Monday, 29 July 2024 15:07

TECHNICAL MATHEMATICS P2 with Memorandum - 2024 Grade 12 June Common Exams

MARKS: 150

TIME: 3 hours

INSTRUCTIONS AND INFORMATION

1. This question paper consists of ELEVEN questions.

3. Clearly show ALL calculations, diagrams, graphs, et cetera which you have used in  determining the answers.

4. Answers only will NOT necessarily be awarded full marks.

5. You may use an approved scientific calculator (non-programmable and non graphical) unless stated otherwise.

6. If necessary, round off your answers to TWO decimal places, unless stated  otherwise.

7. Diagrams are NOT necessarily drawn to scale.

8. An information sheet with formulae is included at the end of the question paper.

9. Write neatly and legibly.

QUESTION 1

In the diagram below, AOB is a triangle with vertices A(15 ; 18); O(0 ; 0) and B(23 ; 5).  β is the angle of inclination of line OB and α is the angle of inclination of line OA.

1.1 Determine the gradients of OA and OB. (4)

1.2 Determine the angle of inclination of line OB. (3)

1.3 Find the size of AOB, correct to the nearest whole number. (4)

1.4 AOBM is a parallelogram. Find the coordinates of M. (5) [16]

QUESTION 2

2.1 The diagram below shows the circle with equation x2 + y2 = 13. The contact point  of a tangent to the circle is at A(− 2 ; − 3).

2.1.1 Write down the radius of the circle in simplified surd form. (1)

2.1.2 Determine the equation of the tangent to the circle at point A in the form  y = . .. (4)

2.1.3 Write down the coordinates of another point where the line AO intersects  with the circle. (2)

2.2 Sketch the graph of . Clearly indicate the intercepts. (3) [10]

QUESTION 3

3.1 In the diagram below A(a ; √3 ) and OA = 3.

Determine the value of the following, without using a calculator:

3.1.1 a (3)

3.1.2 sec θ (1)

3.1.3 cosec (θ + 360°) (3)

3.2 Determine the values of x, if tan(x − 30°) = −0,982 and 0° ≤ x − 30° ≤ 360°. (4) [11]

QUESTION 4

4.1 Simplify: [sin(180°− θ )tan(180° + θ)sin(270°)] ÷ cos(360°− θ)tan(180°− θ) (6)

4.2 Prove that: (cosec B  − cot B )2 = (1 + cos B) ÷ (1 − cos B) (6) [12]

QUESTION 5

Given the functions defined by f(x) = cos(x − 30) and g(x) = 2 sin x for x ∈ (0° ;360°).

5.1 Write down the period of f. (1)

5.2 Write down the amplitude of g. (1)

5.3 On the same axes, given in your SPECIAL ANSWER BOOK, draw the graphs of f and g. Clearly show the turning points, endpoints and the intercepts with the axes. (8)

5.4 Use your graphs to determine for which values of x is:

5.4.1 g(x) ≥ 0 (2)

5.4.2 f(x). g(x) <0 in the second quadrant (2) [14]

QUESTION 6

6.1 Write down the sine rule for ΔABD. (1)

6.2 A ship at sea, observes that the angles of elevation to the top and bottom of a  lighthouse on a cliff are 7° and 9° respectively. It is known that the height of the  lighthouse is 48 m.

6.2.1 The size of BAC, stating a reason (2)

6.2.2 The size of ABD, stating a reason (2)

6.2.3 The length of AC (4)

6.2.4 The distance between the ship and the bottom of the cliff (2)

6.2.5 The height of the cliff (3) [14]

QUESTION 7

In the diagram below, O is the centre of the circle. OS ∥ PQ and PS meet OQ at T. Q S

7.1 If P1 = X, express T1 in terms of x. Give reasons. (6)

7.2 If x = 30°, calculate the sizes of the angles in ΔQST. Give reasons where necessary. (5)

7.3 Show that ΔPQS ≡ ΔSOP. (3) [14]

QUESTION 8

PQRS is a cyclic quadrilateral with PS = PQ. SR is produced to meet V such that PR ∥ QV.  TP and TS are tangents to the circle. Q1 = x .

