Trigonometry Practice (withCalculator) [178 marks]

1. The following diagram shows the chord [AB] in a circle of radius 8 cm,where .

Find the area of the shaded segment.

AB = 12 cm[7 marks]

Markschemeattempt to find the central angle or half central angle (M1)

eg , cosine rule, right triangle

correct working (A1)

eg

correct angle (seen anywhere) eg (A1)

correct sector areaeg (A1)

area of triangle (seen anywhere) (A1) egappropriate approach (seen anywhere) (M1)eg , their sector-their triangle22.5269area of shaded region A1 N4 Note: Award M0A0A0A0A1 then M1A0 (if appropriate) for correct trianglearea without any attempt to find an angle in triangle OAB. [7 marks]

cos θ = , sin−1 ( ) , 0.722734, 41.4096∘, − sin−1 ( )82+82−122

2∙8∙868

π2

68

AOB

1.69612, 97.1807∘, 2 × sin−1 ( )68

(8)(8)(1.70), (64π), 54.275912

97.1807360

(8)(8) sin 1.70, (8)(12) sin 0.722, × √64 − 36 × 12, 31.749012

12

12

Atriangle − Asector

= 22.5 (cm2)

2a.

At Grande Anse Beach the height of the water in metres is modelled by the function , where is the number of hours after 21:00 hours on 10

December 2017. The following diagram shows the graph of , for .

The point represents the first low tide and represents thenext high tide.

How much time is there between the first low tide and the next high tide?

Markschemeattempt to find the difference of -values of A and B (M1)eg6.25 (hours), (6 hours 15 minutes) A1 N2[2 marks]

h(t) = p cos(q × t) + r th 0 ⩽ t ⩽ 72

A(6.25, 0.6) B(12.5, 1.5)

x

6.25 − 12.5

2b. Find the difference in height between low tide and high tide.

Markschemeattempt to find the difference of -values of A and B (M1)eg

A1 N2[2 marks]

y

1.5 − 0.6

0.9 (m)

2c. Find the value of ;p

[2 marks]

[2 marks]

[2 marks]

Markschemevalid approach (M1)

eg A1 N2

[2 marks]

, 0.9 ÷ 2max−min2

p = 0.45

2d. Find the value of ;

MarkschemeMETHOD 1period (seen anywhere) (A1)valid approach (seen anywhere) (M1)eg

0.502654 A1 N2

METHOD 2attempt to use a coordinate to make an equation (M1)egcorrect substitution (A1)eg0.502654

A1 N2

[3 marks]

q

= 12.5

period = , q = , 2πb

2πperiod

2π12.5

q = , 0.503 (or − , − 0.503)4π25

4π25

p cos(6.25q) + r = 0.6, p cos(12.5q) + r = 1.5

0.45 cos(6.25q) + 1.05 = 0.6, 0.45 cos(12.5q) + 1.05 = 1.5

q = , 0.503 (or − , − 0.503)4π25

4π25

2e. Find the value of .r

[3 marks]

[2 marks]

Markschemevalid method to find (M1)

eg A1 N2

[2 marks]

r

, 0.6 + 0.45max+min2

r = 1.05

2f. There are two high tides on 12 December 2017. At what time does thesecond high tide occur?

MarkschemeMETHOD 1attempt to find start or end -values for 12 December (M1)egfinds -value for second max (A1)

23:00 (or 11 pm) A1 N3METHOD 2 valid approach to list either the times of high tides after 21:00 or the -values ofhigh tides after 21:00, showing at least two times (M1) egcorrect time of first high tide on 12 December (A1)eg 10:30 (or 10:30 am) time of second high tide = 23:00 A1 N3METHOD 3attempt to set their equal to 1.5 (M1)egcorrect working to find second max (A1)eg23:00 (or 11 pm) A1 N3[3 marks]

t

3 + 24, t = 27, t = 51

t

t = 50

t

21:00 + 12.5, 21:00 + 25, 12.5 + 12.5, 25 + 12.5

h

h(t) = 1.5, 0.45 cos( t) + 1.05 = 1.54π25

0.503t = 8π, t = 50

[3 marks]

3a.

