case the Object has been translated by 4 to the right and 2 up As described in the vector shown. IMAGE OBJECT

3) Multiply this distance by the scale factor
Method: Draw lines from the centre of enlargement through all the corners of the object 2) Find the distance of the first corner from the centre of enlargement (counting squares is best) 3) Multiply this distance by the scale factor 4) Measure this new distance from the centre of enlargement and mark it as the first corner in the enlarged image Repeat for every corner until the enlarged shape is ready to be drawn in. NOTE: If the Scale Factor is –ve, measure in the opposite direction when plotting the image. The image will be upside down and on the other side of the centre (essentially it has been rotated by 180 degrees at the same time as being enlarged). If the Scale Factor is between -1 and 1 the shape will actually get smaller! Eg Enlargement Scale Factor 3 with (2,1) as the centre of enlargement

Enlarge the shape shown by scale factor -2 using the point B as the centre of enlargement
OBJECT B IMAGE

Enlarge the shape shown by scale factor ½ using the point C as the centre of enlargement
OBJECT IMAGE C

Area and perimeter of 2d shapes
Area of a trapezium = ½ (a + b) × h Remember: This is on the formula page! Sector = a pizza slice from a circle Arc = part of the circumference of a circle Segment = pieces sliced across a circle Area of a circle = ½ π × radius2 Area of a sector = ½ π × radius2 × a h – perpendicular distance between the two parallel sides b Circles: Radius = distance from the middle to a circle to the circumference Diameter = width through the middle of a circle Diameter = 2 × radius radius θ

Circumference of circle = π × diameter
Arc length = π × diameter × diameter θ Eg 1 Find the Perimeter of the sector below which is made from a circle of perimeter 5 cm Eg 2 Find the Area of the segment shown if the radius of the circle is 8 cm 100o 300o Area of segment = Area of sector– Area of triangle Arc of sector = π × 82 × = Area of Triangle = ½ × 8 × 8 × Sin 100 = Area of Segment = – = 24.3 cm2 Perimeter consists of 2 radii (each 5 cm) and an arc Arc length = π × diameter × = Total perimeter = = cm

Triangles: Area of a triangle = ½ base × height OR Area of a triangle= ½ a × b × Sin C Remember to use the vertical height (ie from the base to the tip of the triangle when using this formula) Use this formula when you have two sides of the triangle and the angle in between – it is on the formula page! Eg Find the area Eg 1 3cm 50O 10 cm 12 cm 10 cm Area = ½ 10 × 3 = 15 cm2 Area = ½ 10 × 12 × Sin 50 = 91.9 cm2 Eg 2 4cm 8 cm Area = ½ 8 × 3 = 16 cm2

Eg Answer: The key to this question is that the areas are equal so start by writing down an expression for the area of each shape: (3x +2)(x +1) = Area of rectangle (3x + 5)x + 5x = Area of L shape Areas are equal: (3x +2)(x +1) = (3x + 5)x + 5x 3x2 + 5x + 2 = 3x2 + 5x + 5x 5x + 2 = 10x x = 2/5

Volume and Surface Area of 3d shapes
A prism is a 3d shape where the cross section is uniform (unchanged) Examples of prisms – cuboids, cylinders Examples of non-prisms – pyramids, cones, spheres To find the volume of any prism calculate the area of the cross-section and multiply it by the length of the prism: Eg 1 Find the Volume of the prism below Area of cross-section (blue) = 6 x x 4 = 64 Volume = 64 x 10 = 640 cm3. Eg 2 The prism below has a volume of 200 cm3. If it is 12 cm long what is the area of the cross – section? Volume = Area of cross-section x length = Area x so Area =200 ÷ 12 = cm2 8 cm 4 cm 6 cm 10cm 12 cm

These are in the formula booklet
Volume of a sphere = 4/3 π × radius3 Surface area of a sphere = 4 π × radius2 Volume of a pyramid = 1/3 Area of Base x height Volume of a cone =1/3 π × radius2 x height (because it is a pyramid with a circular base, so the area of the base is π × r2 ) Surface area of a cone = Area of base Area of curved surface π × radius π × radius x slope length These are in the formula booklet Slope length Note: The formula for the curved surface of a cone is the only formula that uses the actual slope length rather than the height. The curved surface part of the formula (π × radius x slope length) is on the formula page. radius Surface area of a cylinder = Area of top and bottom + Area of curved surface 2 x π × radius π × diameter x height (Two circles) (Rectangle that has equal to the circumference of the top/bottom)

Answer Radius of sphere = 3 so height of cylinder = 9cm Cylinder = π x 32 x 9 = 81π Hemisphere = 4/3 π x 33 ÷ 2 = 18 π Volume = 99 π Volume of liquid is 49.5 π 18π in the hemisphere So 31.5π = Volume of cylinder 31.5π = π x 32 x d d= 3.5 cm

Join these edges together
Eg A cone is made by cutting a 60 degree sector from a circle of radius 8 cm as show in in the picture. This is a classic A* grade question. Method: Find the length of the arc from A to B. This length is the same as the circumference of the base of the cone. Use this fact to find the radius of the base of the cone. Arc length = 16 × π × = Then π × diameter = diameter = radius of cone base = Use the fact that the slant height of the cone OA will be equal to the radius of the original circle. Then use Pythagoras with this length and the radius of the base of the cone to find the cone’s height. 300 Join these edges together A O B O What is the volume of the cone? 8 A & B h Volume = 1/3 × π × × = cm3 2

Eg The diagram shows a solid made from a cone and a hemi-sphere.

