Grade 12 Mathematics Revision Questions (Including Solutions) unimaths.co.za Get read for universit mathematics b downloading free lessons taken from Unimaths Intro Workbook. Visit unimaths.co.za for more info. 1
Algebra Questions Eas Algebra Question A1: Solve for : (3 )(5 ) = 3 Epand brackets and rearange the epression into standard form: a 2 + b + c = 0 Question A2: Solve for and simultaneousl: 3 = 2 2 2 + 2 = 1 Let = 2 3 then substitute, in 2 2 + 2 = 1, with 2 3. 2
Question A3: Given the graph of f () = 3 + a 2 + b, below, determine a and b. B(2, 10) A( 1, 3.5) You have two unknowns and ou are given two points on the graph. Solve for a and b using simultaneous equations. 3
Question A4: Determine the average gradient of f () = 3 + a 2 + b, below, between the points A and B. B(2, 10) A( 1, 3.5) The average gradient of f (), between the points A and B, is equal to the gradient of the straight line that runs through A and B. Question A5: Find the gradient of f () = 4 3 9 at = 3 Epress the terms of f () in standard form, ie. a n and then use the polnomial rule to differentiate each term. 4
Question A6: Given the graph of f () = a, find a. A( 1, 0.5) 1 A negitive eponent inverts the base. Question A7: If f () = 2 and g() = 100.3, determine the value of for which f () = g(). Set f () = g() and take logs of both sides. Question A8: Given the following arithmetic sequence: Calculate the value of p. 1 p; 2p 3; p + 5;... Think of two different was to epress the common difference in terms of p. 5
Question A9: Given f () below, determine f ( 7 2 ) : 5 4 T P( 1, 9 4 ) Use the form: f () = a( p) 2 + q to determine the function. Question A10: Find the -coordinate of the inflection point of f () = 3 + 2 + 8 12. The inflection point occurs when the rate of change of the function changes. In other words, find the turning point of the derivative. 6
Question A11: Simplif completel: ( 1 + 2 2 ) 2 8 2 8 = 2 2 Medium Algebra Question A12: Solve for : 3 + 1 = 4 Once ou have rearrange ou trinomial into the standard form of a 2 + b + c, ou will see that a = 3, so our brackets will have to look like this: (3 )( ). Now all ou have to do is split c up in such a wa that, when ou epand the brackets, ou get b. Question A13: Solve for : 15 2 + 19 8 = 0 Once ou have rearrange ou trinomial into the standard form of a 2 + b + c, split a up as follows: (5 )(3 ). Now all ou have to do is split c up in such a wa that, when ou epand the brackets, ou get b. In general ou would use trial and error to see which is the best wa to split a up. 7
Question A14: Solve for : 13 4 < 9 2 Rearrange the epression into standard form: 0 < a 2 +b+c. Find the roots and displa them on a number line. Use the number line to determine which values satisf the epression. Question A15: f () = a and g() = b 2. Find a and b. f () g() (0;1) P ( 1; 1 ) 2 You have a point on each graph and there is onl one unknown in each function. 8
Question A16: Determine the -values for which f 1 () > 0. f () (0;1) To find the inverse ou need to reflect f about the line =. To help visulize what this will look like, turn our phone onto it s left side. The -ais is now the -ais and the -ais is now the -ais. However, take note that the positive and negative values on the new -ais are swopped round (+ on the left, - on the right). Question A17: Given the graph of f () = 2, find f 1 (3.5), where f 1 () is the inverse of f (). 1 9
In general, to find the inverse of a function, swop the s and the s and then rearrange to make the subject of the formula. The inverse of an eponential is a log function. Question A18: Find the 299th term in following sequence: 6; 6; 2; 6; 18;... To find T n in the form of an 2 + bn + c, find the second difference of the sequence(find the difference between each term and then find the difference between those differences). Equate it to 2a. Now ou onl have to find b and c. Question A19: Below is a sketch of g(). If g( 1) = b a + c, determine the values of a,b and c. 2 1 10
Subtracting a value from shifts the function to the right. Adding a value to shifts the function to the left. Subtracting a value from the whole function shifts the function downwards. Adding a value to the whole function shifts the function upwards. You know what a standard hpebola looks like so ou can determine g() b looking at how the standard hperbola has been shifted. You then also need to find a b substituting a point into g(). From there it is one more step to find g( 1). Question A20: Given the graph of f () = 3 + 3 2 2 + 6, below, for which values of p will g() = 3 + 3 2 2 + 6 + p have ONE real root? B(2, 10) A( 1, 3.5) For g() to have one real root, both turning points have to be either above or below the -ais. Difficult Algebra Question A21: Determine the coordinate of the turning point of p if p() = f (3) and f () = 2 2 + 8 + 10. 11
Replace each with 3, in f () and then simplif. Question A22: Calculate the eact value (do not round off) of: 10 2009 10 2011 10 2007 a.b = a. b Question A23: Give the graph of = 2 and a line running from A(t,t 2 ) to B(3,0). Determine the value of t which minimizes the length of AB. A(t,t 2 ) B(3, 0) Use pthogorous to find an epression for AB 2 (the value of t that minimizes AB is the same value that minimizes AB 2 ). Use differentiation to find the value of t that minimizes AB 2. Question A24: Calculate the value of 1234567893 1234567894 1234567895 1234567892. You won t be able to use a calculator for this. 12
