Trigonometric Functions Practice (Calc) 2018 · 2018. 9. 24. · Trigonometric Functions Practice (Calc) 2018 [190 marks] 1a. The following diagram shows the graph of , for . The

Trigonometric Functions Practice (Calc) 2018[190 marks]

1a.

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 .

Markscheme(i) valid approach (M1)

eg

A1 N2

(ii) valid approach (M1)

eg period is 12, per

A1

AG N0

(iii) METHOD 1

valid approach (M1)

eg

, substitution of points

A1 N2

METHOD 2

valid approach (M1)

eg

, amplitude is 6

A1 N2

[6 marks]

f(x) = asin bx + c 0 ⩽ x ⩽ 12

f (3, 5) (9, 17)

c

b = π6

a

5+172

c = 11

= , 9 − 32πb

b = 2π12

b = π6

5 = asin( × 3) + 11π6

a = −6

17−52

a = −6

[6 marks]

1b.

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 ( k0

) g(11.5, 17)

k

g(x)

k = 2.5

g(x) = −6 sin( (x − 2.5)) + 11π6

2a.

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

One sector is OAB, and .

Write down the exact value of in radians.

Markscheme A1 N1

[1 mark]

r

AÔB = θ

θ

θ = 2π5

2b.

The area of sector AOB is .

Find the value of .

Markschemecorrect expression for area (A1)

eg

evidence 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

2c. Find AB.

[3 marks]

[1 mark]

[3 marks]

[3 marks]

MarkschemeMETHOD 1

evidence of choosing cosine rule (M1)

eg

correct substitution of their and into RHS (A1)

eg

11.7557

A1 N2

METHOD 2

evidence of choosing sine rule (M1)

eg

correct substitution of their and (A1)

eg

11.7557

A1 N2

[3 marks]

a2 = b2 + c2 − 2bc cosA

r θ

102 + 102 − 2 × 10 × 10 cos( )2π5

AB = 11.8 (mm)

=sinAasinB

b

r θ

=sin 2π

5

AB

sin( (π− ))12

2π5

10

AB = 11.8 (mm)

3a.

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 TownB, on a bearing of 115°.

Find .

Markschemevalid approach (M1)

eg

A1 N2

[2 marks]

AB̂C

70 + (180 − 115), 360 − (110 + 115)

AB̂C = 135∘

3b. Find the distance from Town A to Town C.

[2 marks]

[3 marks]

Markschemechoosing cosine rule (M1)

eg

correct substitution into RHS (A1)

eg

12.0651

12.1 (km) A1 N2

[3 marks]

c2 = a2 + b2 − 2ab cosC

52 + 82 − 2 × 5 × 8 cos135

3c. Use the sine rule to find .

Markschemecorrect substitution (must be into sine rule) A1

eg

17.0398

A1 N1

[2 marks]

AĈB

=sinAĈB5

sin135AC

AĈB = 17.0

4a.

The following diagram shows a quadrilateral ABCD.

Find BD.

AD = 7 cm, BC = 8 cm, CD = 12 cm, DÂB = 1.75 radians, AB̂D = 0.82 radians.

[2 marks]

[3 marks]

Markschemeevidence of choosing sine rule (M1)

eg

correct substitution (A1)

eg

9.42069

A1 N2

[3 marks]

=asinA

b

sinB

=asin1.75

7sin0.82

BD = 9.42 (cm)

4b. Find .

Markschemeevidence of choosing cosine rule (M1)

eg


eg

1.51271

(radians) (accept 86.7°) A1 N2

[3 marks]

DB̂C

cosB = , a2 = b2 + c2 − 2bc cosBd2+c2−b2

2dc

, 144 = 64 + BD2 − 16BD cosB82+9.420692−122

2×8×9.42069

DB̂C = 1.51

5a.

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

Find the height of the seat when .


eg

A1 N2

[2 marks]

h t

h(t) = −15 cos1.2t + 17, for t ⩾ 0.

t = 0

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

h(0) = 2 (m)

5b. The seat first reaches a height of 20 m after minutes. Find.

k

k

[3 marks]

[2 marks]

[3 marks]

Markschemecorrect substitution into equation (A1)

eg

valid attempt to solve for (M1)

eg ,

1.47679

A1 N2

[3 marks]

20 = −15 cos1.2t + 17, − 15 cos1.2k = 3

k

cos1.2k = − 315

k = 1.48

5c. Calculate the time needed for the seat to complete a full rotation, giving your 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)

eg

5.2 (min) (must be 1 dp) A1 N2

[3 marks]

t h = 20

period = , 5.23598, 6.7 − −1.482π1.2

6a.

