Download GRADE 10.Trigonometry

CAMI Mathematics
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athematics: Grade 10
GRADE 10_CAPS Curriculum
10.9 Trigonometry
1.1 Define the trigonometric ratios sin θ , cos θ and tan θ using rightright-angled
triangle.
triangle.
(a)
(b)
cosA = …
sinC = …
tanA = ...
sinA = …
tanC = …
cosC = …
(c)
(d)
sinZ = …
cosZ = …
tanZ = …
sinY = …
cosY = …
tanY = …
sinQ = …
tanR = …
cosQ = …
1.2
Extend the definitions of sin θ , cos θ and tan θ for 0° ≤ θ ≤ 360° .
(a)
(c)
(e)
(g)
(i)
cos 100°
sin 300°
sin 315°
sin 240°
tan 150°
(b)
(d)
(f)
(h)
(j)
tan 210°
tan 135°
cos 120°
cos 225°
sin 135°
1.3 Define the reciprocals of the trigonometric ratios cosec θ , sec θ and cot θ ,
using rightright-angled triangles.
triangles.
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athematics: Grade 10
sinA = …
cosA =
tanA = …
cosecA =…
secA = …
cot A = …
cosec C = …
sec C = …
cotC = …
sin C = …
cos C = …
tan C = …
1.4 Derive values of the trigonometric ratios for the special cases (without using
a calculator).
calculator).
(a)
tan 225°. sin 135°. tan 300°
cos 315°. cos 225°. cos 150°
(b)
tan 120°
tan 330°
(c)
sin60°.cos30°.tan60°
(d)
sin30°.tan45°.cos45°
(e)
tan 120°. cos 210°
sin 240°. sin 240°
(f)
cos 330°
cos 225°. cos 315°. tan 225°
1.5 Solve twotwo-dimensional problems involving rightright-angled triangles.
(a) The length of a mast is 8.5m, and the length of the shadow of the mast is
7.25m. Calculate the angle of elevation of the sun at the particular moment.
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(b) The angle of elevation of a glider according to a woman on the ground is
43°. If the glider is 2340m from the woman, calculate the altitude of the glider.
(c) Two towers are 12m apart. From B the angle of elevation to DE is 29° and
from D the angle of elevation to BC is 48°. Calculate the difference in the heights
of the towers.
(d) A building (DF) and a tower (CE) are 94m apart. From the roof of the
building the angle of elevation to the top of the tower is 15° and the angle of
depression to the bottom of the tower is 46°. Calculate the height of the tower.
1.6
Solve simple trigonometric equations for angles between 0° and 90°.
(a) sin51° = cos β , β an acute angle.
(b) cos33° = sin α
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(c)
(d)
(e)
(f)
sin75° = cos3 θ
cos 4 α = sin 5 α
cos ( β – 43°) = sin 65°
sin( θ + 54°) = cos ( θ – 8°)
1.7 Use diagrams to determine the numerical values of ratios for angles from 0°
and 360°.
(a)
(b)
(c)
(d)
(e)
If 17sinA = 15, 0°≤ A ≤ 90°, determine tan A.
If 9tan β = 40 and β is an acute angle, determine sin β .
If 6sin α – 5 = 0 and α ∈ [90°;180°], determine cos α .
If -5cos β – 4 = 0 and β ∈ [180°;270°], determine sin β .
If 5sin θ – 4 = 0 and 90° ≤ θ ≤ 180°, determine cos θ .
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athematics:
athematics: Grade 10
MEMO
1.1 Define the trigonometric ratios sin θ , cos θ and tan θ using rightright-angled
angled
triangle. [7.2.1.1
[7.2.1.1]
7.2.1.1
(a)
(b)
c
b
c
sin C =
b
a
tan A =
c
6
x
8
tan C =
6
6
cos C =
x
cos A =
sin A =
(c)
(d)
120
65
; sinY =
x
x
65
120
cosZ =
; cosY =
x
x
120
65
tanZ =
; tanY =
65
120
sinZ =
1.2
q
p
r
tanR =
q
r
cosQ =
p
sinQ =
Extend the definitions of sin θ , cos θ and tan θ for 0° ≤ θ ≤ 360° .
[7.4.2.2;
[7.4.2.2; 7.4.2.3]
7.4.2.3
(a) cos 100° = -cos 80°
(b) tan 210° = tan 30°
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(c)
(e)
(g)
(i)
sin 300° = -sin 60°
sin 315° = -sin 45°
sin 240° = -sin 60°
tan 150° = -tan 30°
(d)
(f)
(h)
(j)
tan 135° = -tan 45°
cos 120° = - cos 60°
cos 225° = -cos 45°
sin 135° = sin 45°
1.3 Define the reciprocals of the trigonometric ratios cosec θ , sec θ and cot θ ,
using rightright-angled triangles. [7.2.1.3; 7.2.1.4; 7.2.1.5; 7.2.1.2]
7.2.1.2]
a
b
c
cos A =
b
a
tan A =
c
sin A =
cosec C =
cosec A =
b
a
a
cot C =
c
c
sin C =
b
a
cos C =
b
c
tan C =
a
b
c
c
cot A =
a
sec A =
b
c
sec C =
b
a
1.4 Derive values of the trigonometric ratios for the special cases (without using
a calculator). [7.3.2.1; 7.3.2.3; 7.3.1.5; 7.3.1.1]
7.3.1.1
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athematics: Grade 10
(a)
tan 225°. sin 135°. tan 300°
cos 315°. cos 225°. cos150°
tan 45°. sin 45°.(− tan 60°)
=
cos 45°.(− cos 45°).( − cos 30°)
1
1.