The following diagram shows a circle with centre O and radius 40 cm.

The points A, B and C are on the circumference of the circle and .

Find the length of arc ABC.

Markschemecorrect substitution into arc length formula (A1)eg

A1 N2[2 marks]

AOC = 1.9 radians

(40)(1.9)

arc length = 76 (cm)

3b. Find the perimeter of sector OABC.

Markschemevalid approach (M1)eg

A1 N2[2 marks]

arc + 2r, 76 + 40 + 40

perimeter = 156 (cm)

3c. Find the area of sector OABC.

[2 marks]

[2 marks]

[2 marks]

Markschemecorrect substitution into area formula (A1)eg

A1 N2[2 marks]

(1.9)(40)212

area = 1520 (cm2)

4a.

The depth of water in a port is modelled by the function , for , where is the number of hours after high tide.

At high tide, the depth is 9.7 metres.At low tide, which is 7 hours later, the depth is 5.3 metres.

Find the value of .

Markschemevalid approach (M1)

eg , sketch of graph, A1 N2

[2 marks]

d(t) = p cos qt + 7.50 ⩽ t ⩽ 12 t

p

max−min2 9.7 = p cos(0) + 7.5

p = 2.2

4b. Find the value of .

Markschemevalid approach (M1)eg , period is

0.448798, (do not accept degrees) A1 N2

[2 marks]

q

B = 2πperiod 14, , 5.3 = 2.2 cos 7q + 7.5360

14

q = ( )2π14

π7

[2 marks]

[2 marks]

4c. Use the model to find the depth of the water 10 hours after high tide.

Markschemevalid approach (M1)eg7.010457.01 (m) A1 N2[2 marks]

d(10), 2.2 cos( ) + 7.520π14

5a.

A ship is sailing north from a point A towards point D. Point C is 175 km north of A.Point D is 60 km north of C. There is an island at E. The bearing of E from A is 055°.The bearing of E from C is 134°. This is shown in the following diagram.

Find the bearing of A from E.

Markschemevalid method (M1)eg235° (accept S55W, W35S) A1 N2[2 marks]

180 + 55, 360 − 90 − 35

[2 marks]

[2 marks]

5b. Finds CE.

Markschemevalid approach to find (may be seen in (a)) (M1)

egcorrect working to find (may be seen in (a)) (A1)

eg , evidence of choosing sine rule (seen anywhere) (M1)egcorrect substitution into sine rule (A1)eg

146.034 A1 N2

[5 marks]

AEC

AEC = 180 − 55 − ACE, 134 = E + 55

AEC

180 − 55 − 46, 134 − 55 AEC = 79∘

=asinA

bsinB

=CEsin55∘

175sinAEC

CE = 146 (km)

5c. Find DE.

Markschemeevidence of choosing cosine rule (M1)egcorrect substitution into right-hand side (A1)eg192.612

A1 N2[3 marks]

DE2 = DC2 + CE2 − 2 × DC × CE × cos θ

602 + 146.0342 − 2 × 60 × 146.034 cos 134

DE = 193 (km)

5d. When the ship reaches D, it changes direction and travels directly to theisland at 50 km per hour. At the same time as the ship changes direction,a boat starts travelling to the island from a point B. This point B lies on (AC),between A and C, and is the closest point to the island. The ship and the boat arriveat the island at the same time. Find the speed of the boat.

[5 marks]

[3 marks]

[5 marks]

Markschemevalid approach for locating B (M1)eg BE is perpendicular to ship’s path, angle correct working for BE (A1)egvalid approach for expressing time (M1)egcorrect working equating time (A1)eg27.269427.3 (km per hour) A1 N3[5 marks]

B = 90

sin 46∘ = , BE = 146.034 sin 46∘, 105.048BE146.034

t = , t = , t =ds

dr

192.61250

= , =146.034sin46∘

r192.612

50s

105.04850

192.612

6a.

