Math Studies Year 1 –Semester II Final Exam IB Review Questions
1.
The diagram shows the straight lines L1 and L2. The equation of L2 is y = x.
(a)
Find
(i)
(ii)
the gradient of L1;
the equation of L1.
(3)
(b)
Find the area of the shaded triangle.
(3)
(Total 6 marks)
2.
The straight line, L1, has equation y = 
(a)
1
x – 2.
2
Write down the y intercept of L1.
(1)
(b)
Write down the gradient of L1.
(1)
The line L2 is perpendicular to L1 and passes through the point (3, 7).
(c)
Write down the gradient of the line L2.
(1)
(d)
Find the equation of L2. Give your answer in the form ax + by + d = 0 where a, b, d 
.
(3)
(Total 6 marks)
3.
The diagram below shows the line PQ, whose equation is x + 2y = 12. The line intercepts the axes at P and Q
respectively.
diagram not to scale
(a)
Find the coordinates of P and of Q.
(3)
(b)
A second line with equation x – y = 3 intersects the line PQ at the point A. Find the coordinates of A.
(3)
(Total 6 marks)
4.
The coordinates of the vertices of a triangle ABC are A (4 , 3) , B (7, –3) and C (0.5, p).
(a)
Calculate the gradient of the line AB.
(2)
(b)
Given that the line AC is perpendicular to the line AB
(i)
write down the gradient of the line AC;
(ii)
find the value of p.
(4)
(Total 6 marks)
5.
The mid-point, M, of the line joining A(s, 8) to B(−2, t) has coordinates M(2, 3).
(a)
Calculate the values of s and t.
(2)
(b)
Find the equation of the straight line perpendicular to AB, passing through
the point M.
(4)
(Total 6 marks)
6.
y
3
2
1
–5 –4 –3 –2 –1
–1
0
1
2
3
4
5
x
–2
–3
–4
–5
–6
(a)
(b)
On the grid above, draw a straight line with a gradient of –3 that passes through the point (–2, 0).
Find the equation of this line.
(Total 8 marks)
7.
A is the point (2, 3), and B is the point (4, 9).
(a)
Find the gradient of the line segment [AB].
(b)
Find the gradient of a line perpendicular to the line segment [AB].
(c)
The line 2x + by – 12 = 0 is perpendicular to the line segment [AB]. What is the value of b?
8.
Three points are given A(0, 4), B(6, 0) and C(8, 3).
(a)
Calculate the gradient (slope) of line AB.
(Total 4 marks)
(2)
(b)
Find the coordinates of the midpoint, M, of the line AC.
(c)
Calculate the length of line AC.
(d)
Find the equation of the line BM giving your answer in the form ax + by + d = 0 where a, b and d 
(e)
State whether the line AB is perpendicular to the line BC showing clearly your working and reasoning.
(2)
(2)
.
(5)
(3)
(Total 14 marks)
9.
The line L1 shown on the set of axes below has equation 4y + 3x = 24. L1 cuts the x-axis at A and cuts the y-axis at B.
Diagram not drawn to scale
y
L1
B
L2
M
O
A
x
C
(a)
Write down the coordinates of A and B.
(2)
M is the midpoint of the line segment [AB].
(b)
Write down the coordinates of M.
(2)
The line L2 passes through the point M and the point C (0, –2).
(c)
Write down the equation of L2.
(2)
(d)
Find the length of
(i)
MC;
(2)
(ii)
AC.
(2)
(e)
The length of AM is 5. Find
(i)
the size of angle CMA;
(3)
(ii)
10.
the area of the triangle with vertices C, M and A.
(2)
(Total 15 marks)
The figure shows a triangular area in a park surrounded by the paths AB, BC and CA, where AB = 400 m, AB̂C = 50
and BĈA = 30.
diagram not to scale
A
400 m
30º
C
B
(a)
Find the length of AC using the above information.
–1
Diana goes along these three paths in the park at an average speed of 1.8 m s .
(b)
Given that BC = 788m, calculate how many minutes she takes to walk once around the park.
(Total 6 marks)
11.
The following diagram shows the side view of a tent. The side of the tent AC is 6 m high. The ground AB slopes
upwards from the bottom of the tent at point A, at an angle of 5° from the horizontal. The tent is attached to the ground
