IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers
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Trig Practice – 08 and Specimen Papers
1.
In triangle ABC, AB = 9 cm, AC =12 cm, and B̂ is twice the size of Ĉ .
Find the cosine of Ĉ .
2.
In the diagram below, AD is perpendicular to BC.
CD = 4, BD = 2 and AD = 3. CÂD =  and BÂD = .
(Total 5 marks)
Find the exact value of cos ( − ).
(Total 6 marks)
3.
The diagram below shows the boundary of the cross-section of a water channel.
 x 
The equation that represents this boundary is y = 16 sec   – 32 where x and y are both measured
 36 
in cm.
The top of the channel is level with the ground and has a width of 24 cm. The maximum depth of the
channel is 16 cm.
Find the width of the water surface in the channel when the water depth is 10 cm. Give your answer
in the form a arccos b where a, b .
(Total 6 marks)
4.
A system of equations is given by
cos x + cos y = 1.2
sin x + sin y = 1.4.
(a) For each equation express y in terms of x.
(b)
Hence solve the system for 0  x < , 0 < y < .
(2)
(4)
(Total 6 marks)
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IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers
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5.
Find, in its simplest form, the argument of (sin + i (1− cos ))2 where  is an acute angle.
6.
In triangle ABC, BC = a, AC = b, AB = c and [BD] is perpendicular to [AC].
(Total 7 marks)
(a)
Show that CD = b − c cos A.
(b)
Hence, by using Pythagoras’ Theorem in the triangle BCD, prove the cosine rule for the
triangle ABC.
(c)
1
2 3 2
If AB̂C = 60, use the cosine rule to show that c = a  b  a .
2
4
(1)
(4)
(7)
(Total 12 marks)
7.
The above three dimensional diagram shows the points P and Q which are respectively west and
south-west of the base R of a vertical flagpole RS on horizontal ground. The angles of elevation of
the top S of the flagpole from P and Q are respectively 35 and 40, and PQ = 20 m.
Determine the height of the flagpole.
(Total 8 marks)
8.
The depth, h (t) metres, of water at the entrance to a harbour at t hours after midnight on a particular
day is given by
 t 
h (t) = 8 + 4 sin  , 0  t  24.
6
(a) Find the maximum depth and the minimum depth of the water.
(b)
(3)
Find the values of t for which h (t)  8.
(3)
(Total 6 marks)
9.
Consider triangle ABC with BÂC = 37.8, AB = 8.75 and BC = 6.
Find AC.
10.
(a)
Sketch the curve f(x) = sin 2x, 0 ≤ x ≤ π.
(b)
Hence sketch on a separate diagram the graph of g(x) = csc 2x, 0 ≤ x ≤ π, clearly stating the
coordinates of any local maximum or minimum points and the equations of any asymptotes.
(c)
Show that tan x + cot x ≡ 2 csc 2x.
(Total 7 marks)
(2)
(5)
(3)
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IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers
(d)
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Hence or otherwise, find the coordinates of the local maximum and local minimum points on
π
the graph of y = tan 2x + cot 2x, 0 ≤ x ≤ .
2
(5)
(e)
π
Find the solution of the equation csc 2x = 1.5 tan x – 0.5, 0 ≤ x ≤ .
2
(6)
(Total 21 marks)
11.
In a triangle ABC, Â = 35°, BC = 4 cm and AC = 6.5 cm. Find the possible values of B̂ and the
corresponding values of AB.
(Total 7 marks)
12.
Solve sin 2x =
2 cos x, 0 ≤ x ≤ π.
(Total 6 marks)
13.
5
The obtuse angle B is such that tan B =  . Find the values of
12
(a) sin B;
(1)
(b)
cos B;
(c)
sin 2B;
(d0
cos 2B.
(1)
(2)
(2)
(Total 6 marks)
3
, find the possible values of tan θ.
4
14.
Given that tan 2θ =
15.
Let sin x = s.
(a) Show that the equation 4 cos 2x – 3 sin x cosec3 x + 6 = 0 can be expressed as
8s4 – 10s2 + 3 = 0.
(Total 5 marks)
(3)
(b)
Hence solve the equation for x, in the interval [0, π].
(6)
(Total 9 marks)
16.
(a)
If sin (x – α) = k sin (x + α) express tan x in terms of k and α.
(3)
(b)
Hence find the values of x between 0° and 360° when k =
1
and α = 210°.
2
(6)
(Total 9 marks)
17.
The angle θ satisfies the equation 2 tan2 θ – 5 sec θ – 10 = 0, where θ is in the second quadrant. Find
the value of sec θ.
(Total 6 marks)
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IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers
18.
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The lengths of the sides of a triangle ABC are x – 2, x and x + 2. The largest angle is 120°.
(a) Find the value of x.
(6)
(b)
15 3
Show that the area of the triangle is
.
4
(3)
(c)
Find sin A + sin B + sin C giving your answer in the form
p q
r
where p, q, r 
.
(4)
(Total 13 marks)
19.
A farmer owns a triangular field ABC. The side [AC] is 104 m, the side [AB] is 65 m and the angle
between these two sides is 60°.
(a) Calculate the length of the third side of the field.
(3)
(b)
Find the area of the field in the form p 3 , where p is an integer.
(3)
Let D be a point on [BC] such that [AD] bisects the 60° angle. The farmer divides the field into two
parts by constructing a straight fence [AD] of length x metres.
65x
(c) (i)
Show that the area o the smaller part is given by
and find an expression for the area
4
of the larger part.
(ii) Hence, find the value of x in the form q 3 , where q is an integer.
(8)
(d)
BD 5
 .
Prove that
DC 8
(6)
(Total 20 marks)
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IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
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Trig Practice – 08 and Specimen Papers – MarkScheme
1.
9
12