.(− 3 )
2
=
1
1
3
.(−
).(−
)
2
2
2
= −2 2
(b)
tan 120°
tan 330°
− tan 60°
=
− tan 30°
3
=
1
3
=3
(c)
sin60°.cos30°.tan60°
3 3
=
.
. 3
2 2
3 3
=
4
(d)
sin30°.tan45°.cos45°
1
1
= .1.
2
2
1
=
2 2
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(e)
tan 120°. cos 210°
sin 240°. sin 240°
(− tan 60°).( − cos 30°)
=
(− sin 60°).(− sin 60°)
3
2
3.
=
(
3 2
)
2
=2
(f)
cos 330°
cos 225°. cos 315°. tan 225°
cos 30°
=
(− cos 45°). cos 45°. tan 45°
3
2
=
−
1
2
.
1
2
.1
=− 3
1.5 Solve twotwo-dimensional problems involving rightright-angled triangles.
[7.7.1.1; 7.7.1.2; 7.7.1.3]
7.7.1.3
(a) The length of a mast is 8.5m, and the length of the shadow of the mast is
7.25m. Calculate the angle of elevation of the sun at the particular moment.
AB
= tan x
BC
8.5
= tan x
7.25
8.5
x = tan −1 (
)
7.25
x = 49.5°
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athematics: Grade 10
(b) The angle of elevation of a glider according to a woman on the ground is
43°. If the glider is 2340m from the woman, calculate the altitude of the glider.
AC
sin x
BC
x
= sin 43°
2340
x = 2340 × sin 43°
x = 1595.9m
(c) Two towers are 12m apart. From B the angle of elevation to DE is 29° and
from D the angle of elevation to BC is 48°. Calculate the difference in the heights
of the towers.
∆BCD :
BC
= tan 48°
BD
x
= tan 48°
12
x = 12 × tan 48°
x = 13.33m
BC − DE = 13.33 − 6.65 = 6.68m
∆BDE :
DE
= tan 29°
BD
y
tan 29°
12
y = 12 × tan 29°
y = 6.65m
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(d) A building (DF) and a tower (CE) are 94m apart. From the roof of the
building the angle of elevation to the top of the tower is 15° and the angle of
depression to the bottom of the tower is 46°. Calculate the height of the tower.
∆BCD :
BC
= tan 46°
DB
x
= tan 46°
94
x = 94 × tan 46°
x = 97.34m
∆BDE :
BE
= tan 15°
DB
y
= tan 15°
94
y = 94 × tan 15°
y = 25.19m
EC = 97.34 + 25.19 = 122.53m
1.6
Solve simple trigonometric equations for angles between 0° and 90°.
[7.6.2.1; 7.6.2.3; 7.6.2.5]
7.6.2.5]
(a) sin51° = cos β , β an acute angle.
sin 51° = cos β
sin 51° = sin(90° − β )
∴ 51° = 90° − β
∴ β = 90° − 51°
∴ β = 39°
(b) cos33° = sin α
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cos 33° = sin α
cos 33° = cos(90° − α )
∴ 33° = 90° − α
∴ α = 90° − 33°
∴ α = 57°
(c)
sin75° = cos3 θ
sin 75° = cos 3θ
sin 75° = sin(90° − 3θ )
∴ 75° = 90° − 3θ
∴ 3θ = 90° − 75°
∴ 3θ = 15°
∴ θ = 5°
(d)
cos 4 α = sin 5 α
cos 4α = sin 5α
cos 4α = cos(90° − 5α )
∴ 4α = 90° − 5α
∴ 9α = 90°
∴ α = 10°
(e)
cos ( β – 43°) = sin 65°
cos( β − 43°) = sin 65°
cos( β − 43°) = cos(90° − 65°)
∴ β − 43° = 90° − 65°
∴ β = 68°
(f)
sin( θ + 54°) = cos ( θ – 8°)
sin(θ + 54°) = cos(θ − 8°)
sin(θ + 54°) = sin(90° − (θ − 8°)
sin(θ + 54°) = sin(90° − θ + 8°)
∴θ + 54° = 90° − θ + 8°
∴ 2θ = 90° − 54° + 8°
∴ 2θ = 44°
∴θ = 22°
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athematics: Grade 10
1.7 Use diagrams to determine the numerical values of ratios for angles from 0°
and 360°. [7.6.3.1;
[7.6.3.1; 7.6.3.3; 7.6.3.5; 7.6.5.1]
7.6.5.1
(a) If 17sinA = 15, 0°≤ A ≤ 90°, determine tan A.
x2 + y2 = r 2
x 2 + (15) 2 = (17) 2
x 2 = 64
x=8
∴ tan A =
15
8
(b) If 9tan β = 40 and β is an acute angle, determine sin β .
x2 + y2 = r 2
(9) 2 + (40) 2 = r 2
81 + 1600 = r 2
r 2 = 1681
r = 41
∴ sin β =
40
41
(c) If 6sin α – 5 = 0 and α ∈ [90°;180°], determine cos α .
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x2 + y2 = r 2
x 2 + (5) 2 = (6) 2
x 2 = 36 − 25
x = − 11
∴ cos α =
− 11
6
(d) If -5cos β – 4 = 0 and β ∈ [180°;270°], determine sin β .
x2 + y2 = r 2
(−4) 2 + y 2 = (5) 2
16 + y 2 = 25
y = −3
∴ sin β =
−3
5
(e) If 5sin θ – 4 = 0 and 90° ≤ θ ≤ 180°, determine cos θ .
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x2 + y2 = r 2
x 2 + (4) 2 = (5) 2
x 2 + 16 = 25
x = −3
∴ cos θ =
−3
5