Let 1234567890 =. Rewrite the epression in terms of. Question A25: Calculate the value of: 1 1 2 + 1 2 3 + 1 3 4 + 1 4 5 +... + 1 2008 2009 Sum the first two terms of the series, then the first three and then the first four. Algebra Solutions Solution A1: (3 )(5 ) = 3 15 5 3 + 2 = 3 2 8 + 12 = 0 ( 6)( 2) = 0 = 6 OR = 2 Solution A2: Part 1: Let = 2 3 2 ( 2 3 )2 + 2 ( 2 3 ) = 1 5 9 2 + 4 3 1 = 0 5 2 + 12 9 = 0 (5 3)( + 3) = 0 = 3 5 OR = 3 = 2 3 when = 3 5 OR = 2 when = 3 (3/5,2/3) or (-3,-2) 13
Solution A3: Substituting A into f (), we get: 3.5 = ( 1) 3 + a( 1) 2 + b( 1) 3.5 = 1 + a b 4.5 = a b Substituting B into f (), we get: 10 = (2) 3 + a(2) 2 + b(2) 10 = 8 + 4a + 2b 18 = 4a + 2b 9 = 2a + b 9 2a = b We sunstitute b = 9 2a back into the first equation to get: 4.5 = a (9 2a) 4.5 = a 9 + 2a 4.5 = 3a a = 3 2 We sunstitute a = 3 2 back into b = 9 2a to get: b = 9 2 b = 6 ( ) 3 2 Solution A4: m ave = m AB 10 ( 3.5) m ave = 2 ( 1) m ave = 13.5 3 m ave = 4.5 14
Solution A5: f () = 4 3 9 f () = 4 1 2 1 9 3 f () = 2 3 2 1 3 2 f (3) = 2(3) 3 2 1 3 (3)2 f (3) = 2.27 1 2 3 f 1 (3) = 2. 3 27 1 2 f 1 (3) = 2. 3 27 f (3) = 3.38 Solution A6: We have a 1 = 1 2, therefore a1 = 2, therefore a = 2. Solution A7: f () = g() 2 = 100.3 log(2 ) = log(100.3 ) log2 = log100 + log3 log2 = log100 + log3 log2 log3 = log100 (log2 log3) = log100 (log2/3) = log100 = log100 log 2/3 = 11.36 Solution A8: We know that the common difference is the same between all terms, hence the name. Therefore, we can epress the common difference, for this this sequence, in terms of p in two different was and equate them: T 2 T 1 = T 3 T 2 (2p 3) (1 p) = (p + 5) (2p 3) 2p 3 1 + p = p + 5 2p + 3 4p = 12 p = 3 15
Solution A9: Using the form f () = a( p) 2 + q, and the co-ordinates of the turning point, (p,q) = ( 1, 9 4 ), we first determine a: f () = a( p) 2 + q 5 4 = a((0) (1))2 + ( 9 ) 4 5 4 = a 9 4 1 = a We now determine the function b substituting the values for a, p and q into the form f () = a( p) 2 +q. f () = ( 1) 2 9 4 ( ) (( ) ) 7 7 2 f = 1 9 2 2 4 ( ) 7 f = 4 2 Solution A10: f () = 3 + 2 + 8 12 f () = 3 2 + 2 + 8 f () = 6 + 2 To find the inflection point, we set the second derivitive to 0 and solve for. 0 = 6 + 2 = 1 3 Solution A11: ( 1 + ) 2 2 2 8 2 ( = 1 + 2 2 2 + 2 2) 8 2 =1 + 2 2 2 + 2 2 2 2 2 =1 + 2 2 16
Solution A12: 3 + 1 = 4 3 2 + 1 = 4 3 2 4 + 1 = 0 (3 1)( 1) = 0 = 1 3 OR = 1 Solution A13: 15 2 + 19 8 = 0 (3 1)(5 + 8) = 0 = 1 3 OR = 8 5 Solution A14: 13 4 < 9 2 0 < 9 2 13 + 4 0 < (9 4)( 1) 0 0 + + 4 9 1 < 4 9 OR > 1 Solution A15: b(1) 2 = 1 2 b = 1 2 a 1 = 1 2 a = 1 2 Solution A16: 0 < < 1 17
Solution A17: To find the inverse of = 2 we swop the s and the s and then rearrange to make the subject of the formula. = 2 log() = log(2 ) log 2 () = log 2 (2 ) log 2 = log 2 (2) How do we know to take log 2 of each side? Because log 2 2 = 1 so we are onl left with the on the right side of the equation. Now we have f 1 () = log 2. Therefore: f 1 (3.5) = log 2 (3.5) f 1 (3.5) = log(3.5) log(2) f 1 (3.5) = 1.81 Solution A18: First we look at at the second difference. If it is constant then our general term can be written in the form T n = an 2 + bn + c. 6 6 2-6 0-4 -8-4 -4 The second difference is equal to 2a.Therefore: 2a = 4 a = 2 T n = 2n 2 + bn + c Now we substitute two know terms into T n to get a pair of simultaneous equations: and 6 = 2(1) 2 + b(1) + c 6 = 2 + b + c 8 = b + c 6 = 2(2) 2 + b(2) + c 6 = 8 + 2b + c 14 = 2b + c 18
B subtracting the first equation from the second we get b = 6, b substituting b back into the first equation, we get c = 2. Therfore: T n = 2n 2 + 6n + 2 T 299 = 2(299) 2 + 6(299) + 2 T 299 = 177006 Solution A19: Let us first focus on finding g(). We can see that it has been shifted 1 unit to the right (subtract 1 from ) and 2 units up (add 2 to the whole function). Therefore: g() = a 1 + 2 To find a we have to substitute a point on the graph, into the function. (0,0) is a point on the graph. Therefore: 0 = a (0) 1 + 2 2 = a g() = 2 1 + 2 g( 1) is simpl g() shifted 1 unit to the right. Therefore: g( 1) = 2 ( 1) 1 + 2 g( 1) = 2 2 + 2 a = 2b = 2c = 2 Solution A20: For g() to have one real root, both turning points have to be either above or below the -ais. For B to be below the -ais, the graph has to be shifted down b an value less than -10 (because B is 10 units above the -ais). For A to be above the -ais, the graph has to be shifted up b an value greater than 1 (because A is 1 unit below the -ais). Therefore: p < 10 OR p > 1 Solution A21: p() = 2(3) 2 + 8(3) + 10 p() = 18 2 + 24 + 10 From here ou can use The turning point formula, completing the square or differentiation to find the turning point, hence the coordinate of the turning point. 19
The turning point formula: = b 2a = 24 36 = 2 3 = 18 b substitution Completing the square: p() = 18 2 + 24 + 10 ( p() = 18 2 2 + 18 3) = 2 3 and = 18 Differentiation: p() = 18 2 + 24 + 10 p () = 36 + 24 p () = 0 at the turning point 0 = 36 + 24 = 2 3 = 18 b substitution Solution A22: 10 2009 10 2011 10 2007 10 2007 10 = 2 10 2007 10 4 10 2007 10 2007 10 = 2 10 2007 ( 10 4 1) 10 2 = 10 4 1 = 10 10 2 1 = 10 99 20