sin C sin B
9
12

sin C sin 2C
(M1)
A1
9
12

sin C 2 sin C cos C
 9(2 sin C cos C) = 12 sin C
 6 sin C (3 cos C  2) = 0 or equivalent
(sin C  0)
2
 cos C =
3
Using double angle formula
M1
(A1)
A1
[5]
2.
METHOD 1
AC = 5 and AB = 13
(may be seen on diagram)
3
4
cos α  and sin α 
5
5
3
2
cos β 
and sin β 
13
13
Note: If only the two cosines are correctly given award (A1)(A1)(A0).
Use of cos (   ) = cos  cos  + sin  sin 
3 3
4 2
 
= 
(substituting)
5 13 5 13
=
 17 13 


65 
5 13 
17
25  13  36
1 

 

2  5  13
 5 13 
Use of cos ( + ) + cos (  ) = 2 cos  cos 
3
3
cos α  and cos β 
5
13
cos α  β  

3 3
1 
  2  


5 13 5 13 
5 13 
17
(A1)
(A1)
(M1)
M1
A1
METHOD 2
AC = 5 and AB = 13 (may be seen on diagram)
AC 2  AB2  BC 2
Use of cos ( + ) =
2ACAB
=
(A1)
 17 13 




65


N1
(A1)
(M1)
A1
(M1)
(A1)
A1
N1
[6]
3.
 πx 
10 cm water depth corresponds to 16 sec    32   6
 36 
 πx 
Rearranging to obtain an equation of the form sec    k or
 36 
equivalent
ie making a trignometrical function the subject of the equation.
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(A1)
M1
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IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
 πx  8
cos   
 36  13
πx
8
=  arccos
13
36
36
8
x=
arccos
13
π
Note: Do not penalize the omission of .
72
8
Width of water surface is
arccos
(cm)
13
π
Note: Candidate who starts with 10 instead of 6 has the potential
to gain the two M1 marks and the R1 mark.
4.
(a)
(b)
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(A1)
M1
A1
R1
N1
[6]
y = arccos (1.2  cos x)
y = arcsin (1.4  sin x)
The solutions are
x = 1.26, y = 0.464
x = 0.464, y = 1.26
A1
A1
A1A1
A1A1
[6]
5.
(sin + i (1  cos)) = sin   (1  cos) + i 2 sin (1  cos)
Let  be the required argument.
2 sin θ 1  cos θ 
tan =
2
sin 2 θ  1  cos θ 
2 sin θ 1  cos θ 
=
2
1  cos θ  1  2 cos θ  cos 2 θ
2 sin θ 1  cos θ 
=
2 cos θ 1  cos θ 
2