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

AÔC = 1.3 radians

ABC

l = 1.3 × 3

l 3.9

6b. Find the area of the shaded region.

[3 marks]

[2 marks]

[4 marks]

MarkschemeMETHOD 1

valid approach (M1)

eg finding reflex angle,

correct angle (A1)

eg


eg

A1 N3

METHOD 2

correct area of small sector (A1)

eg

valid approach (M1)

eg circle − small sector,


eg

A1 N3

[4 marks]

Total [6 marks]

2π − CÔA

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)

7a.

The following diagram shows the quadrilateral .

Find .


eg


eg

A1 N2

[3 marks]

ABCD

AD = 6 cm, AB = 15 cm,AB̂C = 44∘,AĈB = 83∘andDÂC = θ

AC

=ACsinCB̂A

AB

sinAĈB

=ACsin44∘

15sin83∘

10.4981

AC = 10.5 (cm)

[3 marks]

7b. Find the area of triangle .

Markschemefinding (seen anywhere) (A1)

eg

correct substitution for area of triangle A1

eg

62.8813

A1 N2

[3 marks]

ABC

CÂB

180∘ − 44∘ − 83∘,CÂB = 53∘

ABC

× 15 × 10.4981 × sin 53∘12

area = 62.9 (cm2)

7c. The area of triangle is half the area of triangle .

Find the possible values of .

Markschemecorrect substitution for area of triangle (A1)

eg

attempt to equate area of triangle to half the area of triangle (M1)

eg

correct equation A1

eg

,

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), 62.9887sin θ = 62.8813, sin θ = 0.99829412

12

86.6531 93.3468

θ = 86.7∘ , θ = 93.3∘

7d. Given that is obtuse, find.

MarkschemeNote: Note: If candidates use an acute angle from part (c) in the cosine rule, award M1A0A0 in part (d).


eg

correct substitution into rhs (A1)

eg

A1 N2

[3 marks]

Total [14 marks]

θ

CD

CD2 = AD2 + AC2 − 2 × AD × AC × cosθ

CD2 = 62 + 10.4982 − 2(6)(10.498)cos93.336∘

12.3921

12.4 (cm)

[3 marks]

[5 marks]

[3 marks]

8a.

The following diagram shows triangle .

Find .


eg


eg

A1 N2

[3 marks]

ABC

BC = 10 cm,AB̂C = 80∘ and BÂC = 35∘.

AC

=ACsin(AB̂C)

BC

sin(BÂC)

=ACsin80∘

10sin35∘

AC = 17.1695

AC = 17.2 (cm)

8b. Find the area of triangle .

Markscheme(seen anywhere) (A1)


eg

A1 N2

[3 marks]

Total [6 marks]

ABC

AĈB = 65∘

× 10 × 17.1695 × sin 65∘12

area = 77.8047

area = 77.8 (cm2)

9a.

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

Show that the area of the square is .

ABCD OAB O r

R

AÔB = θ, where 0.5 ≤ θ < π.

ABCD 2r2(1 − cosθ)

[3 marks]

[3 marks]

[4 marks]

Markschemearea of (seen anywhere) (A1)

choose cosine rule to find a side of the square (M1)

eg

correct substitution (for triangle ) A1

eg

correct working for A1

eg

AG N0

Note: Award no marks if the only working is .

[4 marks]

ABCD = AB2

a2 = b2 + c2 − 2bc cosθ

AOB

r2 + r2 − 2 × r × rcosθ, OA2 + OB2 − 2 × OA × OB cosθ

AB2

2r2 − 2r2cosθ

area = 2r2(1 − cosθ)

2r2 − 2r2cosθ

9b. When , the area of the square is equal to the area of the sector .

(i) Write down the area of the sector when .

(ii) Hence find .

Markscheme(i) A1 N1

(ii) correct equation in one variable (A1)

eg

A2 N2

Note: Award A1 for and additional answers.