Enlargement Scale Factors
If two objects are similar it means that their side lengths are related by a multiplying factor called the enlargement scale factor. To find the enlargement scale factor divide a length in the image by a corresponding length in the object. Note: If the enlargement scale factor for length is x then the enlargement scale factor for corresponding areas is x2 and the enlargement scale factor for corresponding volumes is x3 If triangles ABC and CED are similar, find: The length BC The area of triangle CED given that the area of triangle ABC is 8 cm2 A B C D E 4 cm 6 cm 12 cm Answer: Enlargement scale factor = 6÷4 = 1.5 So BC = 12 ÷ 1.5 = 8 cm Area scale factor from ABC to CED = 1.52 So area of CED = 8 x = 18cm2

Answer This is a volume scale factor question. Enlargement scale factor for length = 30÷20 = 1.5 So volume scale factor = (1.5)3 Small bottle holds 480 ml so bigger bottle holds 480 x (1.5)3 = 1620 ml Eg Two solids are similar. The area of one of the faces is 20 cm2 on the larger shape and 5 cm2 on the smaller shape. If the smaller shape has a volume of 8 cm3 what is the volume of the larger shape? Area scale factor = 20 ÷ 5 = 4 So length scale factor will be the square root of the area scale factor, so = 2 Volume scale factor is the length scale factor cubed, so is 23 = 8 in this case. Hence volume of larger solid is 8 x 8 = 64 cm3

Answer: You can use enlargement scale factors to make this question a lot quicker. The volume of the cone missing from the top is = 1/3 × π × 22 × 3 = 4π The larger cone is similar to the smaller one. The length scale factor is 3 by comparing the radii. This makes h = 9 cm. So volume of the whole cone including the missing top which shouldn’t really be there is 1/3 × π × 62 × 3 = 36π The volume of the frustum = 36π - 4π = 32 π 5 marks

f a If you have a situation like this The vector between the
Vector Geometry - Important principles: If you have a situation like this The vector between the end points of the vectors f and a: FA = a - f 2) The vector between BA is the negative of the vector from AB BA =- AB They often ask you to find easier paths at the start of the question and you need to use these paths to help you with the harder paths at the end of the question. 4) If you asked to prove that two vectors are parallel you need to show that one of them is a multiple of the other F A f a O

Answer A) OS = OP + ½ of PQ We have OP = p We need PQ in terms of p and q. Looking carefully PQ = -p + q or q – p So OS = p + ½(q – p) = p + ½q - ½p = ½q + ½p B) RS = RP + PS RP = ½p and PS = ½PQ = ½(q – p) from part a So RS = ½p + ½(q – p) = ½p + ½q – ½p = ½q Since OQ = q clearly RS is a multiple of OQ ( ½OQ to be exact), hence parallel

You need to be able to solve quadratic equations by either factorising, using a graph or by using the quadratic formula. The formula is on the formula page. Expect to see quadratics popping up in strange places and don’t be concerned if you see one as part of a question on area or volume – just rearrange to make the equation equal to zero and solve. If you can’t get the quadratic equation to work then check you r arithmetic – especially the negative numbers

Solving simultaneous equations
If, as part of a question you write down an equation with two variables in it you will not be able to solve the equation uniquely unless you find another equation. You need two equations to solve uniquely for a system containing two variables. IF BOTH EQUATIONS ARE LINEAR Rearrange the equations and then add and subtract multiples of them together to eliminate one of the variables. You can then solve. Eg Solve 3x = y 2x + 3y = 11 Rearrange to get letters on the same side in both equations 3x – 2y = 2x + 3y = Multiply equation 1 by 2 and equation 2 by 3 to get the number of xs in both equations to match: 6x – 4y = 6x + 9y = Since the signs are the same on the x’s subtract equation 2 from equation 1 -13y = now solve to get y = 1 and from that we can see that x = 4 Note: Be careful when subtracting/adding the equations.

