New test - March 25, 2014
[37 marks]
The diagram shows a circle of radius 8 metres. The points ABCD lie on the circumference of the circle.
^D = 73∘ .
^ C = 104∘ , and BC
BC = 14 m, CD = 11.5 m, AD = 8 m, AD
1a. Find AC.
[3 marks]
Markscheme
evidence of choosing cosine rule
(M1)
eg c2 = a2 + b2 − 2ab cos C , C D2 + AD2 − 2 × CD × AD cos D
correct substitution
A1
eg 11.52 + 82 − 2 × 11.5 × 8 cos 104 , 196.25 − 184 cos 104
AC = 15.5 (m)
A1
N2
[3 marks]
1b. (i)
(ii)
^D .
Find AC
^B .
Hence, find AC
[5 marks]
Markscheme
(i)
METHOD 1
evidence of choosing sine rule
eg
sin A
a
sin B
b
=
,
sin AĈD
AD
^D
sin AC
8
=
^ D = 30.0∘
AC
sin D
AC
A1
correct substitution
eg
=
(M1)
sin 104
15.516…
A1
N2
METHOD 2
(M1)
evidence of choosing cosine rule
eg
c2
=
a2
2
+ b − 2ab cos C
A1
correct substitution
2
2
e.g. 8 = 11.5 + 15.516…2 − 2(11.5)(15.516 …) cos C
^ D = 30.0∘
AC
(ii)
A1
N2
^ D from 73
subtracting their AC
(M1)
^ D , 70 − 30.017 …
eg 73 − AC
^ B = 43.0∘
AC
A1
N2
[5 marks]
1c. (c)
(d)
Find the area of triangle ADC.
[6 marks]
Hence or otherwise, find the total area of the shaded regions.
Markscheme
(c)
correct substitution
eg area ΔADC =
area = 44.6 (m2)
(A1)
1
(8)(11.5) sin 104
2
A1
N2
[2 marks]
(d)
attempt to subtract
(M1)
eg circle − ABCD , πr2 − ΔADC − ΔACB
area ΔACB = 12 (15.516 …)(14) sin 42.98
correct working
(A1)
A1
2
eg π(8) − 44.6336 … − 12 (15.516 …)(14) sin 42.98 , 64π − 44.6 − 74.1
shaded area is 82.4 (m2)
[4 marks]
Total [6 marks]
A1
N3
The following diagram shows a triangle ABC.
^ C = 50∘ .
The area of triangle ABC is 80 cm2 , AB = 18 cm , AC = x cm and BA
2.
Find x .
[3 marks]
Markscheme
correct substitution into area formula
eg
(A1)
1
(18x) sin 50
2
setting their area expression equal to 80
(M1)
eg 9x sin 50 = 80
x = 11.6
A1
N2
[3 marks]
The following diagram shows a circle with centre O and radius r cm.
^ B = 1.4 radians .
Points A and B are on the circumference of the circle and AO
^O =
The point C is on [OA] such that BC
3.
π
2
radians .
The area of the shaded region is 25 cm2 . Find the value of r .
[7 marks]
Markscheme
correct value for BC
−−−−−−−−−−−−
eg BC = r sin 1.4 , √r2 − (r cos 1.4)2
(A1)
area of ΔOBC = 12 r sin 1.4 × r cos 1.4 (= 12 r2 sin 1.4 × cos 1.4)
area of sector OAB = 12 r2 × 1.4
A1
A1
(M1)
attempt to subtract in any order
eg sector – triangle, 12 r2 sin 1.4 × cos 1.4 − 0.7r2
correct equation
A1
eg 0.7r2 − 12 r sin 1.4 × r cos 1.4 = 25
attempt to solve their equation
eg sketch, writing as quadratic,
r = 6.37
A1
(M1)
25
0.616…
N4
[7 marks]
Note: Exception to FT rule. Award A1FT for a correct FT answer from a quadratic equation involving two trigonometric functions.
The following diagram shows a circular play area for children.
The circle has centre O and a radius of 20 m, and the points A, B, C and D lie on the circle. Angle AOB is 1.5 radians.
4.
Angle BOC is 2.4 radians.
Find the area of the shaded region.
[3 marks]
Markscheme
calculating sector area using their angle AOC
e.g.
1
(2.38 …)(202 )
2
(A1)
, 200(2.38 …) , 476.6370614 …
(M1)
shaded area = their area of triangle AOB + their area of sector
e.g. 199.4989973 … + 476.6370614 … , 199 + 476.637
shaded area = 676.136 … (accept 675.637 … = 676 from using 199)
shaded area = 676 [676, 677]
A1
N2
[3 marks]
The following diagram shows a triangle ABC.
BC = 6 , C ÂB = 0.7 radians , AB = 4p , AC = 5p , where p > 0 .
5.
(i)
Show that p2 (41 − 40 cos 0.7) = 36 .
[4 marks]
(ii) Find p .
Markscheme
(i) evidence of valid approach
(M1)
e.g. choosing cosine rule
correct substitution
(A1)
2
e.g. 6 = (5p) + (4p)2 − 2 × (4p) × (5p) cos 0.7
2
simplification
e.g. 36 =
25p2
A1
+ 16p2 − 40p2 cos 0.7
p2 (41 − 40 cos 0.7) = 36
AG
N0
(ii) 1.85995 …
p = 1.86
A1
N1
Note: Award A0 for p = ±1.86 , i.e. not rejecting the negative value.
[4 marks]
P^Q =
∘
P^R =
∘
^ R = 45∘ .
^ Q = 70∘ and PQ
The following diagram shows ΔPQR , where RQ = 9 cm, PR
6.
Find the area of ΔPQR .
[2 marks]
Markscheme
correct substitution
(A1)
e.g. area = 12 × 9 × 7.02 … × sin 70∘
29.69273008
area = 29.7
A1
N2
[2 marks]
Consider the triangle ABC, where AB =10 , BC = 7 and C ÂB = 30∘ .
7.
Find the two possible values of AĈ B .
[4 marks]
Markscheme
Note: accept answers given in degrees, and minutes.
evidence of choosing sine rule
e.g.
sin A
a
=
sin B
b
correct substitution
e.g.
sin θ
10
=
(M1)
sin 30∘
7
A1
, sin θ =
5
7
AĈ B = 45.6∘ ​ , AĈ B = 134∘
​ A1A1
N1N1
Note: If candidates only find the acute angle in part (a), award no marks for (b).
[4 marks]
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