by a rope at point B, a distance of 8 m from its base.
C
rope
6m
B
8m
5°
(a)
(b)
(c)
12.
A
Calculate the angle BAC.
Calculate the length of the rope, BC.
Calculate the angle CBA that the rope makes with the sloping ground.
(Total 8 marks)
The figure below shows a hexagon with sides all of length 4 cm and with centre at O. The interior angles of the hexagon
are all equal.
F
A
O
E
B
4 cm
D
C
The interior angles of a polygon with n equal sides and n equal angles (regular polygon) add up to (n – 2) × 180°.
13.
(a)
Calculate the size of angle A B̂ C.
(b)
Given that OB = OC, find the area of the triangle OBC.
(c)
Find the area of the whole hexagon.
(Total 8 marks)
In the diagram below A , B and C represent three villages and the line segments AB, BC and CA represent the roads
joining them. The lengths of AC and CB are 10 km and 8 km respectively and the size of the angle between them is
150°.
diagram not to scale
(a)
Find the length of the road AB.
(3)
(b)
Find the size of the angle CAB.
(3)
Village D is halfway between A and B. A new road perpendicular to AB and passing through D is built. Let T be the
point where this road cuts AC. This information is shown in the diagram below.
diagram not to scale
(c)
Write down the distance from A to D.
(1)
(d)
Show that the distance from D to T is 2.06 km correct to three significant figures.
(2)
A bus starts and ends its journey at A taking the route AD to DT to TA.
(e)
Find the total distance for this journey.
(3)
–1
The average speed of the bus while it is moving on the road is 70 km h .
The bus stops for 5 minutes at each of D and T.
(f)
Estimate the time taken by the bus to complete its journey. Give your answer correct to the nearest minute.
14.
(4)
(Total 16 marks)
A path goes around a forest so that it forms the three sides of a triangle. The lengths of two sides are 550 m and 290 m.
These two sides meet at an angle of 115°.
A diagram is shown below.
diagram not to scale
(a)
Calculate the length of the third side of the triangle. Give your answer correct to the nearest 10 m.
(4)
(b)
Calculate the area enclosed by the path that goes around the forest.
(3)
Inside the forest a second path forms the three sides of another triangle named ABC.
Angle BÂC is 53°, AC is 180 m and BC is 230 m.
diagram not to scale
(c)
Calculate the size of angle AĈB.
(4)
(Total 11 marks)
15.
An old tower (BT) leans at 10 away from the vertical (represented by line TG).
The base of the tower is at B so that MB̂T = 100.
Leonardo stands at L on flat ground 120 m away from B in the direction of the lean.
He measures the angle between the ground and the top of the tower T to be BL̂T = 26.5.
T
not to scale
M
MB = 200
90°
100°
26.5°
B G
BL = 120
L
Find the value of angle BT̂L .
Use triangle BTL to calculate the sloping distance BT from the base, B to the top, T of the tower.
(a)
(i)
(ii)
(b)
Calculate the vertical height TG of the top of the tower.
(c)
Leonardo now walks to point M, a distance 200 m from B on the opposite side of the tower. Calculate the
distance from M to the top of the tower at T.
(5)
(2)
16.
(a)
(3)
(Total 10 marks)
A farmer wants to construct a new fence across a field. The plan is shown below. The new fence is indicated by a
dotted line.
Diagram not to scale
75°
40°
410 m
Calculate the length of the fence.
(5)
(b)
The fence creates two sections of land. Find the area of the smaller section of land ABC, given the additional
information shown below.
A
B
Diagram not to scale
245 m
24
C
(3)
(Total 8 marks)
17.
In triangle ABC, AB = 3.9 cm, BC = 4.8 cm and angle AB̂C = 82.
B
82°