2
2
 

= tan
=
6.
(a)
(b)
(c)
=
M1
(M1)
A1
A1
A1
CD = AC  AD = b  c cos A
METHOD 1
BC2 = BD2 + CD2
a2 = (c sin A)2 + (b  c cos A)2
= c2 sin2 A + b2  2bc cos A + c2 cos2 A
= b2 + c2  2bc cos A
METHOD 2
BD2 = AB2  AD2 = BC2  CD2
 c2  c2 cos2 A = a2  b2 + 2bc cos A  c2 cos2 A
 a2 = b2 + c2  2bc cos A
METHOD 1
b2 = a2 + c2  2ac cos 60  b2 = a2 + c2  ac
 c2  ac + a2  b2 = 0
c=
M1A1
a
 a 2  4a 2  b 2 
2
a  4b 2  3a 2
2
4b 2  3a 2
a
 
2
4
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[7]
R1AG
(M1)
(A1)
A1
A1
(M1)(A1)
A1
A1
(M1)A1
M1
(M1)A1
(M1)A1
2 of 8
IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
1
3
a  b2  a2
2
4
Note: Candidates can only obtain a maximum of the first
three marks if they verify that the answer given in
the question satisfies the equation.
METHOD 2
b2 = a2 + c2  2ac cos 60  b2 = a2 + c2  ac
c2  ac = b2  a2
2
2
a
a
c2  ac +    b 2  a 2   
2
2
2
3 2
 a
2
c   b  a
2
4

=
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AG
(M1)A1
(M1)
M1A1
(A1)
a
3
c    b2  a2
2
4
A1
1
3
 c  a  b2  a2
2
4
AG
[12]
7.
PR = h tan 55, QR = h tan 50 where RS = h
Use the cosine rule in triangle PQR.
202 = h2 tan2 55 + h2 tan2 50  2h tan 55 h tan 50 cos 45
400
h2 
2
2
tan 55  tan 50  2 tan55 tan50 cos 45
= 379.9...
h = 19.5 (m)
M1A1A1
(M1)
A1
(A1)
(A1)
A1
[8]
8.
(a)
(b)
πt
Either finding depths graphically, using sin   1
6
or solving h(t) = 0 for t
h (t)max = 12 (m), h (t)min = 4 (m)
πt
Attempting to solve 8 + 4 sin  8 algebraically or graphically
6
t [0, 6]  [12, 18]  {24}
(M1)
A1A1
N3
(M1)
A1A1
N3
[6]
9.
METHOD 1
Attempting to use the cosine rule i.e.
BC2 = AB2 + AC2  2  AB  AC  cos BÂC
62 = 8.752 + AC2  2  8.75  AC  cos 37.8 (or equivalent)
Attempting to solve the quadratic in AC e.g. graphically, numerically or
with quadratic formula
Evidence from a sketch graph or their quadratic formula (AC =…)
that there are two values of AC to determine.
AC = 9.60 or AC = 4.22
Note: Award (M1)A1M1A1(A0)A1A0 for one correct value of AC.
METHOD 2
BC
AB
Attempting to use the sine rule i.e.