Solution A23: AB 2 = (t 2 0) 2 + (t 3) 2 AB 2 = t 4 +t 2 6t + 9 d dt AB2 = 4t 3 + 2t 6 To find the minimum value of t we set 4t 3 + 2t 6 = 0. We also use factor theorem to find that (t 1) is a factor of 4t 3 + 2t 6, b substituting 1 into 4t 3 + 2t 6 and finding that the epression equals 0. 4t 3 + 2t 6 = 0 2t 3 +t 3 = 0 (t 1)(2t 2 + 2t + 3) = 0 t = 1 There is no real solution for 2t 2 + 2t + 3 Solution A24: Let 1234567890 = 1234567893 1234567894 1234567895 1234567892 = ( + 3)( + 4) ( + 5)( + 2) = ( 2 + 7 + 12) ( 2 + 7 + 10) = 2 + 7 + 12 2 7 10 = 2 Solution A25: We first establish a patern b summing the first two, then three and then four terms of the series: Simplifing these results, we get: 1 1 2 + 1 2 3 = 4 6 1 1 2 + 1 2 3 + 1 3 4 = 9 12 1 1 2 + 1 2 3 + 1 3 4 + 1 4 5 = 16 20 Sum of first two terms = 2 3 Sum of first three terms = 3 4 We can see that the pattern is: Sum of first n terms = Sum of first four terms = 4 5 n n+1. Therefore: 1 1 2 + 1 2 3 + 1 3 4 + 1 4 5 +... + 1 2008 2009 = 2008 2008 + 1 = 2008 2009 21
Trigonometr Questions Eas Trigonometr Question T1: Simplif the following: sin(90 + ).cos(180 ).tan( ) cos(270 ) When an angle is given in reference to the -ais, ie. (90 + ), (270 + ), etc, it can alwas be reduced to (90 ). You must also make the appropriate sign change for the quadrant ou are in. Question T2: Given f () = 2sin, find the minimum positive value (in degrees) of θ such that f ( + θ) = 2cos f () must be shifted to the left. Question T3: Epress the following epression in terms of tan onl: sincos 1 sin 2 + cos 2 Use sin 2 + cos 2 = 1 to simplif the epression. 22
Question T4: Given the following graph, where f () = 2sin and g() = cos2, [ 180 ;180 ]: 2 f () g() 1 180 90 90 180 1 2 For which value of will f () g() = 3? Where about on the graph is the distance between f () and g(), equal to three? No working is required, the answer can be read straight off the graph. 23
Question T5: Find theta. ( 1.8, 1.5) θ The length of the opposite and adjacent sides are given b the point (-1.8,1.5). Question T6: Given the following graph, where f () = 2cos and g() = tan2, [ 90 ;90 ]: 2 1 90 90 1 2 24
( Determine the period of f. 2) The period is the number of degrees ( that it takes for a trig function to go through one full ccle. We know this to be 360 for cos. For cos the frequenc has been halved. This means that the period has been 2) doubled (the period is the inverse of the frequenc). Question T7: Given the following graph, where f () = 2cos and g() = tan2, [ 90 ;90 ]: 2 1 90 90 1 2 Determine the equation of the asmptote of g( 25 ), where [0 ;90 ] First ou will need to determine the equation of the asmptote of g() where [0 ;90 ]. Then ou will have to shift it either left or right. 25
Question T8: Given the following graph, where f () = 2cos and g() = tan2, [ 90 ;90 ]: 2 1 90 90 1 2 For which values of will 2cos.tan2 > 0? This occurs over the intervals of for which the signs of f () and g() are equal. The ke is to identif the asmptotes of g(). Question T9: Simplif completel: sin(90 )cos(180 ) + tan.cos( )sin(180 + ) sin() = cos(90 ) sin 2 + cos 2 = 1 26
Question T10: If 13sinθ 5 = 0, find cosθ Draw a right angle triangle with angle θ. You have the lengths of the opposite and hpotenuse, now find the adjacent side. Question T11 Solve for [0 ;360 ] if 1 cos = 0,435. 2 There are two values for on the interval [0 ;360 ] for which this equalit holds. Medium Trigonometr Question T12: Triangle ABC is isosceles with AB = BC. B c a A b C Epress cosb in terms of a and b. Simplif as far as possible. Cosine rule. 27
Question T13: If cos = t, epress tan in terms of t. If cos = t then we can construct a right angle triangle that has an angle, an adjacent side of length t and a hpotenuse of length 1. Then use the Theorem of Pthagoras to find the opposite side. Question T14: Given the following graph, where f () = 2sin and g() = cos2, [ 180 ;180 ]: 2 f () g() 1 180 90 90 180 1 2 For which four values of will g() f () = 1? Where about on the graph is the distance between f () and g(), equal to one? No working is required, the answer can be read straight off the graph. Question T15: Simplif the following epression to one trig ratio of θ: sin( θ).sin(180 θ) + cos(90 + θ) sin(360 θ) tan315 28
If an angle is given with reference to the -ais, eg: sin(180 + θ), then it is eas to reduce. All ou do is change the angle to the acute angle and make the appropriate sign change for the quadrant it is in, eg: we first change sin(180 +θ) to sin(θ), but it is also in the third quadrant so it gets a sign change and becomes sin(θ). If, however, an angle is given with reference to the -ais, we go through the eact same two steps and then we add a third step, which is to change the trig function to its co-ratio. Eg: We change sin(90 + θ) to sin(θ), it is in the second quadrant, so there is no sign change, and for the third step we change the the sin to a cos so that it becomes cos(θ). Question T16: Find the length of OT if OR = 7.5cm. T P(12,5) R O θ S RTO = POS Question T17: If sin23 = p, write the following in terms of p: sin46 Use sin2θ = 2sinθ cosθ then sin 2 θ + cos 2 θ = 1. 29
Question T18: Determine the minimum and maimum values of the following: 1 f () = 3sin 2 + 4cos 2 sin 2 + cos 2 = 1 Difficult Trigonometr Question T19: Determine the value of: tan1 tan2 tan3 tan4... tan87 tan88 tan89 tanθ = sinθ cosθ and sinθ = cos(90 θ) Question T20: Two ships, A and B, are 120km apart. Ship A is at a bearing of 67 from D and 97km awa from D. DN points due north. Ship B is at a bearing of 208 from D. Determine the bearing of Ship A from Ship B, that is M ˆBA. A N 67 97km M D 208 120km B 30