IF ONE QUATIONS IS LINEAR BUT THE OTHER IS QUADRATIC
Rearrange one equation – the linear one! To get an explicit equation for one variable. Then replace this variable in the second equation using the equation. Eg Solve y2 + 16= 8x 3 + y = 2x Answer Rearrange the second equation to get y = 2x – 3 now replace y in the ‘harder’ equation with 2x – 3 - be careful to use brackets if they are necessary. (2x – 3) = 8x (2x – 3)(2x – 3) + 16 = 8x 4x2 – 12x = 8x Its a quadratic equation so rearrange to make it equal to zero and then solve 4x2 – 20x + 25 = 0 (2x – 5)(2x - 5) = 0 x = 2.5 or when x = 2.5 , y = 2 Note: This question could be phrased by asking you where the curves y2 + 16= 8x and 3 + y = 2x meet. Note also that you expect two values for x and 2 corresponding y values, but the quadratic in this case only gives one solution (because it is repeated)

Answer a) Parallel lines so 2x + y = 180 because: This will be y as these are alternate angles b) 6x + 2y – 100 = 360 Since angles in a quad Sum to 360 degrees Adding 100 6x + 2y = 460 Dividing by 2 3x + y = 230 c) Solve simultaneously 2x + y = 180 Subtracting X = 50 so y = 80 (c) Find the values of ‘x’ and ‘y’

Solving Equations by Trial and Improvement

Method: Guess a number that might work Try it out – if the answer is too low make a second guess that is higher - if the answer is too high make a second guess that is lower. Keep guessing, each time going half way between your best two results Note: Your final answer is the guess with the correct number of decimal places that is the closest fit. You must demonstrate for sure that your answer is the best. Solve (to 2 decimal places) x2 + 2x = ) x2 + x = 80 3) x2 – 3x = 100 x3 – x = 75 5) x3 + x2 = 100 1) x2 – x = 60 2) x2 – 3x = 100

straight line going through the origin)
Proportion Two things are in direct proportion if the value of the second can be found from the value of the first by a multiplication. In other words: (This means that the graph of y plotted against x would be a straight line going through the origin) (A quick check is to see if doubling one quantity would lead to an expected doubling in the other quantity. Eg If you travel for twice as long will the distance you travel double?) The other types of direct proportion to watch out for are: A is directly proportional to B A is directly proportional to B 2 A is directly proportional to the square root of B

m per second when it has been falling for 2 seconds.
Eg The speed at which ball falls to the ground when dropped from a tall tower is directly proportional to the square of the time it has been falling for. The ball reaches a speed of 12 m per second when it has been falling for 2 seconds. a) How long will it take for the ball to reach a speed of 100 m per second? b) When the ball has been falling for 20seconds how fast will it be travelling? Answer v = speed t = time v is directly proportional to the time squared so v = k × t2 When t = 2 v = so = k × 22 12 = k × 4 k = 3 v = 3 × t2 a) Now use the formula to answer the questions by solving the relevant equations v = what is t? = 3 × t2 so t = b) When t = 20s what is v v = 3 × v = 1200 m per second METHOD Write down the proportionality equation with the constant k Use the info in the question to find a value for k Write down the formula with k now known

If y and x are inversely proportional then we say:
But think of it like this: y × x = k A good way to think about inverse proportion is that as one thing doubles the other halves. Eg If I run twice as fast will I get there in half the time? If so then it is likely to be inverse proportional Note that: y is inversely proportional to the square of x But think of it as y × x2 = k to the square root of x But think of it as y × = k

So p = number of people h = number of hours
Eg If it take six people 8 hours to complete a job how many hours will it take 4 people? Answer Often proportionality questions mention the words inverse or direct proportion in the question, but with this example you just need to recognise that this is likely to be an inverse proportion situation. Twice as many people mean the job completing in half the time would seem reasonable! So p = number of people h = number of hours p is inversely proportional to h so p × h = k Use info from the question to find the value of the constant ‘k’ 6 × 8 = k so k =48 Formula is p × h = k = 48 So when p = 4 we need to solve for h × h = 48 implies that h = 12 hours METHOD Write down the proportionality equation with the constant k Use the info in the question to find a value for k Write down the formula with k now known

Pythagoras If you see a right angled triangle think Trigonometry and Pythagoras! If you need to find side lengths and can’t see a way forwards sometime drawing in a perpendicular to make a right angled triangle is helpful as you can then use pythagoras. Eg Find the area of this isosceles triangle Dropping a perpendicular from the top of the triangle makes finding the height obvious using pythagoras 132 = h2 + 52 169 = h2 + 52 169 = h2 + 25 144 = h2 12 = h can ignore -12! 13 cm 13 cm 13 cm 13 cm h 10 cm 10 cm Hence area = 12 x 5 = 60 cm2

Eg Find the value of x. Answer
If a right angled triangle then (x + 5)2 + (x – 2)2 = 102 (pythagoras theorem) x2 + 10x x2 - 4x +4 = 100 2x2 + 6x + 29 = 100 This is a quadratic so 2x2 + 6x – 71 = 0 make it equal to zero and solve The positive solution is 4.64 from the formula Expand out each bracket properly: Remember (x + 5)2 (x + 5)(x + 5) = x2 + 10x + 25 Note: You can also use pythagoras to prove that a triangle is right angled. It must be if the three sides fit the rule a2 + b2 = c2

Constructions

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