3.9 cm
Diagram not to scale
4.8 cm
A
C
(a)
Calculate the length of AC.
(3)
(b)
Calculate the size of angle AĈB.
(3)
(Total 6 marks)
18.
The diagram below shows a crane PQR that carries a flat box W. (PQ) is vertical, and the floor (PM) is horizontal.
Diagram not to scale
R
7.8 m
6.5 m
Q 102º
W
11.1 m
h
P
M
Given that PQ = 11.1m, QR = 7.8 m, PQ̂R =102° and RW = 6.5 m, calculate
(a)
PR;
(2)
(b)
angle PR̂Q;
(2)
(c)
the height, h, of W above (PM).
(3)
(Total 7 marks)
19.
The right pyramid shown in the diagram has a square base with sides of length 40 cm.
The height of the pyramid is also 40 cm.
diagram not to scale
(a)
Find the length of OB.
(4)
(b)
20.
Find the size of angle OB̂P .
(2)
(Total 6 marks)
The following diagram shows a sloping roof. The surface ABCD is a rectangle. The angle ADE is 55°. The vertical
height, AF, of the roof is 3 m and the length DC is 7 m.
B
C
A
7m
3m
55°
E
F
(a)
Calculate AD.
(b)
Calculate the length of the diagonal DB.
D
(Total 8 marks)
21.
The following diagram shows the rectangular prism ABCDEFGH. The length is 5 cm, the width is 1 cm, and the height
is 4 cm.
C
H
B
G
Diagram not to scale
D
E
A
(a)
(b)
F
Find the length of [DF].
Find the length of [CF].
(Total 8 marks)
22.
The diagram shows a cuboid 22.5 cm by 40 cm by 30 cm.
H
E
G
F
40 cm
D
C
30 cm
A
23.
(a)
Calculate the length of [AC].
(b)
Calculate the size of GÂC .
22.5 cm
B
(Total 4 marks)
A dog food manufacturer has to cut production costs. She wishes to use as little aluminium as possible in the
construction of cylindrical cans. In the following diagram, h represents the height of the can in cm, and x represents the
radius of the base of the can in cm.
diagram not to scale
3
The volume of the dog food cans is 600 cm .
(a)
Show that h =
(b)
(i)
600
πx 2
.
(2)
(ii)
Find an expression for the curved surface area of the can, in terms of x.
Simplify your answer.
Hence write down an expression for A, the total surface area of the can, in terms of x.
(4)
(c)
Differentiate A in terms of x.
(3)
(d)
Find the value of x that makes A a minimum.
(3)
(e)
Calculate the minimum total surface area of the dog food can.
(2)
(Total 14 marks)
24.
In the diagram below, PQRS is the square base of a solid right pyramid with vertex V. The sides of the square are 8 cm,
and the height VG is 12 cm. M is the midpoint of [QR].
Diagram not to scale
V
VG = 12 cm
P
Q
8 cm
G
S
(a)
(i)
(ii)
M
R
8 cm
Write down the length of [GM].
Calculate the length of [VM].
(2)
(b)
25.
Find
(i)
(ii)
the total surface area of the pyramid;
the angle between the face VQR and the base of the pyramid.
(4)
(Total 6 marks)
An office tower is in the shape of a cuboid with a square base. The roof of the tower is in the shape of a square based
right pyramid.
The diagram shows the tower and its roof with dimensions indicated. The diagram is not drawn to scale.
O
10 m
H
E
G
F
40 m
D
A
(a)
6m
C
B
Calculate, correct to three significant figures,
(i)
the size of the angle between OF and FG;
(3)
(ii)
the shortest distance from O to FG;
(2)
(iii)
the total surface area of the four triangular sections of the roof;
(3)
(iv)
the size of the angle between the slant height of the roof and the plane EFGH;
(2)
(v)
the height of the tower from the base to O.
(2)
A parrot’s nest is perched at a point, P, on the edge, BF, of the tower. A person at the point A, outside the building, measures
the angle of elevation to point P to be 79°.
(b)
Find, correct to three significant figures, the height of the nest from the base of the tower.
(2)
(Total 14 marks)
Math Studies Year 1 –Semester II Final Exam IB Review Questions Mark Scheme
1.
(a)
02
60
1 2

