Technical Mathematics Grade 12 P2 NSC November 2023 QP

Downloaded from hlayiso.com “wie basic education [, aaa Department: WR ALY, Basic Education REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 12 a TECHNICAL MATHEMATICS P2 5 ‘ NOVEMBER 2023 in MARKS: 150 TIME: 3 hours This question paper consists of 16 pages and a 2-page information sheet. Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 2 DBE/November 2023 NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. 2. This question paper consists of 11 questions. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. Clearly show ALL calculations, diagrams, graphs, etc. that you used in determining your answers. Answers only will NOT necessarily be awarded full marks. If necessary, round off answers to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. An information sheet with formulae is included at the end of the question paper. Write neatly and legibly. Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 3 DBE/November 2023 NSC QUESTION 1 The diagram below shows ADEF with vertices bE 4;8), E(4;-4) and F(-2;-8). The angle of inclination of DE with the positive x-axis is @ . E=90° DC4;8) AY x \ er x (45-4) F(-2;-8) 1.1 Determine the gradient of DE. (2) 1.2 Determine the size of angle a. GB) 1.3 Determine whether the line parallel to DE, passing through F, also passes through point (-10;5). (4) 1.4 Calculate the area of ADEF. (5) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 4 DBE/November 2023 NSC QUESTION 2 21 In the diagram below, O is the centre of the circle. JK is a tangent to the circle at point J(12 ; 9). yA 5 (12; 9) 6) <4 2.1.1 Determine the equation of the circle passing through J. (2) 2.1.2 Complete the following: Moy X My = -- (1) 2.1.3 Determine the equation of JK in the form y=... (4) 2 2 2.2 Given: +2 =1 11 64 to x? y? 2.2.1 Express the equation in the form — + > =1 a b r¢5) x? 2 2.2.2 Hence, sketch the graph defined by + Yn] 11 64 (2) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 5 DBE/November 2023 NSC QUESTION 3 3.1 Given: x = 152,4° and y = 24,8° Determine the following: 3.1.1 sin (x — y) (2) 1 x 3.1.2 sec] = + 80° 3 se( 2 } (2) 3.2 Given: sin f =-3 and B (90° ; 270°) Determine the following without the use of a calculator: 3.2.1 cosec # qa 3.2.2 tan B + cos B (5) 33 Determine the value(s) of x if cosx =— sin56,7° and x€ (0° ; 360°) (4) [14] QUESTION 4 4.1 Complete the following: 4.1.1 cosec A =~ qd) 4.1.2 cos (27 +A) =... qd) 4.13 cosec(180° + A) =... (1) 42 Simplify the following: sin(180° + A)-cot (360° — A)-cos (27 — A) + sin?(360° — A) () COSEC X —COSEC X - SEC X 43 Given: 7 = cot x sec x — (tan: xt+ 1) 43.1 Factorise: sec x — sec? x (1) cosec x —COSEC X - SEC X 3. Hence, the identity: = cot x 43.2 ence, prove the identity sec x —(tan?x +1) (4) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 QUESTION 5 6 DBE/November 2023 NSC The graphs below represent the functions defined by f (x)= cosax and g(x)=tanx for xe[0°; 180°] y i 3} —t ae mere ee a i a — — -1 $$ 2—____——_—_ — — H i i | | x 0 45° 90°; i 2} __ wane ot it i =i... L _ i i Use the graphs above to answer the following: 5.1 Write down: 5.1.1 The value of a qd) 3.1.2 The period of g ql) 5.1.3 The value of x for which -tanx+1=0 (2) 5.1.4 The range of g dg) 5.1.5 The value(s) of x for which f(x)< 0 (2) 5.2 Determine (180°) - f(180°) (2) 5.3 Write down the value(s) of x for which f is decreasing. (2) fy Please turn over Copyright reserved

Downloaded from hlayiso.com Technical Mathematics/P2 7 DBE/November 2023 NSC QUESTION 6 The picture and diagram below show a crane PQR lifting a box from point M to point R. PQ and MR are perpendicular to the ground level MP, such that PQRM lies in the same vertical plane. T isapoint on MR. PQ=8m; POR =114° and QPR = 49° R ' Te H M M 6.1 Determine the length of PR. (4) 62 Write down the size of RPM qd) 63 Complete the following ratio with respect to ARPM: sinR P M=— qd) 6.4 If TR =5m, determine MT. @G) (9] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 8 NSC Give reasons for your statements in QUESTIONS 7, 8 and 9. QUESTION 7 In the diagram below, O is the centre of the circle. M is the midpoint of chord AB and OM=3 cm AP isa tangent to the circle at A. AB=OP=8cm 71 Write down, stating a reason, the size of M,. 72 Give areason why A ,=90° 73 Determine the length of AP. Copyright reserved DBE/November 2023 Please turn over 2) qa) (3) [6]

Downloaded from hlayiso.com Technical Mathematics/P2 9 DBE/November 2023 NSC QUESTION 8 8.1 In the diagram below, A, B, E and C are points on the circle with centre O. OCB=33° Determine, stating reasons, the size of the following angles: 8.1.1 B, Q) 812 0, ) 8.1.3 E (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 10 DBE/November 2023 NSC 8.2 In the diagram below, RT is a tangent to circle ADBC at point C such that TCB=37° AC || DB 8.2.1 Write down, stating reasons, FOUR other angles equal to 37° (6) 8.2.2 Hence, show that A AEC ||| A BED Q) 8.2.3 Hence, complete the statement AE x ED=...x ... (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 11 DBE/November 2023 NSC 8.3 In the diagram below, R, Q, P and S are points on the circle. T isapoint on PQ. SQP =32° and STP =68° SQ bisects Q. PS =ST 8.3.1 Write down the size of the following angles: @ a a) () Ps Q) 8.3.2 Hence, show that Pi = 82 (5) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 12 DBE/November 2023 NSC QUESTION 9 The diagram below shows APQR with ST || QR. PS=8cm, SQ=3cm, RT=4cm and QR=21 cm. P Bg a T Ss > ¢ fq A 2 Q * 21 em R 9.1 Give the correct reason for the statement: Pr _ PS (.) TR SQ 9.2 Hence, calculate the length of PT. 9.3 Complete the statement and give the correct reason: ST PS QR (-) 94 Hence, calculate the length of ST. Copyright reserved Please turn over a) (2) (2) (2) {7]