The following diagram shows the graph of , for .

The graph of has a minimum point at and a maximum point at .

(i) Find the value of .(ii) Show that .(iii) Find the value of .

f(x) = a sin bx + c 0 ⩽ x ⩽ 12

f (3, 5) (9, 17)

c

b = π6

a

[6 marks]

Markscheme(i) valid approach (M1)

eg A1 N2

(ii) valid approach (M1)eg period is 12, per

A1 AG N0

(iii) METHOD 1valid approach (M1)eg , substitution of points

A1 N2METHOD 2valid approach (M1)

eg , amplitude is 6 A1 N2

[6 marks]

5+172

c = 11

= , 9 − 32πb

b = 2π12

b = π6

5 = a sin( × 3) + 11π6

a = −6

17−52

a = −6

6b.

The graph of is obtained from the graph of by a translation of . The

maximum point on the graph of has coordinates .

(i) Write down the value of .(ii) Find .

Markscheme(i) A1 N1(ii) A2 N2

[3 marks]

g f ( k

0)

g (11.5, 17)

k

g(x)

k = 2.5

g(x) = −6 sin( (x − 2.5)) + 11π6

[3 marks]

6c.

The graph of changes from concave-up to concave-down when .

(i) Find .(ii) Hence or otherwise, find the maximum positive rate of change of .

Markscheme(i) METHOD 1 Using recognizing that a point of inflexion is required M1eg sketch, recognizing change in concavityevidence of valid approach (M1)eg , sketch, coordinates of max/min on

(exact) A1 N2METHOD 2 Using recognizing that a point of inflexion is required M1eg sketch, recognizing change in concavityevidence of valid approach involving translation (M1)eg , sketch,

(exact) A1 N2(ii) valid approach involving the derivative of or (seen anywhere) (M1)

eg , max on derivative, sketch of derivativeattempt to find max value on derivative M1eg , dot on max of sketch3.14159max rate of change (exact), 3.14 A1 N2[6 marks]

g x = w

w

g

g

g′′(x) = 0 g′

w = 8.5f

x = w − k 6 + 2.5w = 8.5

g f

g′(w), − π cos( x)π6

−π cos( (8.5 − 2.5)), f ′(6)π6

= π

[6 marks]

7a.

The following diagram shows a circle, centre O and radius mm. The circle isdivided into five equal sectors.

One sector is OAB, and .

Write down the exact value of in radians.

Markscheme A1 N1

[1 mark]

r

AOB = θ

θ

θ = 2π5

7b.

The area of sector AOB is .

Find the value of .

Markschemecorrect expression for area (A1)

egevidence of equating their expression to (M1)eg

A1 N2[3 marks]

20π mm2

r

A = r2 ( ) , 12

2π5

πr2

5

20π

r2 ( ) = 20π, r2 = 100, r = ±1012

2π5

r = 10

7c. Find AB.

[1mark]

[3 marks]

[3 marks]

MarkschemeMETHOD 1evidence of choosing cosine rule (M1)egcorrect substitution of their and into RHS (A1)eg11.7557

A1 N2METHOD 2evidence of choosing sine rule (M1)egcorrect substitution of their and (A1)

eg11.7557

A1 N2[3 marks]

a2 = b2 + c2 − 2bc cos A

r θ

102 + 102 − 2 × 10 × 10 cos( )2π5

AB = 11.8 (mm)

=sinAa

sinBb

r θ

=sin 2π

5

AB

sin( (π− ))12

2π

5

10

AB = 11.8 (mm)

8a.

The following diagram shows three towns A, B and C. Town B is 5 km from Town A,on a bearing of 070°. Town C is 8 km from Town B, on a bearing of 115°.

Find .ABC [2 marks]

Markschemevalid approach (M1)eg

A1 N2[2 marks]

70 + (180 − 115), 360 − (110 + 115)

ABC = 135∘

8b. Find the distance from Town A to Town C.