=    ,0.333 
3 6

(i)
(ii)
y= 
(M1)
(A1)
(C2)
1
x+2
3
(A1)(ft)
(C1)
Notes: Follow through from their gradient in part (a)(i).
Accept equivalent forms for the equation of a line.
(b)
area =
6  1.5
2
(A1)(M1)
Note: Award (A1) for 1.5 seen, (M1) for use of triangle formula with 6 seen.
= 4.5
(A1)
(C3)
[6]
2.
(a)
–2
(A1)
(C1)
Note: Accept (0, –2)
(b)

1
2
(A1)
(C1)
(c)
2
(A1)(ft)
(C1)
Note: Follow through from their answer to part (b).
(d)
y = 2x + c (can be implied)
7=2×3+c
c=1
y = 2x + 1
Notes: Award (M1) for substitution of (3, 7), (A1)(ft) for c.
Follow through from their answer to part (c).
OR
y – 7 = 2(x – 3)
Note: Award (M1) for substitution of their answer to part (c), (M1) for substitution
of (3, 7).
2x – y + 1 = 0 or –2x + y – 1 = 0
(M1)
(A1)(ft)
(M1)(M1)
(A1)(ft)
(C3)
Note: Award (A1)(ft) for their equation in the stated form.
[6]
3.
(a)
0 + 2y = 12 or x + 2(0) = 12
P(0, 6)
(accept x = 0, y = 6)
Q(12,0)
(accept x = 12, y = 0)
(C3)
Notes: Award (M1) for setting either value to zero.
Missing coordinate brackets receive (A0) the first time this occurs. Award
(A0)(A1)(ft) for P(0,12) and Q(6, 0).
(M1)
(A1)
(A1)
(b)
x + 2(x – 3) = 12
(6, 3)
(accept x = 6, y =
3)
(A1)(A1)
Note: (A1) for each correct coordinate.
Missing coordinate brackets receive (A0)(A1) if this is the first time it occurs.
(M1)
(C3)
[6]
4.
(a)
m(AB) =
33
=–
74
2
(C2)
(M1)(A1)
Note: Award (M1) for attempt to substitute into correct gradient formula.
(b)
1
2
(i)
m(AC) =
(A1)(ft)
(ii)
p 3 1
 (or equivalent method)
0.5  4 2
(M1)(A1)(
ft)
Note: Award (M1) for equating gradient to
1
. (A1) for correct substitution.
2
p = 1.25
(A1)(ft)
(C4)
[6]
5.
(a)
(b)
s=6
t = –2
(A1)
(A1)
gradient of AB =
 2  8  10 5


 26 8 4
(A1)(ft)
Note: (A1) for gradient of AM or BM =
Perpendicular gradient = 
(C2)
5
4
4
5
(A1)(ft)
Equation of perpendicular bisector is
4
y  xc
5
4
3   (2)  c
5
(M1)
c = 4.6
y = –0.8x + 4.6
or 5y = –4x + 23
(A1)(ft)
(C4)
[6]
6.
y
3
2
1
–5 –4 –3 –2 –1
–1
0
1
2
3
4
5
x
–2
–3
–4
–5
–6
(a)
line passes through (–2, 0)
line is straight
negative gradient (line must be straight for mark to be awarded)
correct gradient (line must be straight for mark to be awarded)
(A1)
(A1)
(A1)
(A1)
(C4)
(b)
y – 0 = –3(x + 2) or 3x + y = 3(–2) + 1(0) or y = –3x + c etc
3x + y = –6 (or equivalent)
(M1)
(A1)(A1)(
(C4)
A1)
Note: Award (C4) ft for y = –3x + candidate’s y-intercept (or equivalent).
Otherwise award:
(A1) for y with = in a linear equation,
(A1) for y = –3x or y + 3x seen or for m = –3
(A1) for candidate’s y-intercept included in a linear expression.
Do not ft candidate’s gradient if it is wrong in the diagram, no mark for stand alone
–3x
[8]
7.
(a)
For the line (AB), m =
=
93
42
6
2
=3
(A1)
1
3
(b)
m=–
(c)
2x + by – 12 = 0
(A1)
2
12
x
b
b
1
2
Therefore,   
3
b
y= 
(M1)
6=b
(A1)
[4]
8.
(a)
(b)
40
06
2

3
gradient 
(M1)
(A1) or (G2)
 08 43 
midpoint  
,

2 
 2
= (4, 3.5)
(A1)(A1)
Note: Award (A1) if x and y coordinates not explicitly made clear.
(c)
AC  (0  8) 2  (4  3) 2
(M1)
Note: Award (M1) for using the distance formula and substituting the correct
numbers.
= 8.06 ( 65 )
(d)
(A1)(G2)
Gradient BM 
3.5  0
46
(M1)
Note: Award (M1) for using values of B and M.