You need to find M ˆBD and D ˆBA separatl and then add them together. M ˆBD can be determined without doing an working. Question T21: Determine the lowest positive value solution of: sin + 2cos 2 = 1 Rewite the equation in terms of sin onl and then solve as a quadratic. Question T22: If 2 = a and 3 = b, epress sin15 in terms of a and b. Think of two special angles such that, when one is subtracted from the other, ou get 15. Then, using the double angle formula, sin( ) = sincos cossin, epress sin15 in terms of trig functions of the special angles that ou have chosen. Question T23: If cos = t, epress sin2 in terms of t. If cos = t then we can construct a right angle triangle that has an angle, an adjacent side of length t and a hpotenuse of length 1. Then use the Theorem of Pthagoras to find the opposite side. Watch out though, this triangle is for the angle, and we are looking for sin2. You will have to use a trig identit to epress sin2 in terms of trig functions of just. 31
Question T24: Solve the following equation for ( 180 ;180 ]: sincos 1 + cos 2 sin 2 = 0 Use sin 2 + cos 2 = 1 to simplif the equation. Find the general solution and use it to find the values of ( 180 ;180 ] that satisf the equation. For ( 180 ;180 ], the round bracket means we don t include 180 and the square bracket means we do include 180. Question T25: Given the following graph, where f () = 2sin and g() = cos2, [ 180 ;180 ]: 2 f () g() 1 180 90 90 180 1 2 For which values of is f () > g()? f () is greater than g() on an interval which is defined b the points of intersection of the two functions. 32
Question T26: AB is a vertical pole, 50m high, with two cables running from the top, down to two securing points on the ground, at C and D. The shaded region, BCD, is the ground (horizontal plane). AĈB = 55, A ˆDB = 48 and CÂD = 71. A 71 50m B 48 D 55 C Find the distance between the two securing points, C and D. Visualisation is ke. You have to view this diagram as 3D, not 2D. All the triangles ou can see, are in different planes. Therefore, none of the angles can be related to each other. Your first step is to find the lines AC and AD. AC is the hpotenuse of the vertical triangle, ABC, and AD is the hpotenuse of the vertical triangle, ABD. 33
Trigonometr Solutions Solution T1: sin(90 + ).cos(180 ).tan( ) cos(270 ) = sin(90 ).( cos).( tan) cos(90 ) sin cos.( cos).( cos = ) sin = cos Solution T2: f () = 2sin needs to be shifted 90 to the left for it to equal 2cos. For an function, if we want to shift it to the left then we have to add a positive value to. Therefore: 2sin( + 90 ) = 2cos 90 Solution T3: sincos 1 sin 2 + cos 2 = sincos cos 2 + cos 2 = sincos 2cos 2 = sin 2cos = 1 sin 2 cos = 1 2 tan Solution T4: To get f () g() = 3, f () has to be greater than g() (obviousl, else our result would be negative). Also, we can see that the maimum value for f () is 2, and the minimum value of g() is -1. So, f () g() will onl equal 3 where both f () is at it s maimum and g() is at it s minimum. It is clear from the graph that this onl happens at = 90. 34
Solution T5: The length of the opposite side is given b the -coordinate of (-1.8,1.5) and the length of the adjacent side is given b the -coordinate of (-1.8,1.5). Therefore we have: Length of opposite side: 1.8 Length of adjacent side: 1.5 So we have: tanθ = 1.8 1.5 1 1.8 θ = tan 1.5 θ = 50.19 Solution T6: ( For cos the frequenc has been halved. This means that the period has been doubled (the period is 2) ( the inverse of the frequenc). We know that the period for cos is 360. Therefore the period for cos 2) must be 720 Solution T7: First ou will need to determine the equation of the asmptote of g() where [0 ;90 ]. It is = 45. Then, if we subtract 25 from in our function: g( 25 ), we will shift g() to the right b 25. Therefore the asmptote will also shift 25 to the right, giving us a new equation of = 70 Solution T8: f () is positive for the whole interval of [ 90 ;90 ], so we onl have to determine the values of for which g() is positive. To do this have to identif the asmtotes of g(). Because the frequenc of g() has been doubled, the period has been halved, therefore the occur at 45 and 45. Therefore g() is positive over the following intervals: 90 < < 45 0 < < 45 Which means that 2cos.tan2 > 0 over those same intervals. 35
Solution T9: First we note that, if sin() = cos(90 ), then we have: Now we simplif: sin(90 ) = cos(90 (90 )) sin(90 ) = cos() sin(90 )cos(180 ) + tan.cos( )sin(180 + ) =cos(). cos() + sin.cos(). sin() cos = cos 2 () + sin. sin() = cos 2 () sin 2 () = (cos 2 () + sin 2 ()) = 1 Solution T10: Draw a right angle triangle with angle θ. You have the lengths of the opposite and hpotenuse. Fill them in and use the theorem of Pthagoras to determine that the adjacent side is 12. 13 5 θ adjacent 2 = 13 2 5 2 adjacent = 13 2 5 2 adjacent = 12 Therefore we have: 13 5 θ 12 Therefore, cosθ = 12 13. 36
Solution T11 1 cos = 0,435 2 cos = 0,87 = cos 1 0,87 = 29.54 B making a rough sketch of cos, it can be seen that the equalit will also hold for 360 29.54 = 330.46. Solution T12: b2 2a 2 2a 2 b 2 = a 2 + c 2 2accosB ABC is isosceles, therefore a = c b 2 = a 2 + a 2 2a 2 cosb = cosb cosb = 1 b2 2a 2 Solution T13: To find the opposite side of the right angle triangle, we use the Theorem of Pthagoras: (opposite side) 2 = (1) 2 ( t) 2 opposite side = (1) 2 ( t) 2 = 1 t We know that tan is equal to opposite over adjacent, therefore: 1 t tan = t 1 t = t Solution T14: For = 180,0 and 180, g() has a value of 1 and f () has a value of 0. Therefore, at these three values, g() f () is equal to 1. At = 90, we can see that g() = 1 and f () = 2. Therefore g() f () is also equal to 1 at = 90. 37
