Download 1. h = r so 2r2 = 100 Þ r2 = 50 (M1) l = 10q = 2pr (M1) Þ q = (A1) = q

1.
h = r so 2r2 = 100  r2 = 50
l = 10 = 2r
2π 50
=
10
2 π5 2
=
10
 =  2 = 4.44 (3s.f.)
Note: Accept either answer.
(M1)
(M1)
(A1)
(A1)
[4]
2.
(a)
(b)
1 2
1
r   (152)(2)
2
2
2
= 225 (cm )
Area =
1 2
15 sin 2 = 102.3
2
Area = 225 – 102.3 = 122.7 (cm2)
= 123 (3 s.f.)
Area ∆OAB =
(M1)
(A1) (C2)
(A1)
(A1) (C2)
[4]
3.
(a)
l = r
(b)
AÔB (obtuse) = 2 – 2
1
1
Area =  r 2 =
(2 – 2) (15)2
2
2
or
ACB = 2 × OA
= 30cm
= 482 cm2 (3 s.f.)
(M1)
(A1) (C2)
(A1)
(M1)(A1)
(A1) (C4)
[6]
4.
Note: Do not penalize missing units in this question.
(a)
AB2
= 122 + 122 – 2 × 12 × 12 × cos 75° (A1)
= 122(2 – 2 cos 75°)
(A1)
2
= 12 × 2(1 cos 75°)
AB = 12 2(1  cos 75)
(AG)
2
Note: The second (A1) is for transforming the initial expression
to any simplified expression from which the given result can be
clearly seen.
1
(b)
PÔB = 37.5°
BP = 12 tan 37.5°
= 9.21 cm
(A1)
(M1)
(A1)
OR
BP̂A = 105°
BÂP = 37.5°
AB
BP

sin105  sin 37.5
ABsin37.5
BP =
= 9.21(cm)
sin 105
(c)
(i)
(ii)
(d)
(e)
1
 1

 12  9.21  or  12  12 tan 37.5 
2
2


= 55.3(cm2) (accept 55.2 cm2)
Area ∆OBP =
1
(9.21)2 sin105°
2
= 41.0(cm2) (accept 40.9 cm2)
Area ∆ABP =
1
π  75

 12 2 75 
 π  12 2 
 or
2
180  360

2
= 94.2 (cm ) (accept 30π or 94.3 (cm2)))
Area of sector =
Shaded area = 2 × area ∆OPB – area sector
= 16.4 (cm2) (accept 16.2 cm2, 16.3 cm2)
(A1)
(M1)
(A1)
3
(M1)
(A1)
(M1)
(A1)
4
(M1)
(A1)
2
(M1)
(A1)
2
[13]
5.
sin A =
5
12
 cos A = 
13
13
But A is obtuse  cos A = –
(A1)
12
13
sin 2A = 2 sin A cos A
5  12 
  
=2×
13  13 
120
=–
169
(A1)
(M1)
(A1) (C4)
[4]
6.
(a)
3sin2 x + 4cos x = 3(1 – cos2 x) + 4cos x
= 3 – 3cos2 + 4 cos x
(A1) (C1)
2
(b)
3sin2 x + 4 cos x – 4 = 0 3 – 3cos2 x + 4 cos x – 4 = 0
 3cos2 x – 4cos x + 1 = 0
(A1)
(3cos x – 1)(cos x – 1) = 0
1
cos x =
or cos x = 1
3
x = 70.5° or x = 0
(A1)(A1) (C3)
Note: Award (C1) for each correct radian answer, i.e. x = 1.23
or x = 0
[4]
7.
(a)
(b)
(i)
AP =
( x  8) 2  (10  6) 2  x 2  16 x  80
(ii)
OP =
( x  0) 2  (10  0) 2  x 2  100
AP 2  OP 2  OA 2
2AP  OP
2
( x  16 x  80)  ( x 2  100 )  (8 2  6 2 )
(A1)
cos OP̂A 
(M1)
=
(M1)
=
2 x 2  16 x  80 x 2  100
2 x 2  16 x  80
2 x 2  16 x  80 x 2  100
cos OP̂A 
(c)
(M1) (AG)
x 2  8 x  40
{( x 2  16 x  80)( x 2  100 )}
For x = 8, cos OP̂A = 0.780869
arccos 0.780869 = 38.7 (3 s.f.)
2
(M1)
(AG)
3
(M1)
(A1)
OR
8
10
OP̂A = arctan (0.8) = 38.7 (3 s.f.)
tan OP̂A 
(d)
(A1)
2
OP̂A = 60°  cos OP̂A = 0.5
0.5 =
x 2  8 x  40
{( x 2  16 x  80)( x 2  100)}
2x2 – 16x + 80 –
x = 5.63
(e)
(M1)
(i)
{( x 2  16 x  80)( x 2  100 )} = 0
f(x) = 1 when cos OP̂A = 1
hence, when OP̂A = 0.
This occurs when the points O, A, P are collinear.
(M1)
(M1)
(G2)
4
(R1)
(R1)
(R1)
3
(ii)
The line (OA) has equation y =
3x
4
(M1)
40
(= 13 13 )
3
When y = 10, x =
(A1)
OR
40
(= 13 13 )
3
Note: Award (G1) for 13.3.
x=
(G2)
5
[16]
8.
(a)
x is an acute angle => cosx is positive.
2
2
cos x + sin x = 1 => cosx =
=>
cos x =
=
(b)
1
1–  
3
8
9
(=
(M1)
1 – sin x
2
(M1)
2
(A1)
2 2
)
3
1
3
(A1) (C4)
2
cos2x = 1 – 2sin2x = 1 – 2  
=
(M1)
7
9
(A1) (C2)
Notes: (a)

Award (M1)(M0)(A1)(A0) for.
 1 
 
cos  sin –1    = 0.943
3

(b)

 1 
 
Award (M1)(A0) for.cos  2 sin –1    = 0.778.
3

4