8.1 Name, with reasons, four other angles equal to x. (8)

8.2 Give a reason for P4 = S3. (1)

8.3 Prove, with reasons, that T = QPS. (5) [14]

QUESTION 9

In the diagram below, B is the midpoint of side QR. C and D are points on PR such that  PC = CD = DR. PR = 15 cm.

9.1 Show that BD ∥ QC. (3)

9.2 Prove that PA = AB. (3)

9.3 Determine the length of QR, if PD : DR = 2 : 1. (6) [12]

QUESTION 10

A fan in a jet engine has a diameter of 340 cm and a circumferential velocity of 568 metres per second.

10.1 Convert 568 m/s to km/h. (2)

10.2 Determine the rotational frequency of the wheel in hours. (5)

10.3 Determine the angular velocity of the wheel in seconds. (3)

10.4 Determine the distance, in km, a point on the fan will cover in 15 seconds. (3)

10.5 Determine how long it will take the fan to make half a revolution. (2) [15]

QUESTION 11

11.1 A pendulum in a clock, FIGURE A, follows the path as depicted in the diagram  below, FIGURE B. There is a radius of 30 cm and the angle formed is 60°.

11.1.1 Determine the length of arc CD, that the pendulum follows. (3)

11.1.2 Determine the area of sector ECD. (3)

11.1.3 Calculate the length of the pendulum. (3)

11.2 An analogue clock has a diameter of 30 cm, and a chord length of 20 cm.

Determine the length of the hour hand. (5)

11.3 The ordinates in the irregular figure are: 7 cm, 6,5 cm, x , 5 cm; 7,5 cm, 6 cm, 4 cm  and 3 cm respectively as indicated in the diagram below. The width of the irregular  figure is 11,55 cm and the area is 63,525 cm2

Determine the length of the unknown ordinate x. (4) [18]

TOTAL: 150

INFORMATION SHEET: TECHNICAL MATHEMATICS

Angular velocity =  ω = 2 π n             where n = rotation frequency

Angular velocity =  ω = 360° n            where n = rotation frequency

Circumferential velocity= v = π Dn      where D = diameter and n = rotation frequency

Circumferential velocity= ν = ωr            where ω= Angular velocity and r = radius

Arc length s = rθ                  where r = radius and θ = central angle in radians

Area of a sector 2r/s      =    where r = radius, s = arc length

MARKING GUIDELINE

NOTE:

• Continuous accuracy (CA) applies only where indicated in this marking guideline.
• Assuming values/answers in order to solve a problem is unacceptable.
 MARKING CODES / NASIENKODES M Method/Metode A Accuracy/Akkuraatheid AO Answer only/Slegs antwoord CA Consistent accuracy/Deurlopende akkuraatheid F Formula/Formule I Identity/Identiteit R Rounding/Afronding S Simplification/Vereenvoudiging ST Statement/Bewering RE Reason/Rede ST RE Statement and correct reason/Bewering en korrekte rede SF Substitution correctly in correct formula/Korrekte vervanging in die korrekte formule NPU No penalty for omitting units/Geen penalisering vir eenhede uitgelaat
 1.1 mOA = (18 − 0) ÷ (15 − 0) mOA =6/5 mOB = (5 − 0) ÷ (23 − 0) mOB =5/23AO: Full marks / Volpunte ✓ M ✓ S A ✓ M ✓ S A   (4) 1.2 tan β = mOB tan β  =5/23 ∴ β = 12,26° ✓ M ✓ Subst./Vervang A ✓ S CA n(3) 1.3 tan α = mOB tan α =65 ∴ α = 50,19° ∴ AOB = 50,19° − 12,26° = 37,93° ≈ 38° ✓ M ✓ Subst. A ✓ S CA ✓ Answer/Antwoord CA (4)
 1.4 x-coordinate of /koördinate van M =23+15 =38 y-coordinate of /koördinate M = 23 + 15 =38 M(38;23) (If learner used this method award full  marks/Indien leerling hierdie metode gebruik  ken volpunte toe.) ✓ M ✓ Subst. A ✓ M CA ✓ S CA ✓ Answer/Antwoord CA (5) [16]