Solution T15: sin( ).sin(180 ) + cos(90 + ) sin(360 ) tan315 sin.sin sin = ( sin) ( tan45 ) sin.sin sin = sin + 1 sin(sin + 1) = sin + 1 = sin Solution T16: OP = 13 (B Pthagoras) TOR = 90 θ (Angle sum of straight line) Therefore RTO = θ (Angle sum of triangle) So we have: OT OR = 13 5 OT = OR 13 5 OT = 7.5 13 5 OT = 19.5 Solution T17: First we notice that 46 is double 23 so the double angle formula, sin2θ = 2sinθ cosθ, comes to mind. Using this formula we get: sin46 = 2sin23 cos23 Now we need to epress cos23 in terms of p. To do this we use sin 2 θ + cos 2 θ = 1: sin 2 θ + cos 2 θ = 1 sin 2 23 + cos 2 23 = 1 p 2 + cos 2 23 = 1 cos 2 23 = 1 p 2 cos23 = 1 p 2 NOTE: Pthagoras could have been used just as effectivel to arrive at cos23 = 1 p 2 : 38
p 1 23 1 p 2 Putting this all together, we get: sin46 = 2sin23 cos23 sin46 = 2p 1 p 2 Solution T18: Rewrite the epression in terms of either sin or cos (we will consider sin), using the trig identit sin 2 + cos 2 = 1: 1 f () = 3sin 2 + 4cos 2 1 f () = 3sin 2 + 4(1 sin 2 ) 1 f () = 3sin 2 + 4 4sin 2 1 f () = 4 sin 2 We know that the maimum value of sin is 1, hence the maimum value of sin 2 is 1. We also know that the minimum value of sin 2 is 0. f () will be at it s maimum when its denominator is at its minimum. This occurs when sin 2 is at it s maimum: f () = 1 4 1 f () = 1 3 f () will be at it s minimum when its denominator is at its maimum. This occurs when sin 2 is at it s minimum: f () = 1 4 0 f () = 1 4 39
Solution T19: tan1 tan2 tan3 tan4... tan87 tan88 tan89 ( )( ) ( ) ( )( ) sin1 sin2 sin45 sin88 sin89 = cos1 cos2... cos45... cos88 cos89 ( )( ) ( ) ( sin1 sin2 sin45 sin(90 2 )( ) sin(90 1 ) ) = cos1 cos2... cos45... cos(90 2 ) cos(90 1 ) ( )( ) ( ) ( )( ) sin1 sin2 sin45 cos2 cos1 = cos1 cos2... cos45... sin2 sin1 =tan45 =1 Solution T20: We can see that N ˆDB is equal to 152 (360 208 ). We know that the sum of interior angles equal 180, therefore M ˆBD = 28 To determine D ˆBA, we use the sin rule: sina a sind ˆBA 97 Therefore M ˆBD = 28 + 30.58 = 58.58. = sinb b = sin141 120 D ˆBA = sin 1 ( 97 sin141 120 D ˆBA = 30.58 ) Solution T21: Using the identit sin 2 + cos 2 = 1, rewite the equation in terms of sin onl: sin + 2cos 2 = 1 sin + 2(1 sin 2 ) = 1 0 = 2sin 2 sin 1 If we let sin = k, we get: 2k 2 k 1 = 0 (2k + 1)(k 1) = 0 k = 1 2 sin = 1 2 OR k = 1 OR sin = 1 We now have to consider the general solution for sin = 1 2 and sin = 1. For sin = 2 1, we have: 40
( = sin 1 1 ) + n.360 OR 2 ( (180 sin 1 1 )) + n.360 n Z 2 For sin = 1, we have: = 30 + n.360 OR 210 + n.360 = sin 1 (1) + n.360 OR ( 180 sin 1 (1) ) + n.360 n Z = 90 + n.360 OR 90 + n.360 = 90 + n.360 So could be an value that satisfies an one of the these three general solutions, where n could be an interger: = 30 + n.360 = 210 + n.360 = 90 + n.360 We are interested in finding the lowest positive value of. B inspection we can see that we get the lowest positive value when we take n = 0 for = 90 + n.360. Therefore 90 is the lowest positive solution for our equation. Solution T22: Solution T23: sin15 = sin(45 30 ) = sin45 cos30 cos45 sin30 = 1 3. 2 2 1. 1 2 2 3 = 2 2 1 2 2 3 1 = 2 2 = b 1 2a To find the opposite side of the right angle triangle, we use the Theorem of Pthagoras: (opposite side) 2 = (1) 2 ( t) 2 opposite side = (1) 2 ( t) 2 = 1 t Now we use a trig identit to epress sin2 in terms of trig functions of just : sin2 = 2sincos We have cos = t. We know that sin is equal to opposite over hpotenuse, therefore: 41
1 t sin = 1 = 1 t Therefore we have: sin2 = 2. 1 t. t = 2. t(1 t) = 4t(1 t) Solution T24: Now we find the general solution for tan: sincos 1 + cos 2 sin 2 = 0 sincos 1 sin 2 + cos 2 = 0 sincos cos 2 + cos 2 = 0 sincos 2cos 2 = 0 sin 2cos = 0 1 sin 2 cos = 0 1 2 tan = 0 tan = 0 = tan 1 (0) + k.180 k Z = 0 + k.180 Now we use our general solution to find the values of such that ( 180 ;180 ], b selecting different values for k: For k=0: = 0 + 0.180 = 0 0 ( 180 ;180 ]. Therefore = 0 is a solution. For k=1: = 0 + 1.180 = 180 180 ( 180 ;180 ]. Therefore = 180 is a solution. 42
For k=-1: = 0 + ( 1).180 = 180 180 / ( 180 ;180 ]. Therefore = 180 is not a solution. We can see that for k > 1 or k < 1 there will not be a solution as the result will fall outside of the interval ( 180 ;180 ]. Solution T25: Looking at the graph, it is clear where f () is greater than g(), between the two points of intersection. However, we must find the points of intersection of the two graphs so that we can define the interval precisel. To do this, we equate the two functions: 2sin = cos2 2sin = 1 2sin 2 2sin 2 + 2sin 1 = 0 If we let sin = k, we can see that we have a quadratic epression: We have to use the formula to solve this quadratic: 2k 2 + 2k 1 = 0 k = 2 ± 2 2 4(2)( 1) 2(2) Therefore k = 0.366 or -1.366. Therefore sin = 0.366. We reject -1.366 because sin cannot be less than -1. We find the general solution for sin: Therefore: = sin 1 0.366 + n.360 OR = 180 sin 1 0.366 + n.360 n Z = 21.47 + n.360 OR = 158.53 + n.360 n Z We get the values for the two intersects when n = 0 for each part of the general solution. Therefore we have: f () > g() for (21.47;158.53) Note the use of round brackets, we do not include the intersection points because f () = g() at these points and we were not asked to find the values of for which f () g(). If we were, then we would have used square brackets to include the intersection points. 43
