Aufgaben zu trigonometrischen Funktionen - Poenitz

Aufgaben zu trigonometrischen Funktionen - Poenitz

4.8. Exercises on trigonometric functions
Exercise 1: Right-angled triangles
Find the missing lengths and angles.
The lengths are given in cm.

b
a


c
part
a
b
c
α
β
a)
b)
4
8,61
c)
d)
36
4,5
13,20
7,6
e)
f)
2,5
g)
8,6
h)
5
2
7,2
13,2
48°
i)
3.9
j)
27,2
k)
17,3
4.6
l)
35,2
54°
23°
64°
56°
36°
53°
Exercise 2: Applications
a) An observer stands 150 from the base of a church tower and sees its top at an
elevation of 9°. His eyes are 1,5 m above ground. How high is the tower?
b) A ladder with length 7,5 m leans to a wall and meets it 6,6 m above the floor.
How far is its base from the wall and what is its angle of inclination to the
40 °
floor?
15 0 m
c) A radio mast is to be secured with 20 m ropes which are to have an elevation
of 65°. How far above ground will they join the mast? How far from the mast
will they be fastened in the ground?
d) How high is a fir, when its shadow is 27,5 m long and the sun stands at an elevation of 38,5° above the horizon?
e) How far comes a paraglider who has started from a height of 25 m at a straight flightpath with a depression of 8°?
f) From the top of a 28,6 m high tower the adjacent river appears at an angular width of 17°. Its nearest bank is 6 m from the
foot of the tower. How broad is the river?
g) For an observer 12 m away a flagpole on top of a 15 m high tower appears at an angular length of 6,5°. How long is the
flagpole?
Exercise 3: pyramids
Find the missing measures. All lengths are given in cm
part
g
s
h
hs
αs
αh
a)
5
4
b)
6
4
c)
d)
8
e)
f)
g)
h)
10
i)
9
4
5
5
s
5
h
7
70°
45°
45°
60°
60°
50°
hs
• •
αs
Exercise 4: Resolution of forces on an inclined plane
A boy with mass m = 20 kg sits on a slide with depression α = 30°. His
acceleration is caused by the parallel component Fpar of the gravitational
force Fg = m∙g with the gravitational field constant g = 9,81 m/s2. Find the
magnitude of Fpar.
αh
g
Fpar
Exercise 5: Intersection angle of two lines
Find the intersection angle of each pair of lines:
1
a) g1(x) = x − 1 und g2(x) = x + 1
2
b) g1(x) = 2x − 3 und g2(x) = x
2
c) g1(x) = − x + 1 und g2(x) = −2x + 4
3
d) g1(x) = −x + 5 und g2(x) = 3x − 2
α
α
Fg
Fn
1
Exercise 6: Measuring angles in radians
Fill in the gaps:
degree
0°
45°