7
4
(A1)(ft)(G2)
y = mx + c
7
0   6  c
4
Note: Award (M1) for using the equation of a straight line.
(M1)
21
2
7
21
y x
4
2
c
(A1)(ft)(G1)
Note: Can award (G3) for this with no working.
7x + 4y – 42 = 0
Note: This step can (ft) within part (d)
(e)
gradient AB  
gradient BC 
(A1)(ft)(G4)
2
3
3
2
(M1)
Note: Award (M1) for attempting to find the gradient of BC.
2 3
   1
3 2
(M1)
Note: Award (M1) for multiplying their two gradients.
Yes, they are perpendicular.
Note: Accept any other valid mathematical method with working shown.
(A1)(ft)
[14]
9.
(a)
A;
B;
(b)
y = 0, 3x = 24  x = 8
A(8, 0)
x = 0, 4y = 24  y = 6
B(0, 6)
M; xm =
(A1)
80
06
= 4, ym =
=3
2
2
M(4, 3)
(c)
(d)
2
(A1)
2
(A1)
3  2 5

40 4
(A1)
y=
5
x – 2 (or equivalent)
4
(A1)
(i)
M(4, 3), C(0, –2)
L2: gradient =
MC =
(4  0) 2  (3  (2)) 2
(ii)
2
(M1)
= 41
= 6.40
(A1)
A(8, 0), C(0, –2)
AC =
8 2  (2) 2
(M1)
= 68
= 8.25
(e)
(A1)
(A1)
(i)
M
 41
C
cos M =
5
 68
2
5  ( 41 ) 2  ( 68 ) 2
2  5 41
A
(M1)
4
=
25  41  68
(M1)
10 41
CM̂A = 91.8° (3 s.f.)
(ii)
Area of CMA =
(A1)
1
41 × 5 sin 91.8°
2
(M1)
= 15.99991171...
= 16.0 (3 s.f.)
(A1)
5
[15]
10.
(a)
sin 50 sin 30

AC
400
(M1)(A1)
Note: Award (M1) for using sine rule with values from the problem, (A1) for
correct substitution.
AC = 613 (3 s.f.)
(b)
(A1)
(C3)
Perimeter = 400 + 613 + 788 = 1801m
Time in seconds =
1801
1000
1.8
(A1)(ft)(A1)
Note: Award (A1) for the perimeter, (A1) for finding the time in seconds, and last
(A1)(ft) for finding the time in minutes. The time in minutes follow through from the
time in seconds.
Time in minutes =
1000 50
 (16.7 to 3 s. f .)
60
3
(A1)(ft)
(C3)
[6]
11.
(a)
(b)
Angle A = 90 – 5 =
85°.
(C2)
2
2
2
BC = 6 + 8 – 2 × 8 × 6 cos(85°)
so BC =
(c)
91.6330487 = 9.57 (3 s.f.)
BC
AC

sin (A) sin (B)
6 sin (85)
sin (B) =
= 0.6244093654
9.572515275
–1
Angle B = sin (0.6244093654) = 38.6°
(M1)(A1)
(M1)(A1)
(A1)
(C3)
(M1)
(A1)
(A1)
(C3)
Note: Allow 38.7° if obtained using 9.57.
[8]
12.
(a)
Sum of internal angles is (n – 2) × 180 = (6 – 2) × 180 = 720°
(M1)(A1)
720
Hence AB̂C 
= 120°
6
(A1)
(C3)
(b)
OB̂C = 60°
(M1)
1
2
so AOBC =
× 4 × 2 × tan (60°) = 6.93 cm (units not
2
(c)
required)
(C3)
(A1)(A1)
Total area of hexagon is 6 × AOBC
(M1)
2
= 6 × 6.93 = 41.6 cm (units not required)
(A1)
(C2)
[8]
13.
UP
Unit penalty (UP) applies in parts (a), (c) and (e).
2
2
2
(a)
AB = 10 + 8 – 2 × 10 × 8 × cos 150°
AB = 17.4 km
Note: Award (M1) for substitution into correct formula, (A1) for correct substitution,
(A1) for correct answer.
8
(b)