Solution T26: To find AC: sin55 = 50 AC AC = 50 sin55 AC = 61.04m To find AD: sin48 = 50 AD AD = 50 sin48 AC = 67.28m To find CD, we use the cosine rule: CD 2 = AC 2 + AD 2 2AC.AD.cos(71 ) CD 2 = 61.04 2 + 67.28 2 2.61.04.67.28.cos(71 ) CD 2 = 5578.47 CD = 74.69m 44
Geometr Questions Eas Geometr Question G1: Determine the coordinates of M, the midpoint of BC. A(3; 7) C( 3; 4) M B(1; 6) M is half wa between the and vales of B and C. 45
Question G2: Determine the coordinates of point A. A(, ) B(3, 4) A is the 180 rotation transformation of B. 46
Question G3: D is a point such that AD is parallel to to BC. Determine the equation of the line AD in the form = m+c. B(1, 4) A( 3, 1) C(5, 2) AD and BC have equal gradients. 47
Question G4: Find the gradient of the tangent to the circle centred at M. M(-2,3) 1 The line from M to the point of tangenc is perpendicular to the tangent. 48
Question G5: The line LP, with equation + 2 = 0, is a tangent at L to the circle with centre M(-4,4). Determine the equation of line NQ, in the form = m + c. L M(-4,4) N P Q NQ is parallel to LP and the are smmetrical about the line running from the centre of the circle to the origin. 49
Question G6: If AD is parallel to BC, determine the value of t for point D. D(7,t) A(1,6) C(6,1) B(3,0) Start b finding the gradient of BC. 50
Question G7: Find the angle, θ, that line AB makes with the -ais. B(7, 1) θ A( 1, 5) Start b finding the gradient of the line. 51
Question G8: Determine the perimeter of PQR. P( 1, 2) R(3, 0) Q( 2, 2) Use the distance formula: length = ( 1 2 ) 2 + ( 1 2 ) 2 52
Question G9: Determine the coordinates of point B. The equation of the circle is ( 3) 2 + ( + 2) 2 = 25. B You have the -value of B. Question G10: Calculate the distance between the centres of of the two circles with equations ( 3) 2 +(+2) 2 = 25 and ( 12) 2 + ( 10) 2 = 100. The coordinates of the centres of the circles can be determined directl from their equations. For the first circle it s (3, 2). Question G11: Determine the equation of the right side vertical tangent to the circle with equation ( 12) 2 +( 10) 2 = 100. This question does not require an calculation, drawing a rough labelled sketch ma help. 53
If the tangent is vertical then it is parallel to the -ais, so it will have an equation of the form = c, where c is a constant. To determine the value of, think about how far the tangent is from the centre of the circle and the -value of the coordinate of the centre of the circle. Then add those two values together. Medium Geometr Question G12: If G(a;b) is a point such that A, G and M lie on the same straight line, find an epression that gives b in terms of a. A(3; 7) C( 3; 4) G(a;b) M( 1; 1) B(1; 6) Find the equation of the straight line that runs through M and A. 54
Question G13: A, B and C are the vertices of a triangle that lies on the circumfrence of a circle with centre M. Determine the size of angle θ. C(0, 8) A( 8, 2) M θ B( 2, 6) If BC is running through the centre of the circle and A is on the circumfrence, then BÂC = 90. This is true no matter where A is on the circumfrence. 55
Question G14: Determine the equation of the circle with centre B and radius AB, in the form A 2 +B+C 2 +D+E = 0. A B(3, 4) Start b determining the equation of the circle in the form ( 0 ) 2 + ( 0 ) 2 = r 2, where ( 0, 0 ) is the centre of the circle. You have ( 0, 0 ), now all ou need is r. 56
Question G15: Given N = (1,6), calculate the area of triangle ABC. C(3, 9) N B(7, 2) A( 5, 3) AC is the base of the triangle and BN is the height. 57
Question G16: Calculate θ. D(2, 9) B(6, 6) θ A( 6, 1) C( 6, 3) Find the angle that each line makes with the horizontal, b finding the gradient of each line. Question G17: Determine the area of a triangle with sides of length 6cm, 9cm and 13cm. Cosine rule and sine rule. 58
Question G18: The circle centred at K(-2,3) touches the -ais. If the length of its diameter is doubled and it is shifted 5 units to the right and 1 unit down, determine the equation of the new circle. K(-2,3) Subtract from to shift to the right. Add to to shift to the left. Subtract from to shift upwards. Add to to shift downwards. 59
Difficult Geometr Question G19: Given that the equation of the line running from A to M is = 2 + 1 and G(a;b) is a point anwhere on the line, calculate the two possible values of b if GC = 17. A(3; 7) C( 3; 4) G(a;b) M( 1; 1) B(1; 6) Epress the distance between C and G in terms of a and b. Use the equation of the line between A and M to rewite this epression in terms of one variable. 60
Question G20: If the line BC is a tangent to the circle, with centre (0,0), at B, determine the value of k. A C(k, 1) B(3, 4) You need to do is find the equation of the line BC. You have a point, now all ou need is a gradient. What is the angle between a tangent to a point on the circumfrence of a circle and the radius that etends to that same point? 61
Question G21: Determine the value of the coordinate of point B. A(3, 4) B(5, ) 22.16 C( 4, 2) You need to find the equation of line CB. Start b determining the angle that line AC makes with the horizontal. 62