Markschemechoosing cosine rule (M1)egcorrect substitution into RHS (A1)eg12.065112.1 (km) A1 N2[3 marks]

c2 = a2 + b2 − 2ab cos C

52 + 82 − 2 × 5 × 8 cos 135

8c. Use the sine rule to find .

Markschemecorrect substitution (must be into sine rule) A1

eg17.0398

A1 N1[2 marks]

ACB

=sinACB5

sin135AC

ACB = 17.0

[3 marks]

[2 marks]

9a.

The following diagram shows a quadrilateral ABCD.

Find BD.

Markschemeevidence of choosing sine rule (M1)egcorrect substitution (A1)eg9.42069

A1 N2[3 marks]

AD = 7 cm, BC = 8 cm, CD = 12 cm, DAB = 1.75 radians, ABD = 0.82 radians.

=asinA

bsinB

=asin1.75

7sin0.82

BD = 9.42 (cm)

9b. Find .DBC

[3 marks]

[3 marks]

Markschemeevidence of choosing cosine rule (M1)

egcorrect substitution (A1)

eg1.51271

(radians) (accept 86.7°) A1 N2[3 marks]

cos B = , a2 = b2 + c2 − 2bc cos Bd2+c2−b2

2dc

, 144 = 64 + BD2 − 16BD cos B82+9.420692−122

2×8×9.42069

DBC = 1.51

10a.

The height, metros, of a seat on a Ferris wheel after minutes is given by

Find the height of the seat when .

Markschemevalid approach (M1)eg

A1 N2[2 marks]

h t

h(t) = −15 cos 1.2t + 17, for t ⩾ 0.

t = 0

h(0), − 15 cos(1.2 × 0) + 17, − 15(1) + 17

h(0) = 2 (m)

10b. The seat first reaches a height of 20 m after minutes. Find .k k

[2 marks]

[3 marks]

Markschemecorrect substitution into equation (A1)egvalid attempt to solve for (M1)

eg ,

1.47679 A1 N2

[3 marks]

20 = −15 cos 1.2t + 17, − 15 cos 1.2k = 3

k

cos 1.2k = − 315

k = 1.48

10c. Calculate the time needed for the seat to complete a full rotation, givingyour answer correct to one decimal place.

Markschemerecognize the need to find the period (seen anywhere) (M1)eg next value when correct value for period (A1)eg5.2 (min) (must be 1 dp) A1 N2[3 marks]

t h = 20

period = , 5.23598, 6.7 − −1.482π1.2

3

[3 marks]

11a.

The following diagram shows a circle with centre and radius cm.

Points A, B, and C lie on the circle, and .

Find the length of arc .

Markschemecorrect substitution (A1)eg = (cm) A1 N2

[2 marks]

O 3

AOC = 1.3 radians

ABC

l = 1.3 × 3l 3.9

11b. Find the area of the shaded region.

[2 marks]

[4 marks]

MarkschemeMETHOD 1valid approach (M1)

eg finding reflex angle, correct angle (A1)egcorrect substitution (A1)eg

A1 N3METHOD 2correct area of small sector (A1)egvalid approach (M1)eg circle − small sector, correct substitution (A1)eg

A1 N3[4 marks]Total [6 marks]

2π − COA

2π − 1.3, 4.98318

(2π − 1.3)3212

22.4243

area = 9π − 5.85 (exact), 22.4 (cm2)

(1.3)32, 5.8512

πr2 − θr212

π(32) − (1.3)3212

22.4243

area = 9π − 5.85 (exact), 22.4 (cm2)

12a.

The following diagram shows the quadrilateral .

Find .