sin CÂB
17.4
sin 150 
(M1)(A1)
(A1)(G2)
(M1)(A1)
CÂB = 13.3°
(A1)(ft)(G
2)
Notes: Award (M1) for substitution into correct formula, (A1) for correct
substitution, (A1) for correct answer. Follow through from their answer to part (a).
UP
UP
(c)
AD = 8.70 km (8.7 km)
Note: Follow through from their answer to part (a).
(A1)(ft)
(d)
DT = tan(13.29...°) × 8.697... = 2.0550...
= 2.06
Notes: Award (M1) for correct substitution in the correct formula, award (A1) for
the unrounded answer seen. If 2.06 not seen award at most (M1)(A0).
(M1)(A1)
(AG)
8.70 2  2.06 2 + 8.70 + 2.06
(e)
(A1)(M1)
= 19.7 km
(A1)(ft)(G
2)
Note: Award (A1) for AT, (M1) for adding the three sides of the triangle ADT,
(A1)(ft) for answer. Follow through from their answer to part (c).
(f)
19.7
× 60 + 10
70
(M1)(M1)
= 26.9
(A1)(ft)
Note: Award (M1) for time on road in minutes, (M1) for adding 10, (A1)(ft) for
unrounded answer. Follow through from their answer to (e).
= 27 (nearest minute)
(A1)(ft)(G
3)
Note: Award (A1)(ft) for their unrounded answer given to the nearest minute.
[16]
14.
UP
UP applies in this question
2
2
2
(a)
l = 290 + 550 – 2 × 290 × 550 × cos 115°
Note: Award (M1) for substituted cosine rule formula, (A1) for correct substitution.
l = 722
= 720 m
Note: If 720 m seen without working award (G3).
The final (A1) is awarded for the correct rounding of their answer.
(b)
Area =
1
× 290 × 550 × sin 115
2
(M1)(A1)
(A1)(G2)
(A1)
(M1)(A1)
Note: Award (M1) for substituted correct formula (A1) for correct substitution.
UP
2
= 72 300 m
(A1)(G2)
(c)
180
230

sin B sin 53
(M1)(A1)
Note: Award (M1) for substituted sine rule formula, (A1) for correct substitution.
B = 38.7°
(A1)(G2)
AĈB = 180 – (53° + 38.7°)
= 88.3°
(A1)(ft)
[11]
15.
(a)
(i)
(ii)
UP
Angle BT̂L = 180 – 80 – 26.5 or 180 – 90 – 26.5 + 10
= 73.5
BT
120

sin (26.5) sin (73.5)
BT = 55.8 m (3s.f.)
(b)
(c)
UP
TG = 55.8 sin (80) or 55.8 cos (10)
= 55.0 m (3s.f.)
Note: Apply (AP) if 0 missing
(M1)
(A1)(G2)
(M1)(A1)(ft)
(A1)(ft)
5
(M1)
(A1)(ft)(G2)
2
2
2
2
MT = 200 + 55.8 – 2  200  55.8  cos (100)
(M1)(A1)(ft)
MT = 217 m (3s.f.)
(A1)(ft)
Notes: Follow through only from part (a)(ii).
Award marks at discretion for any valid alternative method.
If radian mode has been used throughout the question, award (A0) to the first
incorrect answer then follow through, but negative lengths are always awarded
(A0)(ft).
The answers are (all 3s.f.)
(a)(ii) 124 m (A0)(ft)
(b) 123 m (A1)(ft)
(c) 313 m (A1)(ft)
3
[10]
16.
Note on use of radians:
In (a) the answer will be –874. Award (A0) at the last step for either + or – 874.
In (b) follow through with either sign from (a) can receive (M1) (A1) ft, but in both cases the final answer of  947000
receives (A0) for unrealistic sign and/or for unrealistic magnitude.
(a)
Third angle of triangle =180 – (75 + 40)
(M1)
= 65°
(A1)
Notes: Award (A2) for 65 seen.
For use of 40° or 75° in an otherwise correct sine rule award (M1)(A0)(A0)
Length of fence:
x
410
(sine rule)