Question G22: Find the point P if the equation of the circle centred at point M is (+2) 2 +( 1) 2 = r and MS:MP = 1:3. M S(1,-2) P(a,b) The ratio between the lengths of two line segments is equal to the ratio between the lengths of their corresponding horizontal or vertical components. 63
Question G23: The line LP, with equation + 2 = 0, is a tangent at L to the circle with centre M(-4,4). LN is a diameter. Determine the equation of the circle. L M(-4,4) N P You are looking for the radius of the circle. LN is perpendicular to LP. Start b finding the equation of LN b first finding its gradient. 64
Geometr Solutions Solution G1: ( ( 3) + (1) M = ; 2 =( 1, 1) ) (4) + ( 6) 2 Solution G2: A is the 180 rotation transformation of B. Therefore A = ( 3,4). Solution G3: AD and BC have equal gradients, so we first we find AD s gradient b finding gradient of BC: m = 1 2 1 2 m = 4 ( 2) 1 5 m = 1.5 So far we have = 1.5 + c as the equation of line AD. To find c we substitute point A into the equation: = 1.5 + c 1 = 1.5( 3) + c 3.5 = c = 1.5 3.5 Solution G4: The angle between a radius and a tangent is 90. Also, if two lines are perpendicular to one another then the gradient of one is equal to the negative inverse of the gradient of the other. So, if we find the gradient of radius then we can take it s negative inverse as the gradient of the tangent. Gradient of the radius: Gradient of the tangent: m rad = 1 2 1 2 m = 3 0 ( 2) 1 m = 1 65
m tan = 1 m rad m = 1 1 m = 1 Solution G5: NL is a diameter. Therefore, lines LP and NQ are both perpendicular to NL. Therefore, LP is parallel to NQ. Therefore LP and NQ have the same gradient, which is -1. To find the -intercept of NQ we notice that LP and NQ are both equidistant from the line MO (which runs from the circle centre to the origin). This is true because the centre of the circle is at (-4,4), which means that MO has a gradient of -1 and is therefore parallel to LP and NQ. Therefore, the distance of the -intercepts of LP and NQ, from the origin, must be equal. Therefore, the intercept of NQ must be -2. Putting this together we get the following equation for NQ: = 2 Solution G6: AD is parallel to BC. Therefore, the gradient of BC is equal to that of AD. If we have AC s gradient, we can easil find t. Finding BC s gradient: m = 1 2 1 2 m BC = 1 0 6 3 m BC = 1 3 m AD = 1 3 We can now use the gradient formula in reverse to find t: m AD = 1 3 t 6 7 1 = 1 3 t 6 = 1 6 3 t 6 = 2 t = 8 66
Solution G7: We have that tanθ = m AB. Here s the reason for this: The gradient of a line is equal to opposite. tanθ is equal to. We can see that, for an angle θ that a line adjacent makes with the -ais, an triangle containing θ has opposite side equal to a change in and the adjacent side equal to a change in. Therefore tanθ is equal to. Therefore tanθ is equla to the gradient First we find AB s gradient: m = 1 2 1 2 m AB = 7 ( 1) 1 ( 5) m AB = 8 6 Then we use this gradient to solve for θ. m AB = 4 3 tanθ = m AB θ = tan 1 (m AB ) ( ) 4 θ = tan 1 3 θ = 53.13 Solution G8: Length PQ: ( 1 2 ) 2 + ( 1 2 ) 2 = (( 1) ( 2)) 2 + (2 ( 2)) 2 = (1) 2 + (4) 2 = 17 Length PR: ( 1 2 ) 2 + ( 1 2 ) 2 = (( 1) 3) 2 + (2 0) 2 = ( 4) 2 + (2) 2 = 20 Length QR: 67
( 1 2 ) 2 + ( 1 2 ) 2 = (( 2) 3) 2 + (( 2) 0) 2 = ( 5) 2 + ( 2) 2 = 29 Perimeter: Perimeter =PQ + PR + QR = 17 + 20 + 29 =13.98 Solution G9: Length PQ: We have that the -value of B is 0, because B sits on the -ais. All we need to do is substitute 0 into the equation of the circle to find the -coordinate. ((0) 3) 2 + ( + 2) 2 = 25 9 + ( + 2) 2 = 25 Clearl the -value of B is -6. Therefore B = (0, 6). ( + 2) 2 = 16 + 2 = ±4 = 2 OR = 6 Solution G10: The coordinates of the centres of the circles can be determined directl from their equations. For the first circle it s (3, 2) and for the second circle it s (12,10). Now we need onl find the distance between these two ponts using the distance formula. ( 1 2 ) 2 + ( 1 2 ) 2 = (3 12) 2 + (( 2) 10) 2 = (9) 2 + ( 12) 2 = 81 + 144 = 225 =15 68
Solution G11: The circle has centre (12,10) and radius 10 units. A right side vertical tangent to the circle will be 10 units to the right of its centre and perpendicular to the -ais. Therefore it will have equation = 22. (12, 10) 10 units 12 22 Solution G12: We first have to find the equation of the straight line that runs through M and A. m = 7 ( 1) 3 ( 1) =2 =2 + c 7 =2(3) + c c =1 =2 + 1 We know that (a;b) is a point on this straight line so we can epress b in terms of a b substituting (a;b) into our equation, and we get: b = 2a + 1 69
Solution G13: We have that BÂC = 90. Therefore we need onl find AC and AB to then find θ using the inverse tan function, θ = tan 1 ( AC AB ). AC = (( 8) 0) 2 + (2 8) 2 AC = 100 AC = 10 AB = (( 8) ( 2)) 2 + (2 ( 6)) 2 AB = 100 AB = 10 ( ) AC θ = tan 1 AB ( ) 100 θ = tan 1 100 θ = tan 1 (1) θ = 45 Solution G14: Start b determining the equation of the circle in the form ( 0 ) 2 + ( 0 ) 2 = r 2, where ( 0, 0 ) is the centre of the circle. We are given ( 0, 0 ) = (3, 4). We are also given r = AB. To find AB we need onl find the length of the radius, OB, of the given circle and multipl it b 2. Therefore: OB = (0 (3)) 2 + (0 ( 4)) 2 OB = 5 AB = 10 r = 10 We now plug all our information into the standard equation and then epand it to get the required form: ( 0 ) 2 + ( 0 ) 2 = r 2 ( 3) 2 + ( + 4) 2 = 10 2 2 6 + 9 + 2 + 8 + 16 = 100 2 6 + 2 + 8 75 = 0 A = 1B = 6C = 1D = 8E = 75 70