sin BÂC sin AĈB
8.75sin 37.8
 0.8938...
sin C 
6
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(M1)
A1
M1A1
(A1)
A1A1
N4
(M1)
(A1)
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IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
C = 63.3576... 
C = 116.6423...  and B = 78.842...  or B = 25.5576... 
EITHER
AC
6

or
Attempting to solve
sin 78.842... sin 37.8
AC
6

sin 25.5576... sin 37.8
OR
Attempting to solve AC2 = 8.752 + 62  2  8.75  6  cos 25.5576... or
AC2 = 8.752 + 62  2  8.75  6  cos 78.842... 
AC = 9.60 or AC = 4.22
Note: Award (M1)(A1)A1A0M1A1A0 for one correct value of AC.
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A1
A1
M1
M1
A1A1
N4
[7]
10.
(a)
A2
Note: Award A1 for shape.
A1 for scales given on each axis.
(b)
A5
(c)
π
Asymptotes x = 0, x = , x = π
2
 3π

π 
Max ,1, Min , 1
 4

4 
Note: Award A1 for shape
A2 for asymptotes, A1 for one error, A0 otherwise.
A1 for max.
A1 for min.
sin x cos x

tanx + cot x ≡
cos x sin x
sin 2 x  cos 2 x
≡
sin x cos x
1
≡
1
sin 2 x
2
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M1
A1
A1
4 of 8
IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
≡ 2 csc 2x
tan 2x + cot 2x ≡ 2 csc 4x
 3

Max is at  ,2
 8

(d)
π 
Min is at  , 2 
8 
(e) csc 2x = 1.5 tan x – 0.5
1
1
3
1
tan x  cot x  tan x 
2
2
2
2
tan x + cot x = 3 tan x – 1
1
2 tan x –
–1=0
tan x
2 tan2 x – tan x – 1 = 0
(2 tan x + 1)(tan x – 1) = 0
1
tan x = – or 1
2
π
x=
4
Note: Award A0 for answer in degrees or if more than one value given for x.
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AG
(M1)
A1A1
A1A1
M1
M1
A1
M1
A1
A1
[21]
11.
sin B sin 35

6.5
4
B̂ = 68.8° or 111°
Ĉ = 76.2° or 33.8° (accept 34°)
AB
BC

sinC sinA
AB
4

sin76.2 sin 35
AB = 6.77 cm
AB
4

sin33.8 sin 35
AB = 3.88cm (accept 3.90)
M1
A1A1
A1
(M1)
A1
A1
[7]
12.
2 sin x cos x – 2 cos x = 0
cos x (2 sin x – 2 ) = 0
cos x = 0
x=
π
2
2
2
π 3π
x= ,
4 4
sin x =
(M1)
(A1)
A1
A1A1A1
[6]
13.
(a)
(b)
(c)
5
13
12
cos B = 
13
sin 2B = 2sin B cos B
5
12

=2×
13
13
sin B =
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A1
A1
(M1)
5 of 8
IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
120
169
cos2B = 2cos2 B – 1
 144 
= 2
 1
 169 
119
=
169
= 
(d)
14.
Using tan 2θ =
2 tan
A1
(M1)
A1
[6]
2 tan
(M1)
1  tan 2 
3
1  tan  4
3 tan2 θ + 8 tan θ – 3 = 0
Using factorisation or the quadratic formula
1
tan θ = or –3
3
2
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
A1
(M1)
A1A1
[5]
15.
2
4(1 – 2s ) – 3s
(a)
(b)
1
+6=0
s3
4s2 – 8s4 + 6s2 – 3 = 0
8s4 – 10s2 + 3 = 0
Attempt to factorise or use the quadratic formula
1
3
sin2 x = or sin2 x =
2
4
2
π
3π
 x  or x 
sin x =
2
4
4
3
π
2π
 x  or x 
sin x =
2
3
3
Note: Penalise A1 if extraneous solutions given.
M1A1
sinx cos α – cosx sin α = k sin x cos α + k cos x sin α
 tan x cos α – sin α = k tan x cos α + k sin α