sin 65 sin 75
x = 385 m (3 s.f.)
(b)
1
ab sin c
2
1
area =
× 385 × 245 sin 24°
2
(M1)(A1)
(A1)
or (G2) 5
Area =
2
= 19 200 (m ) (3 s.f.)
(M1)(A1)
(A1)
or (G2) 3
[8]
17.
(a)
2
2
2
AC = 3.9 + 4.8 – 2  3.9  4.8  cos 82
2
AC = 33.04
AC = 5.75
(M1)(A1)
(A1)
3
(b)
3 .9
33.04

sin C sin 82
(M1)(A1)
sin C = 0.671889
C = 42.2
(A1)
3
[6]
18.
(a)
2
2
2
PR = 7.8 + 11.1 – 2 × 7.8 × 11.1 × cos 102°
= 60.84 + 123.21 – (–36.00)
= 220.05
220.05 )
PR = 14.8 m (or
(b)
11.1
(M1)
(A1)
14.8
(Follow through with candidate’s answer to part (a))
sin R̂ sin 102 
11.1 sin 102
 sin R̂ 
= 0.7336
14.8
 R̂ = 47.2° (or 47.0° from 220.05 )
2

(M1)
(A1)
2
(c)
R
Q
H = h + 6.5
14.8
H
59.2º
P
M
Angle QP̂R = 180° – (102° + 47.2°)
= 30.8° (or 31.0°)
(M1)
 RP̂M = 90° – 30.8° = 59.2° (or 59.0°)
sin 59.2° =
H
14.8
(M1)
H
14.8
(M1)
 H = 14.8 sin 59.2° = 12.7 m
OR
cos 30.8° =
 H = 14.8 cos 30.8° = 12.7 m
Therefore, h = 12.7 – 6.5
= 6.2 m
(A1)
3
[7]
19.
Note: Unit penalty (UP) applies in part (a)
(a)
PB =
1
40 2  40 2  800 = 28.28 (28.3)
2
(M1)(A1)
Note: Award (M1) for correct substitutions, (A1) for correct answer.
UP
OB =
40  28.28 2 = 49.0 cm ( 2400 cm)
2
(M1)(A1)(
(C4)
ft)
Note: Award (M1) for correct substitution, can (ft) from any answer to PB.
(b)
 40 
sin 1  
 49 
OR
 28.28 
cos 1 