Solution G15: AC is the base of the triangle and BN is the height. We know that the area of the triangle is equal to base height. Therefore we need onl find the lengths of AC and BN to calculate the area. 1 2 AC = (3 ( 5)) 2 + (9 ( 3)) 2 AC = 208 BN = (7 1) 2 + (2 6) 2 BN = 52 Area ABC = 1 2 208 52 = 52 Solution G16: The angle that line CD makes with the horizontal: tanθ = m CD tanθ = 9 ( 3) 2 ( 6) tanθ = 12 8 ) The angle that line AB makes with the horizontal: θ = tan 1 ( 3 2 θ = 56.31 tanθ = m AB tanθ = 6 (1) 6 ( 6) tanθ = 5 12 ) θ = tan 1 ( 5 12 θ = 22.62 B subtracting line AB s angle from line CD s angle, we arrive at θ: 56.31 22.62 = 33.69 71
Solution G17: You need to determine one of the angles in the triangle. For eample, I have chosen the angle between the sides of length 6cm and 9cm (ou ma have chosen to determine either of the other two angles): c 2 = a 2 + b 2 2abcosC 13 2 = 6 2 + 9 2 2.6.9.cosC ( 13 C = cos 1 2 6 2 9 2 ) 2.6.9 C = 118.78 Now we can use the sine rule to determine the area of the triangle: area area ABC = 1 2 absinc ABC = 1 2.6.9sin118.78 = 23.66cm 2 Solution G18: First we determine the equation of the circle centred at K. The general equation for a circle is ( a) 2 + ( b) 2 = r 2, where a and b are the and coordinates of the centre point, respectivel. We have the centre point, it s (-2,3). To find the radius we use the fact that the circle touches the -ais. This means that the -ais is tangential to the circle. Therefore, the radius is equal to the the distance from the centre of the circle to the -ais, which is 2 (which is the magnitude of the coordinate of the centre of the circle). Putting this together we get the following equation for the circle centred at K: ( ( 2)) 2 + ( (3)) 2 = 2 2 ( + 2) 2 + ( 3) 2 = 4 Now we can make the changes. To shift the circle 5 units to the right, we subtract 5 from. To shift the circle 1 unit downwards, we add 1 unit to. To double the diameter is to double the radius, we multipl the radius b 2 BEFORE we square it. Putting this all together we get the folowing equation for the new circle: (( 5) + 2) 2 + (( + 1) 3) 2 = (2 2) 2 ( 3) 2 + ( 2) 2 = 16 Solution G19: Distance between C and G: 17 = (a ( 3)) 2 + (b (4)) 2 72
We have = 2 + 1 b = 2a + 1 17 = (a ( 3)) 2 + (2a + 1 (4)) 2 17 = (a + 3) 2 + (2a 3) 2 17 = a 2 + 6a + 9 + 4a 2 12a + 9 17 = a 2 + 6a + 9 + 4a 2 12a + 9 0 = 5a 2 6a + 1 0 = (5a 1)(a 1) We square both sides to get rid of the square roots. a = 1 5 b = 7 5 or 1 or 3 (1/5, 7/5)or(1, 3) Solution G20: To find k we first need to find the equation of the line BC. To do this, we need to find the gradient of BC. We are given that BC is a tangent to the circle at B and the circle has its centre at (0,0), therefore OB is a radius from (0,0) to B. We have a theorem which states that the angle, between a tangent to a point on the circumfrence of a circle and the radius that etends to that same point, is equal to 90. This means that OB is perpendicular to BC. Therefore, if we can find the gradient of OB then we can take its negative inverse as the gradient of BC. m OB = m OB = 4 3 (0 ( 4)) (0 3) m BC = 3 4 We can set up an equation for BC: = m + c = 3 4 + c To find c, we substitute a known point, on BC, into the equation: = 3 4 + c 4 = 3 4 (3) + c 25 4 = c = 3 4 25 4 73
To find k, we substitute (k,1) into our equation: = 3 4 25 4 1 = 3 4 k 25 4 29 3 = k Solution G21: Determine the angle that line AC makes with the horizontal: tanθ = m AC tanθ = 4 ( 2) 3 ( 4) tanθ = 6 7 ) θ = tan 1 ( 6 7 θ = 40.60 Subtract 22.16 from 40.60 to get the angle that CB makes with the horizontal: Using this angle, find the gradient of line CB: 40.60 22.16 = 18.44 m CB = tan18.44 m CB = 0.33 Use the gradient of line CB and the point C to determine the equation of line CB: = m + c = 0.33 + c ( 2) = 0.33( 4) + c c = 0.67 = 0.33 0.67 B substituting the coordinate of B into the equation of line CB, we get the value of the coordinate: = 0.33 0.67 = 0.33(5) 0.67 = 1 Solution G22: From the equation for the circle centred at M, we know that the coordinates of point M are (-2,1). 74
So, the horizontal component of MS is 3 units and its vertical component is 3 units. If MS:MP = 1:3 then the horizontal component of MP must be 9 units and it s vertical component must also be 9 units. We know that MSP forms a straight line as M and P are the centres of circles that touch at S. Therefore, a will be 9 units from the component of M and b will be 9 units from the component of M. Therefore P = (7,-8). Solution G23: We know that LN is perpendicular to LP because LN is a diameter, LP is a tangent and L is the point of tangenc. Therefore, the gradient of LN is the negative inverse of the gradient of LP. We are given the equation of LP and, putting this equation into standard form, we can see that its gradient is -1. Therefore, the gradient of LN is 1. Now that we have a gradient (m=1) and a point (-4,4) for LN, we can determine it s equation: =m + c (4) =(1)( 4) + c 8 =c = + 8 With this equation we can find the point L, b finding the intersection of LP and LN: = + 8 AND + 2 = 0 ( + 8) + 2 = 0 2 + 6 = 0 = 3 = 5 L = ( 3,5) Now we can find the radius, r, of the circle b finding the distance between L and M: r = ( 1 2 ) 2 + ( 1 2 ) 2 r = ( 3 ( 4)) 2 + (5 4) 2 r = (1) 2 + (1) 2 r = 2 We now have the centre and the radius of the circle. Given the standard form of the equation of a circle, ( a) 2 + ( b) 2 = r 2, we can determine the equation of the circle as follows: ( ( 4)) 2 + ( (4)) 2 = ( 2) 2 ( + 4) 2 + ( 4) 2 = 2 75
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