Download MEI C2 Trigonometry Assessment solutions

MEI Core 2 Trigonometry
Topic assessment
1. Find the angle  and the length x in the triangle shown below.
B
52°
12 cm
x cm
A

15 cm
C
[7]
2. Chang walks 5 km on a bearing of 140°, and then walks 3 km on a bearing of
025°.
(i) How far is Chang from his starting point?
[3]
(ii) On what bearing should Chang walk to get back to his starting point?
[4]
3. A belt is wrapped around a cylinder of radius 2.5 m as shown.
5m
Find the length of the belt.
[5]
4. Find the perimeter and area of the shaded sections of these shapes.
(i)
3 cm
60°
[7]
(ii)
8 cm
45°
10 cm
[7]
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MEI C2 Trigonometry Assessment solutions
5. Solve these equations for 0    360
(i) cos  0.5
(ii) sin   0.5
(iii) tan   2
[6]
6. Solve these equations for 0    2 .
Give your answers as a multiple of  .
3
(i) cos  
2
(ii) sin   0.5
(iii) tan   3
[6]
7. Solve these equations for 0    2 .
Give your answers as a multiple of  .
3
(i) cos 2  
4
2
(ii) 3tan   1
[6]
8. Solve these equations for 0    360
(i) sin 2 x   12 3
(ii) cos 12 x  0.3
(iii) tan 3x  0.5
[9]
9. Solve these equations for 0    360
(i) cos2   sin   1
(ii) 2sin  cos  sin   0
(iii) 3 sin   cos 
[4]
[4]
[3]
Total 70 marks
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MEI C2 Trigonometry Assessment solutions
Trigonometry
Chapter Assessment Solutions
sin  sin 52

12
15
12 sin 52
sin  
15
  39.1 or 140.9 
The value of  cannot be 149.0° as the total of the angles would be greater than
180°.
So   39.1 .
1. Using the sine rule:
Angle A  180  52  39.08  88.92
x
15
Using the sine rule:

sin 88.92 sin 52
15 sin 88.92
x
 19.0 cm
sin 52
[7]
2.
x
B

O
25°
40°
5
25° 3
40°
A
(i) Using the cosine rule: x 2  5 2  32  2  3  5 cos 65
x  4.62 km
[3]
4.6175  3  5
2  3  4.6175
  78.9
Bearing  180  25  78.9
2
(ii) Using the cosine rule: cos  
2
2
 283.9
[4]
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MEI C2 Trigonometry Assessment solutions
3. cos  
Adjacent
2.5 1


Hypotenuse
5
2
x
2.5
  60

5
Angle of arc with belt on = 360  60  60  240

4
240  240 

180
3
4  10
Arc length  r   2.5 

3
 3 
tan  
x
2.5
x  2.5 tan 60
10
 2x
3
10

 5 tan 60
3
 19.1 m(3 s.f.)
Total length of belt 
[5]
4. (i) If the triangle has an angle of 60 at the centre of the circle then it must be
an equilateral triangle and so part of the perimeter is 3cm.
60  60 

180


3

Arc length  r   3    
3
Perimeter  (  3) cm
 6.14 cm (3 s.f.)
3
3
2
Area  Area of sector  Area of triangle
3 1

  3  3 sin 60
2 2
 0.815 cm 2(3 s.f.)
Area of sector  21 r 2  0.5  3 2 


[7]
(ii) Arc length = r
Sector area = 21 r 2
45  45 
Sector 1:

180


4
Arc length  10 

4
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
5
2
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MEI C2 Trigonometry Assessment solutions
Sector area 
Sector 2:
1
2
Arc length  8 
 10 2 

4

4

25 
2
 2
Sector area  21  8 2 

4
 8
Shaded area: Perimeter  2  2  Arc length 1  Arc length 2
5
 4
 2
2
 18.1 cm (3 s.f.)
Area  Area of sector 1  Area of sector 2
25 

 8
2
 14.1 cm 2(3 s.f.)
[7]
5. (i)
cos  0.5
Solutions are in the 1st and 4th quadrants.
  60 or   360  60  300
  60, 300
(ii) sin   0.5
Solutions are in the 3rd and 4th quadrants
  180  30  210 or   360  30  330
  210, 330
(iii) tan   2
Solutions are in 1st and 3rd quadrants.
  63.4 or   180  63.4  243.4
  63.4, 243.4
[2]
[2]
[2]
6. (i)
3
2
Solutions are in 1st and 4th quadrants.

 11
  or   2  
6
6
6
 11
 ,
6 6
cos  
(ii) sin   0.5
Solutions are in 1st and 2nd quadrants
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[2]
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MEI C2 Trigonometry Assessment solutions



or    
6
 5
6
,

6

5
6
6
[2]
(iii) tan   3
Solutions are in 1st and 3rd quadrants

 4
  or     
3
3
3
 4
 ,
3 3
7. (i)
cos 2  
[2]
3
4
3
2
3
has solutions in the 1st and 4th quadrants
cos  
2

 11
  or   2  
6
6
6
3
has solutions in the 2nd and 4th quadrants
cos   
2
 5
 7
   
or     
6
6
6
6
 5  7 11
 ,
,
,
6 6 6
6
cos   
(ii) 3 tan   1
1
tan 2  
3
1
tan   
3
1
tan  
has solutions in the 1st and 3rd quadrants
3

 7
  or     
6
6
6
1
tan   
has solutions in the 2nd and 4th quadrants
3
 5
 11
   
or   2  
6
6
6
6
 5  7 11
 ,
,
,
6 6 6
6
[3]
2
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[3]
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MEI C2 Trigonometry Assessment solutions
8. (i)
sin 2 x   21 3
Solutions for 2x are in the 3rd and 4th quadrants.
2 x  180  60  240 or 2 x  360  60  300
or 2 x  360  240  600 or 2 x  360  300  660
x  120, 150, 300, 330
[3]
(ii) cos x  0.3
Solutions for 21 x are in the 1st and 4th quadrants, but the one in the 4th
quadrant will give a value for x which is out of the range.
1
2 x  72.5 
1
2
x  145
(iii) tan 3 x  0.5
Solutions for 3x are in the 1st and 3rd quadrants
3 x  26.6 or 3 x  180  26.6  206.6
or 3 x  360  26.6  386.6 or 3 x  360  206.6  566.6
or 3 x  360  386.6  746.6 or 3 x  360  566.6  926.6
x  8.9, 68.9, 128.9, 188.9, 248.9, 308.9
[3]
[3]
9. (i)
cos 2   sin   1
 1  sin 2    sin   1
sin 2   sin   0
sin   sin   1   0
sin   0
sin   1  0
or
  0, 180, 360
sin   1
  90
  0, 90, 180,360
[4]
(ii) 2 sin  cos   sin   0
sin   2 cos   1   0
sin   0
2 cos   1  0
or
  0, 180, 360
cos    21
  0, 120, 180,240,360
  120, 240
[4]
(iii)
3 sin   cos 
3 tan   1
1
tan  
3
  30,210
[3]
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