 2.1 2.1.1 r = √13 ✓ A  (1) 2.1.2 ✓ F ✓ SF A✓ S ✓ equation / vergelyking CA OR/OF ✓ grad. radius A ✓ grad. tan / raaklyn CA✓ SF A ✓ equation / verglyking CA  (4) 2.1.3 (3 ; 2) ✓ x-value/waarde A✓ y-value/ waarde A (2)

 2.2 ✓ elliptical shape /   elliptiese vorm A ✓ x-intercepts /   afsnitte A ✓ y-intercepts /   afsnitte A (3) [10]

 3.1 3.1.1 a2 + (√3)2= (3)2 a2 = 9 − 3 a2 = 6 a = ±√6 ∴ a = −√6 ✓ M  ✓ S ✓ value of / waarde van a  (3) 3.1.2 −√6 / 3 ✓ Answer/Antwoord (1) 3.1.3 cosec (θ + 360°) = cosec θ∴ cosec θ =3 / √3 ✓ Reduction/ Reduksie A✓✓ cosec ratio / verh. CA (3) 3.2 tan (x − 30°) = −0,982 Ref. /Verw. < = tan-1(0,982) Ref. /Verw. < = 44,48° ∴ I: x − 30° = 180° − 44,48° ∴ x = 165,52° AND/EN ∴ IV: x − 30° = 360° − 44,48° ∴ x = 345,52° ✓ Ref. / Verw. ∠ CA✓ Quadrants / kwadrante A✓✓ values of x / waardes van x CA (4) [11]

 4.1 ✓ sin(θ) A  ✓ tan(θ) A ✓ −1 A ✓ cos(θ) A ✓ − tan(θ) A ✓ −tan(θ) A (6) 4.2 [12]

QUESTION/VRAAG 5

 f(x) = cos(x − 30) and /en g(x) = 2 sin x for/ vir x ∈ (0° ; 360°) 5.1 periodf = 360° ✓ A  (1) 5.2 Amplitudeg = 2 ✓ A (1) 5.3 f: ✓ y-intercept at / y-afsnit   by 0,87 ✓ x-intercepts at 120° and   300° x-afsnitte by 120° en 300°✓✓ turning point at / draaipunt by (30°; 1)   and/en (210° ; -1) g: ✓ y-intercept at / y-afsnit by 0 ✓ x-intercept at 180° and  360° / x-afsnit by 180° en  360° ✓✓ turning point at / draaipunt by (90°; 1) and/en (270° ; -1) (8) 5.4.1 0° ≤ x ≤ 180° ✓0° ≤ x CA✓x ≤ 180° CA (2) 5.4.2 120° < x < 180° and /en 300° < x < 360° ✓120° < x < 180° CA✓300° < x < 360° CA (2) [14]

QUESTION/VRAAG 6

 6.1 a/sin A = b/sinB = d/sinD ✓ A (1) 6.2 6.2.1 BAC = 9° − 7° = 2° (given/gegee) ✓ ST A ✓ RE   (2) 6.2.2 ABD = 180° − 9° − 90° = 81° (int. <’s of Δ / binne <’e van Δ) ✓ ST A ✓ RE  (2) 6.2.3 ✓ F ✓ SF A ✓ S CA ✓ AC value / waarde (4) 6.2.4 ✓ ratio / verh. A ✓ value / waarde CA (2) 6.2.5 AC2 = AD2 + CD2 (1 358,44)2 = (1 348,31)2 + CD2 27419,3775 = CD2165,59 m = CD ✓ F ✓ SF CA ✓ height / hoogte CA (3) [14]

QUESTION/VRAAG 7

 7.1 S2 = X (alt. <’s = / verw. <’e =; PQ∥SO) O2 = 2x (< at centre = 2×< at circumf. /midpts< = 2× omtreks<)T1 = 3x (ext. < of Δ / buite < van Δ) ✓ ST ✓ RE A ✓ ST ✓ RE A ✓ ST ✓ RE A (6) 7.2 T1 = 3xT1 = 3(30°) = 90° O2 = 2xO2 = 2(30°) = 60° ∴ Q2 = 60° (OQ = OS; Radii) ∴ S1 = 30° ✓ T1  CA✓ O2 CA✓ Q2 ✓RE CA✓ S1 CA(5) 7.3 In Δ PQS and/en Δ SOP: 1. S2 = S1 (proven / bewys) 2. QS = PO (given / gegee) 3. P1 = P2 (given / gegee) ∴ Δ PQS ≡ Δ SOP (<

