Download AM Trigonometry Assessment solutions

Additional Mathematics Trigonometry
Topic assessment
1
The figure shows the cross section of a roof of a building. The length of the roof AC is
7 metres and the angles at B and C are 52° and 38° respectively.
A
7
C
h
38°
x
52°
B
Find the height, h, above the base and the length of the roof, AB.
[5]
2
Find the value of x in the range 0°  x  360° for which sin x  53 and tan x   34 .
[2]
3
You are given that sin y   54 and cos y   53 . Find the exact value of tan y.
[2]
4
Solve the equation sin2 x = 0.9 in the range 0°  x  360°.
[5]
5
(i)
Show that cos x  2sin2 x = 2cos2 x + cos x  2.
[2]
(ii)
Hence, or otherwise, solve the equation cos x  2 sin2 x + 2 = 0 in the
range 0°  x  360°.
[5]
6
Solve the following equations in the range 180°  x  180°.
(i)
sin x = 5cos x
[2]
(ii)
3tan2 x + 2tan x 5 = 0
[5]
(iii) 4cos2 x + sin x  3 = 0
[5]
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AM Trigonometry Assessment solutions
7
In the triangle ABC shown, AC = 8, angle C = 38° and angle B = 52°.
A
8
38°
C
8
9
52°
B
(i)
Find the length of the side AB.
[3]
(ii)
Calculate the area of the triangle.
[3]
In the triangle ABC, AB = 6, BC = 8 and CA = 9.
Calculate the size of the largest angle in this triangle.
[4]
In the quadrilateral ABCD shown below, AB = 8, BC = 5, angle B = 70°,
angle DAC = 40° and angle DCA = 50°.
B
70°
5
8
C
50°
A
45°
D
(i)
Calculate the length of the diagonal AC.
[4]
(ii)
Calculate the length of the side AD.
[4]
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AM Trigonometry Assessment solutions
10
A coast line runs East-West and on the coast there are two harbours, A and B.
A boat sails from A on a bearing of 040° for 5 miles to point C before altering course to
135° to sail directly to B.
C
135°
N
5
040°
B
A
Find
(i)
the distance AB,
[5]
(ii)
the distance sailed.
[4]
Total: 60
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AM Trigonometry Assessment solutions
Solutions to topic assessment
1. Using left-hand triangle:
Using right-hand triangle:
sin 38 
h
7
h  7 sin 38  4.31 metres (3 s.f.)
sin 52 
x
h
x
h
sin 52
M1 A2
 5.47 metres (3 s.f.)
2. Since sin x is positive and tan x is negative, x lies in the 2nd quadrant.
The acute angle for which sin x  53 is 36.9°
The corresponding angle in the second quadrant is 180° - 36.9° = 143.1°.
3. tan y 
sin y
cos y

 54 4

 53 3
M1 A1
B1
B1
M1 A1
4. sin 2 x  0.9  sin x  0.94868... , so there is a value of x in all four quadrants.
B2
The acute angle for which sin x  0.94868... is x = 71.6°
The equivalent angle in the second quadrant is 180° - 71.6° = 108.4°.
The equivalent angle in the second quadrant is 180° + 71.6° = 251.6°.
The equivalent angle in the second quadrant is 360° - 71.6° = 288.4°.
The solution is x = 71.6°, 108.4°, 251.6° and 288.4°.
5. (i)
Using the identity sin 2 x  cos2 x  1
cos x  2 sin 2 x  cos x  2(1  cos 2 x )
 cos x  2  2 cos 2 x
B3
M1
A1
 2 cos 2 x  cos x  2
(ii)
cos x  2 sin 2 x  2  0
2 cos 2 x  cos x  2  2  0
2 cos 2 x  cos x  0
cos x(2 cos x  1)
B1
cos x  0 or cos x   21
M1 A1
cos x  0  x  90 or 270
If cos x   21 , x is in the 2nd or 3rd quadrant
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AM Trigonometry Assessment solutions
The acute angle with cosine of 21 is 60°, so the equivalent angle in the 2nd
quadrant is 180° - 60° = 120°, and the equivalent angle in the 3rd quadrant is
180° + 60° = 240°.
The solution is x = 90°, 120°, 240°, 270°.
6. (i)
(ii)
sin x  5 cos x
sin x
5
cos x
tan x  5
tan is positive in the 1st and 3rd quadrants.
The acute angle for which tan x  5 is 78.7°.
The equivalent angle in the 3rd quadrant for -180°  x  180° is 78.7° - 180° =
-101.3°
M1 A1
The solution is x = -101.3°, 78.7°.
3 tan 2 x  2 tan x  5  0
(3 tan x  5 )(tan x  1)  0
M1 A1
M1
tan x   53 or 1
If tan x   53 , x is in the 2nd or 4th quadrants, so x = 180° - 59.0° = 121.0°
or x = -59.0°.
If tan x  1 , x is in the 1st or 3rd quadrants, so x = 45° or x = 45° - 180° = 135°
A2
The solution is x = -135°, -59.0°, 45°, 121.0°.
(iii)
4cos 2 x  sin x  3  0
4(1  sin 2 x )  sin x  3  0
4  4 sin 2 x  sin x  3  0
4 sin 2 x  sin x  1  0
1  1  4  4  1 1  17

8
8
1  17
If sin x 
, x is in the 1st or 2nd quadrant, so x = 39.8°
8
or x = 180° - 39.8° = 140.2°
sin x 
1  17
, x is in the 3rd or 4th quadrant,
8
so x = 23.0° - 180° = -157.0° or x = -23.0°
If sin x 
The solution is x = -157.0°, -23.0°, 39.8°, 140.2°.
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A2
AM Trigonometry Assessment solutions
7. (i)
(ii)
Using the sine rule:
AB
8
sin 38 sin 52
8 sin 38
AB 
 6.25 (3 s.f.)
sin 52

M1 A2
A  180  38  52  90 , so the triangle is right-angled
Taking AB as the base means AC is the height
Area  21  8  6.250  25.0 square units.
B1
M1 A1
B1
8. The largest angle is opposite the longest side, so the largest angle is B.
6 2  8 2  92
Using the cosine rule: cosB 
2 68
M1 A2
B  78.6 (3 s.f.)
9. (i)
Using the cosine rule on triangle ABC: b 2  a 2  c 2  2ac cosB
B1
b 2  5 2  8 2  2  5  8 cos 70
b  7.85 (3 s.f.)
M1 A2
(ii)
Using triangle ACD: D = 180° - 45° - 50° = 85°
AD
Using the sine rule:
sin 50

AC
B1
B1
sin 85
AC sin 50
AD 
 6.04 (3 s.f.)
sin 85
M1 A1
10.
C
1350
400 450
N
5
0400
A
450
500
B
B1 for angle C
M1 A1 for angle B
(i)
Using the sine rule:
5
AB

sin 45 sin 85
5 sin 85
AB 
 7.04 miles (3 s.f.)
sin 45
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AM Trigonometry Assessment solutions
(ii)
5
BC

sin 45 sin 50
5 sin 50
BC 
 5.42 (3 s.f.)
sin 45
Distance sailed = AC + CB = 5 + 5.42 = 10.42 miles.
Using the sine rule:
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M1 A1
M1 A1
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