[4 marks]

θ = α ABCD OAB

θ = α

α

αr2 (accept 2r2(1 − cosα))12

2(1 − cosα) = α12

α = 0.511024

α = 0.511 (accept θ = 0.511)

α = 0.511

10a.

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 .

O 8

A B C AÔB

ACB

[4 marks]

[2 marks]

Markschemecorrect substitution into formula (A1)

eg

A1 N2

[2 marks]

l = 1.2 × 8

9.6 (cm)

10b. Find .

MarkschemeMETHOD 1


eg

correct substitution into right hand side (A1)

eg

A1 N2

METHOD 2


eg

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

eg

A1 N2

[3 marks]

AB

2r2 − 2 × r2 × cos(AÔB)

82 + 82 − 2 × 8 × 8 × cos(1.2)

9.0342795

AB = 9.03 [9.03, 9.04] (cm)

=ABsin(AÔB)

OB

sin(OÂB)

OAB OBA

, 0.970796π−1.22

AB = 9.03 [9.03, 9.04] (cm)

10c. Hence, find the perimeter of the shaded segment .

Markschemecorrect working (A1)

eg

A1 N2

[2 marks]

Total [7 marks]

ABC

P = 9.6 + 9.03

18.6342

18.6 [18.6, 18.7] (cm)

[3 marks]

[2 marks]

11a.

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

;


eg

A1 N2

[2 marks]

y = p sin(qx) + r

A ( , 2)π6

B( , 1)π6

p

, 2 − 1.52−12

p = 0.5

11b. ;


eg

A1 N2

[2 marks]

r

1+22

r = 1.5

11c. .q

[2 marks]

[2 marks]

[2 marks]

MarkschemeMETHOD 1

valid approach (seen anywhere) M1

eg

period (seen anywhere) (A1)

A1 N2

METHOD 2

attempt to substitute one point and their values for and into M1

eg

correct equation in (A1)

eg

A1 N2

METHOD 3

valid reasoning comparing the graph with that of R1

eg position of max/min, graph goes faster

correct working (A1)

eg max at not at , graph goes times as fast

A1 N2

[3 marks]

Total [7 marks]

q = , 2πperiod

2π

( )2π3

= 2π3

q = 3

p r y

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

π

2

q

q = , q =π6

π

2π

23π2

q = 3

sin x

π

6π

23

q = 3

12a.

The following diagram shows triangle ABC.

Find AC. [3 marks]


eg

correct substitution into the right-hand side (A1)

eg

A1 N2

[3 marks]

AC2 = AB2 + BC2 − 2(AB)(BC)cos(AB̂C)

62 + 102 − 2(6)(10)cos100∘

AC = 12.5234

AC = 12.5 (cm)

12b. Find.

Markschemeevidence of choosing a valid approach (M1)

eg sine rule, cosine rule


eg

A1 N2

[3 marks]

BĈA

= , cos(BĈA) =sin(BĈA)

6sin100∘

12.5

(AC)2+102−62

2(AC)(10)

BĈA = 28.1525

BĈA = 28.2∘

13a.

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) + 990

t

t = 1

t = 5

210 sin(0.5 × 5 − 2.6) + 990, P(5)

969.034982…

13b. Find the rate of change of the deer population on 1 May 2014.

[3 marks]

[3 marks]

[2 marks]

Markschemeevidence of considering derivative (M1)

eg

(deer per month) A1 N2

[2 marks]

P ′

104.475

104

13c. Interpret the answer to part (i) with reference to the deer population size on 1 May 2014.

Markscheme(the deer population size is) increasing A1 N1

[1 mark]

14a.

Let

Sketch the graph of.

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]

f(x) = cos( x) + sin( x), for − 4 ⩽ x ⩽ 4.π4 π4

f

14b. Find the values of where the function is decreasing.x

[1 mark]

[3 marks]

[5 marks]

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 N5

eg

[5 marks]

f ′(x) < 0

x = −3, (1, 1.4)

−4 < x < −3, 1 ⩽ x ⩽ 4; x < −3, x ⩾ 1

14c. 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) = asin( (x + c))π4

a ∈ R0 ⩽ c ⩽ 2a

a

a = 1.41421

a = √2, (exact), 1.41,

14d. The function can also be written in the form

, where

, and. Find the value of

.

f

f(x) = asin( (x + c))π4

a ∈ R0 ⩽ c ⩽ 2c

[3 marks]

[4 marks]

MarkschemeMETHOD 1

recognize that shift for sine is found at x-intercept (R1)

attempt to find x-intercept (M1)

eg

(A1)

A1 N4

METHOD 2

attempt to use a coordinate to make an equation (R1)

eg

attempt to solve resulting equation (M1)

eg sketch,

(A1)

A1 N4

[4 marks]

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

π

4

x = −1

c = 1

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

π

4

x = 3 + 4k, k ∈ Z

x = −1

c = 1

15a.