QUESTION/VRAAG 8

 8.1 S1 = x (tan-chord thm / stelling) P2 = x (<’s in same segment / <’e in dieselfde segment)R2 = x (alt. <’s = / verw. <’e =; VQ∥RP) S2 = x (<’s in same segment / <’e in dieselfde segment) ✓ ST ✓ RE A ✓ ST ✓ RE A ✓ ST ✓ RE A ✓ ST ✓ RE A (8) 8.2 Tangents from same point / Raaklyne vanuit dieselfde punt ✓ RE A  (1) 8.3 P4 = R3 = x (tan-chord thm / stelling) ∴ T = 180° − 2x (int. <’s of Δ / binne <’e van Δ) P2 + P3 = 180° − 2x (int. <’s of Δ / binne <’e van Δ) ∴ T= QPS ✓ ST    ✓ RE   A✓ ST   ✓ RE   A✓ ST A(5) [14]

QUESTION/VRAAG 9

 9.1 In ΔCRQ: RD = DC (given / gegee) RB = BQ (given / gegee) ∴ BD ∥ CQ and/en BD = ½ CQ (midpt thm / stelling) ✓ ST A ✓ ST A ✓ RE A (3) 9.2 In ΔPBD: PC = CD (given / gegee) BD ∥ CQ (given / gegee) ∴ PA = AB and/en AC = ½ BD (midpt thm / stelling) ✓ ST A ✓ ST A ✓ RE A (3) 9.3 Δ PAC and/en Δ PBD: 1. P = P (common < / gemene <) 2. PAC = PBD (Corresp. <’s = / ooreenk. <’e =; AC∥BD)3. PCA = PDB (3rd < of Δ / 3de < van Δ) ∴ Δ PAC lll Δ PBD (<<<) ∴ BD : AC = 2 : 1 ✓ ST A ✓ ST A ✓ ST A ✓ ST A ✓ Subst./Vervang A ✓ Answer/Antwoord A  (6) [12]

QUESTION/VRAAG 10

 10.1 ✓ conversion factors /  herleidingsfaktore ✓ answer / antwoord (2) 10.2 ✓F ✓ conversion / herleiding ✓ SF A ✓ S ✓ answer / antwoord (5) 10.3 ω π = 2 n = × 2 38,20 π ≈ 240,02 rad/s ✓F ✓ SF CA ✓ answer / antwoord (3) 10.4 s = vt OR / OF D = ST s = 0,568 × 15 s = 8,52 km ✓F ✓ SF CA ✓ answer / antwoord (3) 10.5 n = number of revolutions / time53,18 =1/2  ⁄  t∴ t ≈ 0,009 sec/sek ✓ SF CA ✓ answer / antwoord (2) [15]

QUESTION/VRAAG 11

 11.1.1 s = rθs = (30) (60° ×π/180°) ∴ CD = 10π ≈ 31,42 cm ✓F ✓ SF A ✓ RT length / lengte (3) 11.1.2 ✓F ✓ SF A ✓ Area OR / OF ✓F ✓ SF A ✓ Area (3) 11.1.3 D = 2r D = 2(30)  ∴ D = 60 cm ✓M ✓Subst./Vervang ✓ Answer/Antwoord (3)
 11.2 4ℎ2 − 4dℎ + x2 = 0 4ℎ2 − 4(30)ℎ + (20)2 = 0 4ℎ2 − 120ℎ + 400 = 0 ℎ2 − 30ℎ + 100 = 0  ∴ ℎ = 3,82 cm ∴ h hour hand length/lengte van uur aanwyser = radius−height of minor segment/hoogte van kleiner segment ∴ hour hand length/lengte van uur aanwyser = (30/2) − 3,82∴ hour hand length/lengte van uuraanwyser = 11,2 cm ✓F ✓SF A ✓S ✓ value of / van ℎ✓ length/lengte (5)
 11.3 3 cm ✓F ✓SF A ✓S ✓ value of/waarde van a OR/ OF ✓F ✓SF A ✓S ✓ value of/ waarde van a (4) [18] TOTAL/TOTAAL: 150