 49 
OR
 40 
tan 1 

 28.28 
(M1)
= 54.7 (54.8)
(A1)(ft)
(C2)
Note: Award (M1) for any correct trig. ratio.
In radians = 0.616, award (M1)(A0).
Note: Common error: (a) OB = 40  20
Award (M0)(A0)(M1) (A1)(ft), and (b) angle
OBP = 63.4° (63.5°)(M1)(A1)(ft).
2
2
= 44.7 cm.
[6]
20.
(a)
3
AD
3
AD =
sin (55)
sin (55°) =
(M1)(A1)
(M1)
AD = 3.66232 = 3.66 m to 3 s.f. (units not required).
(b)
(A1)
(C4)
2
2
2
2
2
DB = AD + DC = 3.66232 + 7
2
DB = 62.4126 hence DB = 7.90 m (units not
required).
(C4)
Note: Use of 3.662 makes no difference to final answer.
Award at most (M0)(A0)(A0)(A1)ft for an incorrect cosine rule formula. Award at
most (M1)(A0)(A0)(A1)ft for incorrect substitution into correct cosine rule formula.
(M1)(A1)
(A1)(A1)
[8]
21.
(a)
5 2  12 =
(b)
4 2  26
26 (or 5.10 (3 s.f.))
2
(M2)(A2)
(C4)
(M2)
= 42 = 6.48 (3 s.f.)
(A2)
(C4)
[8]
22.
(a)
AC =
(22.5) 2  30 2
= 37.5 cm
(b)
tan GÂC 
(M1)
(A1)
40
37.5
GÂC = 46.8° (or 0.818 radians)
(M1)
(A1)
[4]
23.
(a)
2
600 = πx h
600
πx 2
(M1)(A1)
=h
(AG)
Note: Award (M1) for correct substituted formula, (A1) for correct substitution. If
answer given not shown award at most (M1)(A0).
(b)
(i)
C = 2πx
600
πx 2
(M1)
C=
1200
–1
(or 1200x )
x
(A1)
Note: Award (M1) for correct substitution in formula, (A1) for correct
simplification.
(ii)
2
–1
A = 2πx + 1200x
(A1)(A1)(
ft)
Note: Award (A1) for multiplying the area of the base by two, (A1) for adding on
their answer to part (b)(i).
For both marks to be awarded answer must be in terms of x.
(c)
dA
1200
= 4πx –
dx
x2
(A1)(ft)(A
1)(ft)(A1)(ft)
–2
Notes: Award (A1) for 4πx, (A1) for –1200, (A1) for x . Award at most (A2) if any
extra term is written. Follow through from their part (b)(ii).
(d)
4πx –
1200
=0
(M1)(M1)
x2
3 1200
x =
(or equivalent)
4π
x = 4.57
(A1)(ft)(G
2)
Note: Award (M1) for using their derivative, (M1) for setting the derivative to zero,
(A1)(ft) for answer.
Follow through from their derivative.
Last mark is lost if value of x is zero or negative.
(e)
2
–1
A = 2π(4.57) + 1200(4.57)
A = 394
(M1)
(A1)(ft)(G
2)
Note: Follow through from their answers to parts (b) (ii) and (d).
[14]
24.
(a)
(i)
(ii)
GM = 4 cm
2
2
2
VM = 4 + 12
= 16 + 144
= 160
VM =
(b)
(i)
160 = 12.6 cm (3 s.f.)
(A1)
SA = area of square base + 4 (area of triangular face)
=8×8+4×
1
× 8 × 160
2
= 64 + 202.4
2
= 266 cm (3 s.f.)
Note: Using VM = 12.6 gives same final answer to 3
significant figures.
(ii)
(A1)
(M1)
(A1)
2
V
160
12
x
G
4
M
12
tan x =
=3
4
(M1)
x = 71.6° (or 1.25 radians)
OR
sin x =
(A1)
12
(M1)
160
 x = 71.6° (or 1.25 radians)
(A1)
OR
cos x =
4
(M1)
160
 x = 71.6° (or 1.25 radians)
OR
sin x =
(A1)
12
12.6
(M1)
 x = 72.2° (or 1.26 radians)
OR
cos x =
(A1)
4
12.6
(M1)
 x = 71.5° (or 1.25 radians)
(A1)
4
[6]
25.
Note: Throughout this question watch out for and accept other alternative approaches e.g. in part (a)(i), the cosine
formula may not necessarily be used.
O
10
F
(a)
(i)
 10 2  6 2  10 2
 = OF̂G = arccos 
 (2)(10)(6)
10
6
G




(M2)
Notes: Award (M1) for any correct method (formulae).
Award (M1) for substituting correctly in the formula used.
  = 72.5° (3 s.f.)
Note: Award (A1) for correct answer only.
(ii)
(iii)
h = slant height or shortest distance from O to FG = 3 tan
= 9.53939... = 9.54 m (3 s.f.)
Note: Follow through with candidate’s 
Area of OFG =
1
2
(10)(6)(sin )
(A1)
3
(M1)
(A1)
2
(M2)
therefore total surface area of roof = 4 ×
1
(10)(6)(sin )
2
2
= 114.4727... = 114 m (3 s.f.)
(A1)
3
Notes: Award (M1) for using any correct method (formulae).
Award (M1) for substituting correctly in the formulae used.
2
Follow through with candidate’s . Accept 115 m .
(iv)
3
h
(M1)
= 71.7° (3 s.f.)
(A1)
2
(M1)
(A1)
2
Angle between slant height (line) and plane EFGH = arccos  
Note: Follow through with candidate’s h.
(v)
H = Height of tower from base to O
= 40 + h 2  3 2
= 49.055385... = 49.1 m (3 s.f.)
Note: Follow through with candidate’s h
(b)
6 sin 79
sin( 90  79)
(M1)
= 30.9 m (3 s.f.)
(A1)
Height (BP) =
2
[14]