Downloaded from hlayiso.com Technical Mathematics/P2 13 DBE/November 2023 NSC QUESTION 10 10.1 The picture and diagram below show two circular pulleys, A and B, connected with a belt and moving anti-clockwise. Pulley A has aradius of 50cm and pulley B has a radius of 20 cm. The belt covers : of the arch length of pulley A. The belt forms tangents to the pulleys at points C, D, E and F. The arch length of DE = “ em. Pulley A Pulley B Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 14 DBE/November 2023 NSC 10.1.1 Show that reflex CAF=200° () 10.1.2 Convert reflex CAF = 200° to radians. qd) 10.1.3 Hence, determine the major arc length of CF. (3) 10.1.4 Pulley A rotates at 500 revolutions per minute (r/min). (a) Calculate the circumferential velocity (cm/min) of a particle on the belt at point F. (3) (b) Hence, determine, in revolutions per second, the rotational frequency of pulley B. (4) 10.1.5 Determine the area of the shaded minor sector DBE. ()) 10.2 A rectangular advertisement board with a length of 4,6 metres and a breadth of 1,8 metres will be positioned on the semicircular wall as indicated in the diagram below. The height 4 from the top part of the rectangular board to the top part of the semicircular wall is 0,72 metres longer than the breadth of the rectangular advertisement board. 10.2.1 Determine the value of height A. qd) 10.2.2 Hence, calculate the length of the diameter of the semicircular wall. (4) (20} Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 15 DBE/November 2023 NSC QUESTION 11 11.1 The irregular figure below has a horizontal straight side, 12 cm long, which is divided into 6 equal parts. The ordinates dividing the parts are 8 cm, 4.cm, 6cm, h, 7,22 cm, 3,42 cm and 2cm. The length of A is the average of the third and fifth ordinates. 2cm 12cm > 11.1.1 Write down the width of each of the equal parts. dd) 11.1.2 Determine the value of A. Q) 11.1.3 Hence, determine the area of the irregular figure by using the mid-ordinate rule. (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 16 DBE/November 2023 NSC 11.2 The pictures below show spherical soccer balls. Picture A represents ball A with radius = 11 cm and picture B represents a smaller ball B with radius = x cm. Soccer ball A with radius r = 11cm Soccer ball B with radius r = x cm Picture B Calculate x, the radius of ball B, if the volume of ball B is half the volume of ball A. 4 The volume of a sphere is given by V = read (5) 11.3 The diagram below shows a closed cone. The radius of the cone is 5 cm. It has a height of 6 cm and a slant height (J) of 7,81 cm. 11.3.1 Calculate the surface area of the cone, where: 7 Surface area = mr? +ur/ Q) 11.3.2 The radius of the cone is increased by 20% and the height of the cone is decreased by 10%. Determine whether the new surface area is greater than the surface area of the original cone. (5) [18] TOTAL: 150 Copyright reserved

Downloaded from hlayiso.com Technical Mathematics/P2 DBE/November 2023 NSC INFORMATION SHEET: TECHNICAL MATHEMATICS —b+.4b? -4ac b yo fae nb? = ees =-— 2a * 2a 4a a*=box=log,b, a>0,a¥land b>0 A=P(1+ni) A=P(l-ni) A=P(1+i)" A=P(1-i)” ig =(14+ EY" 1 / _ f(x+h)— f(x) = lim 2@ EU SY f° @)= jim, h xr yn Ju" dx = +C ,n#-1 kx" dx =k. +¢ ,n#-1 n+i ntl 1 k f =&=Inx+C, x>0 [ <a&=kinx+C, x>0 x x x a® nx qa’™ Ja w=T +e ,a>0 Jka de=ko +€ ,a>0 + + d= V(x, - 4% +02. -HY M|= Ay eH bal 2 2 yomtre y-y, =m(x—x) m = 27 tanO=m XX x? y? ate! In AABC: —2— = a == @ =b? +c? — 2be.cosA snA sinB sinC Area of A ABC = 5 ab. sin C sin? 6 +cos?@=1 1+ tan? @ =sec’ 0 1+ cot’? @= cosec?@ Copyright reserved

Downloaded from hlayiso.com Technical Mathematics/P2 DBE/November 2023 NSC mrad = 180° Angular velocity =o = 217 where n = rotation frequency Angular velocity = @ = 360° n where n = rotation frequency Circumferential velocity = v= 2 Dn where D = diameter and 7 = rotation frequency Circumferential velocity = v= ar where @ = angular velocity and r = radius Arc length = s=r@ where r = radius and @ = central angle in radians Area of a sector = rs where r = radius, s = arc length 2 Area of a sector = me where r = radius and @ = central angle in radians 4h? -4dh+x? =0 where h = height of segment, d= diameter of circle and x = length of chord . 0, + 0, A, =a(m,+m,+m,+...+ m,) where a = width of equal parts, m, = = o, =n" ordinate and n= number of ordinates OR a+ Ay no{ 2224 0 + 05+ wt on} where a = width of equal parts, 0, = n” ordinate and n = number of ordinates Copyright reserved