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.

O

5 cm

A

rmB

rmC

AÔC = 0.7

ABC[2 marks]

Markschemecorrect substitution into arc length formula (A1)

eg

arc length (cm) A1 N2

[2 marks]

0.7 × 5

= 3.5

15b. Find the perimeter of the shaded sector.


eg

perimeter (cm) A1 N2

[2 marks]

3.5 + 5 + 5, arc + 2r

= 13.5

15c. Find the area of the shaded sector.

Markschemecorrect substitution into area formula (A1)

eg

A1 N2

[2 marks]

(0.7)(5)212

area = 8.75 (cm2)

16a.

In triangle

,

and

. The area of the triangle is

.

Find the two possible values for.

ABC

AB = 6 cm

AC = 8 cm

16 cm2

Â

[2 marks]

[2 marks]

[4 marks]


eg


eg

;

; A1A1 N3

(accept degrees ie;

)

[4 marks]

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

1624

A = arcsin( )23

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

A = 0.7302.41

41.8∘

138∘

16b. Given that is obtuse, find.


eg

correct substitution into RHS (angle must be obtuse) (A1)

eg ,

A1 N2

[3 marks]

Â

BC

BC2 = AB2 + AC2 − 2(AB)(AC) cosA, a2 + b2 − 2ab cosC

BC2 = 62 + 82 − 2(6)(8)cos2.41, 62 + 82 − 2(6)(8)cos138∘

BC = √171.55

BC = 13.09786

BC = 13.1 cm

[3 marks]

17a.

Let

, for

. The following diagram shows the graph of

.

The graph has a maximum at

and a minimum at

.

Write down the value of.

Markscheme A2 N2

Note: Award A1 for.

[2 marks]

f(x) = p cos(q(x + r)) + 10

0 ⩽ x ⩽ 20f

(4, 18)

(16, 2)

r

r = −4

r = 4

17b. Find.

Markschemeevidence of valid approach (M1)

eg

, distance from

A1 N2

[2 marks]

p

maxy value -- y value

2y = 10

p = 8

17c. Find.q

[2 marks]

[2 marks]

[2 marks]


eg period is,, substitute a point into their

,

(do not accept degrees) A1 N2

[2 marks]

2436024

f(x)

q = ( , exact)2π24

π

12

0.262

17d. Solve.


eg line on graph at

(accept) A1 N2

[2 marks]

Note: Do not award the final A1 if additional values are given. If an incorrect value of leads to multiple solutions, award the final A1 only if all solutions within the domain are given.

f(x) = 7

y = 7, 8 cos( (x − 4)) + 10 = 72π24

x = 11.46828

x = 11.5(11.5,7)

q

18a.

Consider a circle with centre

and radius

cm. Triangle

is drawn such that its vertices are on the circumference of the circle.

cm, cm and

radians.

Find.

O

7

ABC

AB = 12.2BC = 10.4

AĈB = 1.058

BÂC

[2 marks]

[3 marks]

MarkschemeNotes: In this question, there may be slight differences in answers, depending on which values candidates carry through in subsequentparts. Accept answers that are consistent with their working.

Candidates may have their GDCs in degree mode, leading to incorrect answers. If working shown, award marks in line with themarkscheme, with FT as appropriate.

Ignore missing or incorrect units.


eg


eg

A1 N2

[3 marks]

=sinÂasinB̂

b

=sinÂ10.4

sin1.05812.2

BÂC = 0.837

18b. Find.