Markschemeevidence of choosing sine rule (M1)eg

correct substitution (A1)eg

A1 N2[3 marks]

ABCD

AD = 6 cm, AB = 15 cm, ABC = 44∘, ACB = 83∘andDAC = θ

AC

=ACsinCBA

ABsinACB

=ACsin44∘

15sin83∘

10.4981

AC = 10.5 (cm)

12b. Find the area of triangle .ABC

[3 marks]

[3 marks]

Markschemefinding (seen anywhere) (A1)

egcorrect substitution for area of triangle A1eg62.8813

A1 N2[3 marks]

CAB

180∘ − 44∘ − 83∘, CAB = 53∘

ABC

× 15 × 10.4981 × sin 53∘12

area = 62.9 (cm2)

12c. The area of triangle is half the area of triangle .Find the possible values of .

Markschemecorrect substitution for area of triangle (A1)egattempt to equate area of triangle to half the area of triangle (M1)egcorrect equation A1eg

, A1A1 N2

[5 marks]

ACD ABC

θ

DAC

× 6 × 10.4981 × sin θ12

ACD ABC

area ACD = × area ABC; 2ACD = ABC12

× 6 × 10.4981 × sin θ = (62.9), 62.9887 sin θ = 62.8813, sin θ = 0.99829412

12

86.6531 93.3468θ = 86.7∘ , θ = 93.3∘

12d. Given that is obtuse, find .θ CD

[5 marks]

[3 marks]

MarkschemeNote: Note: If candidates use an acute angle from part (c) in the cosine rule,award M1A0A0 in part (d). evidence of choosing cosine rule (M1)egcorrect substitution into rhs (A1)eg

A1 N2[3 marks]Total [14 marks]

CD2 = AD2 + AC2 − 2 × AD × AC × cos θ

CD2 = 62 + 10.4982 − 2(6)(10.498) cos 93.336∘

12.3921

12.4 (cm)

13a.

The following diagram shows triangle .

Find .

ABC

BC = 10 cm, ABC = 80∘ and BAC = 35∘.

AC [3 marks]

Markschemeevidence of choosing sine rule (M1)eg

correct substitution (A1)eg

A1 N2[3 marks]

=ACsin(ABC)

BCsin(BAC)

=ACsin80∘

10sin35∘

AC = 17.1695

AC = 17.2 (cm)

13b. Find the area of triangle .

Markscheme(seen anywhere) (A1)

correct substitution (A1)eg

A1 N2[3 marks]Total [6 marks]

ABC

ACB = 65∘

× 10 × 17.1695 × sin 65∘12

area = 77.8047

area = 77.8 (cm2)

[3 marks]

14a.

The following diagram shows a square , and a sector of a circle centre , radius . Part of the square is shaded and labelled .

Show that the area of the square is .

Markschemearea of (seen anywhere) (A1)choose cosine rule to find a side of the square (M1)egcorrect substitution (for triangle ) A1egcorrect working for A1eg

AG N0 Note: Award no marks if the only working is .[4 marks]

ABCD OABO r R

AOB = θ, where 0.5 ≤ θ < π.

ABCD 2r2(1 − cos θ)

ABCD = AB2

a2 = b2 + c2 − 2bc cos θ

AOB

r2 + r2 − 2 × r × r cos θ, OA2 + OB2 − 2 × OA × OB cos θ

AB2

2r2 − 2r2 cos θ

area = 2r2(1 − cos θ)

2r2 − 2r2 cos θ

14b. When , the area of the square is equal to the area of thesector .

(i) Write down the area of the sector when .(ii) Hence find .

θ = α ABCDOAB

θ = α

α

[4 marks]

[4 marks]

Markscheme(i) A1 N1(ii) correct equation in one variable (A1)eg

A2 N2 Note: Award A1 for and additional answers.[4 marks]

αr2 (accept 2r2(1 − cos α))12

2(1 − cos α) = α12

α = 0.511024

α = 0.511 (accept θ = 0.511)

α = 0.511

15a.

The following diagram shows a circle with centre and radius cm.

The points , and are on the circumference of the circle, and radians.

Find the length of arc .