6
radian
90°
135°
2
3

3
180°
360°
5
6
57,29°
70°

9
3
2
2
Exercise 7: Transforming the sine curve
Find the angular velocity ω, the period T, the amplitude A, the phase t0 and the vertical shift y0 of the following functions.
Draw the graphs of f1 - f3 in a common coordinate system with the domain −4 ≤ t ≤ 4
1
3
a) f1(t) = sin[2πt]
b) f1(t) = sin[2πt]
c) f1(t) = sin[2π(t − 1)] + 2
d) f1(t) = sin[ (t − 2)] + 2
3
2
2
2
1
1
f2(t) = sin [πt]
f2(t) = 2sin [ t]
f2(t) = sin[π(t − )]
f2(t) = sin[
(t − 1)]
3
3
2
2
2
3
f3(t) = sin[ t]
f3(t) = 3sin[
t]
f3(t) = sin[ (t + 1)] − 2
f3(t) = sin[π (t + 1)] − 2
2
2
3
2
Exercise 8: Transforming the sine curve
Find the formulae of the functions f1 - f5:
a)
b)
y
y
3
3
f1
f1
2
2
f2
f2
1
-2
-1
f3
f3
0
-3
1
0
1
2
x
3 -3
0
-2
-1
f4
-1
0
1
2
x
3
-1
f4
f5
-2
-2
f5
-3
-3
Exercise 9: Sine rule
Find the missing data of a triangle with sides a, b, c and angles α, β, γ with values given as follows:
a) a = 14,3 m; c = 27,9 m und γ = 82,1°
b) a = 13 m; b = 27 m und α = 27°
Exercise 10: Sine rule
Show that the bisector in a triangle divides the opposite side in the ratio of the adjacent sides.
Exercise 11: Sine and Cosine rule
Find the missing measures of a triangle with sides a, b, c and angles α, β, γ with values given as follows:
a) α = 30°,β = 60°, a = 3 cm
b) a = 6 cm, b = 4 cm, γ = 40°
c) a = 3 cm; b = 4 cm; c = 5 cm
Exercise 12: Cosine rule
To avoid an inaccessible rock, a land surveyor starts at point A, walks 85 m due north to point B and continues in a straight
line for 102 m on a bearing of 052° until point C. Find the distance between A and C.
2
4.8. Solutions tot he exercises on trigonometric functions
Exercise 1: Right-angled triangles
part
a
b
c
α
β
a)
4
3,6
5,38
48°
42°
b)
3,77
7,74
8,61
26°
64°
c)
36
13,2
38,34
69,86
20,13
d)
4,5
6,12
7,6
36,31°
53,69°
e)
9,91
7,2
12,25
54°
36°
f)
2,5
3,71
4,47
34°
56°
g)
8,6
10,01
13,20
40,66
49,34
h)
5
2
5,38
68,20
21,80
i)
3,9
2,44
4,6
57,98°
32,02°
j)
27,2
19,76
33,62
54°
36°
k)
17,3
40,75
44,28
23°
67°
l)
21,18
28,11
35,2
37°
53°
Worked example to a):
β = 90° − 48° = 42°
a
4cm
c=
=
≈ 5,38 cm
sin()
0,74
b = c∙cos(β) = 5,38 cm∙0,67 ≈ 3,6 cm
Exercise 2: Applications
a) Height h = 1,5 m + 150 m∙tan(40°) = 126,5 m.
6, 6
b) Inclination angle α = sin–1(
) ≈ 61,64° and distance d ≈ 7,52  6,62 ≈ 3,56 m
7,5
c) Height h = 20 m∙sin(65°) ≈ 18,12 m und distance d = 20 m∙cos(65°) ≈ 8,45 m
d) Height h = 27,5 m∙tan(38,5°) ≈ 21,87 m
25 m
e) distance d =
≈ 1777,88 m
tan(8)
6m
) ≈ 11,84° to the vertical and has the distance d1 = 6 m from the
28, 6 m
tower. The farther shore appears at the angle α2 = 17° + 11,84° = 28.84° to the vertical and has the distance d2 = 28,6
f) The nearest shore appears at the angle α1 = tan–1(
m∙tan(28,84°) ≈ 15,75 m from the tower. So the river has a width of d2 – d1 = 9,75 m.
13,4 m
g) The lower end of the flagpole appears at the angle α1 = tan–1(
) ≈ 48,15° to the horizontal and is h1 = 13,4 m above
12 m
the eyes of the observer. The upper end appears at the angle α2 = 6,5° + 48,15° = 54,65° und is h2 = 12 m∙tan(54,65°) ≈
16,92 m above the eyes of the observer. Therefore the flagpole has a height of h2 – h1 = 3,52 m.
Exercise 3: pyramids
part
g
s
h
hs
αs
αh
a)
5
4
1,87
3,12
27,87°
36,82°
b)
6,32
6
4
5,10
41,81°
51,66°
c)
6
5,83
4
5
43,32°
53,13°
d)
8
16,53
15,54
16,05
70°
75,52°
e)
10
8,66
5
7,07
35,37°
45°
f)
7
7,83
6,06
7
50,71°
60°
g)
7,07
7,07
5
6,12
45°
54,78°
h)
10
9,25
5,96
7,78
40,12°
50°
i)
6,36
9
7,79
8,42
60°
67,70°
Worked example to a)
The base quadrangle has the diagonal d =
2
Height h =
d
s2    ≈
2
2g=
2 ∙4 ≈ 5,65 cm
2
 5, 65 
42  
 ≈ 1,87 cm (light shaded triangle)
 2 
2
Face height hs =
g2  g2 =
g
h2    ≈
2
s
2
5
1,872    ≈ 3,12 cm (dark shaded trangle)
2
h
 1,87 cm 
Corner angle αs = sin–1   ≈ sin–1 
 ≈ 27,87° (light shaded triangle)
s
 4 cm 
 