METHOD 1

evidence of subtracting angles from (M1)

eg

correct angle (seen anywhere) A1

attempt to substitute into cosine or sine rule (M1)


eg

A1 N3

METHOD 2

evidence of choosing cosine rule M1

eg


eg

A2 N3

[5 marks]

AC

π

AB̂C = π − A − C

AB̂C = π − 1.058 − 0.837, 1.246, 71.4∘

12.22 + 10.42 − 2 × 12.2 × 10.4cos71.4, =ACsin1.246

12.2sin1.058

AC = 13.3 (cm)

a2 = b2 + c2 − 2bc cosA

12.22 = 10.42 + b2 − 2 × 10.4b cos1.058

AC = 13.3 (cm)

[5 marks]

18c. Hence or otherwise, find the length of arc.




METHOD 1

valid approach (M1)

eg

,


eg

(A1)

EITHER

correct substitution for arc length (seen anywhere) A1

eg

subtracting arc from circumference (M1)

eg

OR

attempt to find reflex (M1)

eg

correct substitution for arc length (seen anywhere) A1

eg

THEN

A1 N4

METHOD 2

valid approach to find or (M1)

eg choosing cos rule, twice angle at circumference

correct working for finding one value, or (A1)

eg

,

two correct calculations for arc lengths

eg (A1)(A1)

adding their arc lengths (seen anywhere)

eg M1

ABC

cosAÔC = OA2+OC2−AC2

2×OA×OC

AÔC = 2 × AB̂C

13.32 = 72 + 72 − 2 × 7 × 7 cosAÔC, O = 2 × 1.246

AÔC = 2.492 (142.8∘)

2.492 = , l = 17.4, 14π ×l7

142.8360

2πr − l, 14π = 17.4

AÔC

2π − 2.492, 3.79, 360 − 142.8

l = 7 × 3.79, 14π × 217.2360

arc ABC = 26.5

AÔBBÔC

AÔBBÔC

cosAÔB = 72+72−12.22

2×7×7

AÔB = 2.116,BÔC = 1.6745

AB = 7 × 2 × 1.058 (= 14.8135), 7 × 1.6745 (= 11.7216)

rAÔB + rBÔC, 14.8135 + 11.7216, 7(2.116 + 1.6745)

[6 marks]

A1 N4

Note: Candidates may work with other interior triangles using a similar method. Check calculations carefully and award marks in linewith markscheme.

[6 marks]

arc ABC = 26.5 (cm)

19a.

A Ferris wheel with diameter metres rotates clockwise at a constant speed. The wheel completes rotations every hour. The bottom of the wheel is

metres above the ground.

A seat starts at the bottom of the wheel.

Find the maximum height above the ground of the seat.


eg ,

maximum height (m) A1 N2

[2 marks]

1222.413

13 + diameter13 + 122

= 135

19b.

After t minutes, the height metres above the ground of the seat is given by

(i) Show that the period of is minutes.

(ii) Write down the exact value of .

h

h = 74 + a cos bt.

h

25

b

[2 marks]

[2 marks]

Markscheme(i) period

A1

period minutes AG N0

(ii)

A1 N1

[2 marks]

= 602.4

= 25

b = 2π25

(= 0.08π)

19c. Find the value of .

MarkschemeMETHOD 1

valid approach (M1)

eg ,

,

(accept ) (A1)

A1 N2

METHOD 2

attempt to substitute valid point into equation for h (M1)

eg

correct equation (A1)

eg ,

A1 N2

[3 marks]

a

max − 74|a| = 135−13

274 − 13

|a| = 61a = 61

a = −61

135 = 74 + acos( )2π×12.525

135 = 74 + acos(π)13 = 74 + a

a = −61

19d. Sketch the graph of , for

.h

0 ≤ t ≤ 50

[3 marks]

[4 marks]

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct domain, A1 for approximately correct range,

A1 for approximately correct sinusoidal shape with cycles.

Only if this last A1 awarded, award A1 for max/min in approximately correct positions.

[4 marks]

2

19e. In one rotation of the wheel, find the probability that a randomly selected seat is at least metres above the ground.

Markschemesetting up inequality (accept equation) (M1)

eg ,

, sketch of graph with line

any two correct values for t (seen anywhere) A1A1

eg ,

, ,

valid approach M1

eg ,

,

,

A1 N2

[5 marks]

105

h > 105105 = 74 + acosbty = 105

t = 8.371…t = 16.628…t = 33.371…t = 41.628…

16.628−8.37125

t1−t225

2×8.25750

2(12.5−8.371)

25

p = 0.330

[5 marks]

20a.