Markschemecorrect substitution into formula (A1)eg

A1 N2[2 marks]

O 8

A B C AOB

ACB

l = 1.2 × 8

9.6 (cm)

15b. Find .AB

[2 marks]

[3 marks]

MarkschemeMETHOD 1evidence of choosing cosine rule (M1)

egcorrect substitution into right hand side (A1)eg

A1 N2METHOD 2evidence of choosing sine rule (M1)eg

finding angle or (may be seen in substitution) (A1)

eg A1 N2

[3 marks]

2r2 − 2 × r2 × cos(AOB)

82 + 82 − 2 × 8 × 8 × cos(1.2)

9.0342795

AB = 9.03 [9.03, 9.04] (cm)

=ABsin(AOB)

OBsin(OAB)

OAB OBA

, 0.970796π−1.22

AB = 9.03 [9.03, 9.04] (cm)

15c. Hence, find the perimeter of the shaded segment .

Markschemecorrect working (A1)eg

A1 N2[2 marks]Total [7 marks]

ABC

P = 9.6 + 9.0318.6342

18.6 [18.6, 18.7] (cm)

= sin( ) +

[2 marks]

16a.

The following diagram shows part of the graph of .

The point is a maximum point and the point is a minimum point.Find the value of

;

Markschemevalid approach (M1)

eg A1 N2

[2 marks]

y = p sin(qx) + r

A ( , 2)π6 B ( , 1)π

6

p

, 2 − 1.52−12

p = 0.5

16b. ;

Markschemevalid approach (M1)

eg A1 N2

[2 marks]

r

1+22

r = 1.5

16c. .q

[2 marks]

[2 marks]

[2 marks]

MarkschemeMETHOD 1valid approach (seen anywhere) M1eg

period (seen anywhere) (A1)

A1 N2METHOD 2attempt to substitute one point and their values for and into M1

egcorrect equation in (A1)

eg A1 N2

METHOD 3valid reasoning comparing the graph with that of R1eg position of max/min, graph goes fastercorrect working (A1)eg max at not at , graph goes times as fast

A1 N2[3 marks]Total [7 marks]

q = , 2πperiod

2π

( )2π

3

= 2π3

q = 3

p r y

2 = 0.5 sin(q ) + 1.5, = 0.5 sin(q1) + 1.5π6

π2

q

q = , q =π6

π2

π2

3π2

q = 3

sin x

π6

π2 3

q = 3

17a.

The following diagram shows triangle ABC.

Find AC.

Markschemeevidence of choosing cosine rule (M1)

eg correct substitution into the right-hand side (A1)eg

A1 N2[3 marks]

AC2 = AB2 + BC2 − 2(AB)(BC) cos(ABC)

62 + 102 − 2(6)(10) cos 100∘

AC = 12.5234

AC = 12.5 (cm)

17b. Find .BCA

[3 marks]

[3 marks]

Markschemeevidence of choosing a valid approach (M1)eg sine rule, cosine rulecorrect substitution (A1)

eg

A1 N2[3 marks]

= , cos(BCA) =sin(BCA)6

sin100∘

12.5(AC)2+102−62

2(AC)(10)

BCA = 28.1525

BCA = 28.2∘

18a.

The population of deer in an enclosed game reserve is modelled by the function, where

is in months, and corresponds to 1 January 2014.

Find the number of deer in the reserve on 1 May 2014.

Markscheme (A1)

correct substitution into formula (A1)eg

969 (deer) (must be an integer) A1 N3[3 marks]

P(t) = 210 sin(0.5t − 2.6) + 990t

t = 1

t = 5

210 sin(0.5 × 5 − 2.6) + 990, P(5)

969.034982 …

18b. Find the rate of change of the deer population on 1 May 2014.

[3 marks]

[2 marks]

Markschemeevidence of considering derivative (M1)eg

(deer per month) A1 N2[2 marks]

P ′

104.475104

18c. Interpret the answer to part (i) with reference to the deer population sizeon 1 May 2014.

Markscheme(the deer population size is) increasing A1 N1[1 mark]

19a.