 h 
 1,87 cm 
Edge angle αh = sin–1   ≈ sin–1 
 ≈ 36,82° (dark shaded trangle)
h
 3,12 cm 
 s
αs
h
• •
hs
αh
g
3
Exercise 4: Resolution of forces on an inclined plane
Fh = Fg∙sin(α) = mg∙sin(α) = 98,1 N (corresponds to the gravitational pull of 10 kg)
Exercise 5: Intersection angles of lines
a) α = 45° − 26,5° = 18,5°
c) α = −33,69° − (−63,43°) = 29,74°
b) α = 63,43° − 45° = 18,43°
d) α = 71,57° − (−45°) = 116,57° bzw 63,43°
Exercise 6: Measuring angles in degrees and radians
degree
0°
radian
0
30°

6
45°

4
60°

3
90°

2
120°
2
3
135°
3
4
150°
5
6
Exercise 7: Transforming the graph of the sine function
a)
360°
2
2
1
1
0
1
2
x
3
-2
-1
-2
-2
-3
d)
y
3
3
2
2
1
1
0
-2
-1
0
1
2
x
3
1
114,59°
2
x
0
-1
-3
-3
70°
7
18
0
-3
-1
c)
57,29°
y
3
-1
20°

9
2π
3
0
-2
π
270°
3
2
b)
y
-3
180°
1
2
3
1
2
x
3
y
0
-3
-2
-1
0
-1
-1
-2
-2
-3
-3
Exercise 8: Transforming the graph of the sine function
1
a)
f1(t) = sin[2πt] + 2
b)
f1(t) = sin[πx] + 2
2
π
1
f2(t) = sin[π (t − 1)] + 1
f2(t) = sin[ (t + 3)] + 1
3
2
π
π
1
f3(t) = sin[ (t + 1]
f3(t) = sin[ (t + 2]
2
2
2
2π
3
π
1
f4(t) = sin[ (t + 2)] − 1
f4(t) = sin[
(t + 2)] −
3
5
2
2
π
1
f5(t) = sin[ (t − 2)] − 2
f5(t) = sin[2πt] − 2
2
2
4
Exercise 9: Sine rule
sin(γ)
a) sin(α) = a∙
= 0,51 ⇒ α = 30,5°
c
(Sine rule)
β = 180° − α − γ = 67,39°
sin(β)
b = a∙
= 26,00 m
sin(α)
b) sin(β) = b∙
(angle sum)
(Sine rule)
sin(α)
= 0,94 ⇒ β = 70,54°
a
(Sine rule)
γ = 180° − α − β = 82,46°
sin(γ)
c = a∙
= 28,38 m
sin(α)
(angle sum)
(Sine rule)
Exercise 10: Sine rule
In the lower triangle we have
sin(δa )
a

sin(γ / 2) ca
sin(δb )
b

In the upper trangle we have
sin(γ / 2) cb
Because of δa + δb = 180° we can write
sin(δa) = sin(δb) so that
c
b
a
a

⇔ a  , qed.
c b ca
cb b
cb
δb
ca
b
γ/2
δa
γ/2
a
Exercise 11: Sine and Cosine rules
a) γ = 180° − α − β = 90°
sin(α)
b = a∙
= 5,2 cm
sin(β)
c = a∙
b) c =
(angle sum)
(Sine rule)
sin(γ)
= 6 cm
sin(α)
(Sine rule)
a 2  b2  2a  b  cos(γ) = 3,9 cm
(Cosine rule)
sin(β) = b∙
sin(γ)
= 0,66 ⇒ β = 41,24°
c
(Sine rule)
β = 180° − β − γ = 98,76°
(angle sum)
sin(γ)
Beware the obtuse angle α: sin(α) = a∙
= 0,99.
c
The calculator gives the acute neighbor angle sin−1(0.99) = 180° − α = 81,36°!
c) cos(α) =
cos(β) =
a2
b2
b 2 c2
4
= ⇒ α = 36,8°
2bc
5
a 2 c2
3
= ⇒ β = 53,1°
2ac
5
γ = 180° − α − β = 90°
Exercise 12: Cosine rule
β = 180° - 52,2° = 127,8°
2
(Cosine rule)
(Cosine rule)
(angle sum)
(neighbor angle)
2
AC  AB  BC  2AB  BC  cos(β) = 167,88 m. (Cosine rule)
5