The diagram shows a circle of radius metres. The points ABCD lie on the circumference of the circle.

BC = m, CD =

m, AD = m,

, and .

Find AC.


eg ,

correct substitution A1

eg ,

AC (m) A1 N2

[3 marks]

8

1411.58AD̂C = 104∘

BĈD = 73∘

c2 = a2 + b2 − 2ab cosCCD2 + AD2 − 2 × CD × ADcosD

11.52 + 82 − 2 × 11.5 × 8 cos104196.25 − 184 cos104

= 15.5

20b. (i) Find .

(ii) Hence, find .

AĈD

AĈB

[3 marks]

[5 marks]

Markscheme(i) METHOD 1


eg ,


eg

A1 N2

METHOD 2


eg


e.g.

A1 N2

(ii) subtracting their from

(M1)

eg ,

A1 N2

[5 marks]

=sinAasinB

b

=sinAĈDAD

sinDAC

=sinAĈD8

sin10415.516…

AĈD = 30.0∘

c2 = a2 + b2 − 2ab cosC

82 = 11.52 + 15.516…2 − 2(11.5)(15.516…)cosC

AĈD = 30.0∘

AĈD73

73 − AĈD70 − 30.017…

AĈB = 43.0∘

20c. Find the area of triangle ADC.

Markschemecorrect substitution (A1)

eg area

area (m ) A1 N2

[2 marks]

ΔADC = (8)(11.5)sin 10412

= 44.6 2

20d. Hence or otherwise, find the total area of the shaded regions.

[2 marks]

[4 marks]

Markschemeattempt to subtract (M1)

eg ,

area (A1)

correct working A1

eg ,

shaded area is (m ) A1 N3

[4 marks]

Total [6 marks]

circle − ABCDπr2 − ΔADC − ΔACB

ΔACB = (15.516…)(14)sin 42.9812

π(8)2 − 44.6336… − (15.516…)(14)sin 42.9812

64π − 44.6 − 74.1

82.4 2

21a.

The following diagram shows a triangle ABC.

The area of triangle ABC is cm , AB

cm , AC cm and

.

Find .


eg

setting their area expression equal to (M1)

eg

A1 N2

[3 marks]

80 2

= 18= xBÂC = 50∘

x

(18x)sin 5012

80

9x sin 50 = 80

x = 11.6

21b. Find BC.

[3 marks]

[3 marks]


eg

correct substitution into right hand side (may be in terms of) (A1)

eg

BC A1 N2

[3 marks]

c2 = a2 + b2 + 2ab sin C

x

11.62 + 182 − 2(11.6)(18)cos50

= 13.8

22a.

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

Points A and B are on the circumference of the circle and radians .

The point C is on [OA] such that radians .

Show that .

Markschemeuse right triangle trigonometry A1

eg

AG N0

[1 mark]

r

AÔB = 1.4

BĈO = π2

OC = rcos1.4

cos1.4 = OCr

OC = rcos1.4

22b. The area of the shaded region is cm . Find the value of

.25 2

r

[1 mark]

[7 marks]

Printed for Uplift Education

© International Baccalaureate Organization 2018

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemecorrect value for BC

eg ,

(A1)

area of

A1

area of sector A1

attempt to subtract in any order (M1)

eg sector – triangle,

correct equation A1

eg

attempt to solve their equation (M1)

eg sketch, writing as quadratic,

A1 N4

[7 marks]

Note: Exception to FT rule. Award A1FT for a correct FT answer from a quadratic equation involving two trigonometric functions.

BC = rsin 1.4

√r2 − (rcos1.4)2

ΔOBC = rsin 1.4 × rcos1.412

(= r2sin 1.4 × cos1.4)12

OAB = r2 × 1.412

r2sin 1.4 × cos1.4 − 0.7r212

0.7r2 − rsin 1.4 × rcos1.4 = 2512

250.616…

r = 6.37

Trigonometric Functions Practice (Calc) 2018 [190 marks]MarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkscheme

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Trigonometric Functions Practice (Calc) 2018 · 2018. 9. 24. · Trigonometric Functions Practice (Calc) 2018 [190 marks] 1a. The following diagram shows the graph of , for . The