Let

Sketch the graph of .

f(x) = cos( x) + sin( x), for − 4 ⩽ x ⩽ 4.π

4π

4

f

[1mark]

[3 marks]

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct sinusoidal shape. Only if this A1 is awarded, award the following: A1 for correct domain, A1 for approximately correct range. [3 marks]

19b. Find the values of where the function is decreasing.

Markschemerecognizes decreasing to the left of minimum or right of maximum,eg (R1)x-values of minimum and maximum (may be seen on sketch in part (a)) (A1)(A1)eg two correct intervals A1A1 N5eg [5 marks]

x

f ′(x) < 0

x = −3, (1, 1.4)

−4 < x < −3, 1 ⩽ x ⩽ 4; x < −3, x ⩾ 1

[5 marks]

19c. The function can also be written in the form ,where , and . Find the value of ;

Markschemerecognizes that is found from amplitude of wave (R1)y-value of minimum or maximum (A1)eg (−3, −1.41) , (1, 1.41)

A1 N3[3 marks]

f f(x) = a sin( (x + c))π4

a ∈ R 0 ⩽ c ⩽ 2 a

a

a = 1.41421

a = √2, (exact), 1.41,

19d. The function can also be written in the form ,where , and . Find the value of .

MarkschemeMETHOD 1recognize that shift for sine is found at x-intercept (R1)attempt to find x-intercept (M1)eg

(A1) A1 N4

METHOD 2attempt to use a coordinate to make an equation (R1)eg attempt to solve resulting equation (M1)eg sketch,

(A1) A1 N4

[4 marks]

f f(x) = a sin( (x + c))π4

a ∈ R 0 ⩽ c ⩽ 2 c

cos( x) + sin( x) = 0, x = 3 + 4k, k ∈ Zπ4

π4

x = −1c = 1

√2 sin( c) = 1, √2 sin( (3 − c)) = 0π4

π4

x = 3 + 4k, k ∈ Zx = −1c = 1

[3 marks]

[4 marks]

20a.

The following diagram shows a circle with centre and radius

.

The points,

and lie on the circumference of the circle, and

radians.

Find the length of the arc .

Markschemecorrect substitution into arc length formula (A1)eg arc length (cm) A1 N2[2 marks]

O5 cm

ArmB

rmC

AOC = 0.7

ABC

0.7 × 5= 3.5

20b. Find the perimeter of the shaded sector.

Markschemevalid approach (M1)eg perimeter (cm) A1 N2[2 marks]

3.5 + 5 + 5, arc + 2r

= 13.5

20c. Find the area of the shaded sector.

[2 marks]

[2 marks]

[2 marks]

Markschemecorrect substitution into area formula (A1)eg

A1 N2[2 marks]

(0.7)(5)212

area = 8.75 (cm2)

21a.

In triangle,

and. The area of the triangle is

.

Find the two possible values for .

Markschemecorrect substitution into area formula (A1)eg correct working (A1)eg

; ; A1A1 N3

(accept degrees ie ; )[4 marks]

ABCAB = 6 cmAC = 8 cm16 cm2

A

(6)(8) sin A = 16, sin A =12

1624

A = arcsin( )23

A = 0.729727656 … , 2.41186499 … (41.8103149∘, 138.1896851∘)

A = 0.730 2.4141.8∘ 138∘

21b. Given that is obtuse, find .A BC

[4 marks]

[3 marks]

Printed for UpliftEducation

© International Baccalaureate Organization 2019 International Baccalaureate® - Baccalauréat International® - BachilleratoInternacional®

Markschemeevidence of choosing cosine rule (M1)eg correct substitution into RHS (angle must be obtuse) (A1)eg ,

A1 N2[3 marks]

BC2 = AB2 + AC2 − 2(AB)(AC) cos A, a2 + b2 − 2ab cos C

BC2 = 62 + 82 − 2(6)(8) cos 2.41, 62 + 82 − 2(6)(8) cos 138∘

BC = √171.55

BC = 13.09786BC = 13.1 cm