 (k  1) sin    (k  1)
 
tan 
 tan x =
(k  1) cos   (k  1)

(M1)
M1
A1
AG
(M1)
(A1)
A1A1
A1A1
[9]
16.
(a)
(b)
tan x =
3
sin 210
2
1
 cos 210
2

Now sin 210° = –sin 30° = 
A1
(M1)
1
3
and cos 210° = –cos 30° = 
2
2
1
2  3 2  3  3
tan x =
2
3
3
3

2
 x = 60°, 240°
A1A1
3 
A1
A1A1
[9]
17.
2
2 tan θ – 5 sec θ – 10 = 0
Using 1 + tan2 θ = sec2 θ,  2(sec2 θ – 1) – 5 sec θ – 10 = 0
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(M1)
6 of 8
IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
2 sec2 θ – 5 sec θ – 12 = 0
Solving the equation e.g. (2 sec θ + 3)(sec θ – 4) = 0
3
sec θ =  or sec θ = 4
2
θ in second quadrant  sec θ is negative
3
 sec θ = 
2
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A1
(M1)
A1
(R1)
A1
N3
[6]
18.
(a)
(b)
(c)
(x + 2)2 = (x – 2)2 + x2 – 2(x – 2) xcos120°
x2 + 4x + 4 = x2 – 4x + 4 + x2 + x2 – 2x
0 = 2x2 – 10x
0 = x(x – 5)
x=5
1
Area = × 5 × 3 × sin 120°
2
1
3
=  15 
2
2
15 3
=
4
3
sin A =
2
15 3 1
3 3
 × 5 × 7 × sin B  sin B 
4
2
14
5 3
Similarly sin C =
14
15 3
sin A + sin B + sin C =
14
(M1)
M1A1
(M1)
A1
A1
M1A1
A1
AG
M1A1
A1
A1
[13]
19.
(a)
Using the cosine rule (a2 = b2 + c2 – 2bc cos A)
Substituting correctly
BC2 = 652 + 1042 – 2 (65) (104) cos 60°
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(M1)
A1
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IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
(b)
(c)
(d)
= 4225 + 10 816 – 6760 = 8281
 BC = 91m
1
Finding the area using = bc sin A
2
1
Substituting correctly, area = (65) (104) sin 60°
2
= 1690 3 (accept p = 1690)
1
Smaller area A1 =   (65) (x)sin 30°
2
65x
=
4
1
Larger area A2 =   (104) (x) sin 30°
2
= 26x
(ii) Using A1 + A2 = A
65x
 26x  1690 3
Substituting
4
169x
 1690 3
Simplifying
4
4 1690 3
Solving x =
169
 x  40 3 (accept q = 40)
Using sin rule in ∆ADB and ∆ACD
BD
65
BD
sin 30



Substituting correctly
sin30 sinAD̂B
65 sin AD̂B
DC
104
DC
sin 30



and
sin30 sin AD̂C
104 sin AD̂C
(i)
Since AD̂B  AD̂C = 180°
It follows that sin AD̂B  sinAD̂C
BD DC
BD 65



65 104
DC 104
BD 5


DC 8
Alei - Desert Academy
A1
N2
(M1)
A1
A1
N2
(M1)A1
AG N0
M1
A1 N1
(M1)
A1
A1
A1 N1
(M1)
A1
A1
R1
R1
A1
AG N0
[20]
C:\Users\Bob\Documents\Dropbox\Desert\HL1\3 Trig\HLTrigPractice08.docx on 04/17/2016 at 9:51 AM
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IB Math – High Level Year 1: Trig: Practice 08 and Spec Papers - MarkScheme
C:\Users\Bob\Documents\Dropbox\Desert\HL1\3 Trig\HLTrigPractice08.docx on 04/17/2016 at 9:51 AM
Alei - Desert Academy
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