WISKUNDE
GRAAD 10
NOG OEFENINGE
Definisies van die trigonometriese funksies : antwoorde.
MATHEMATICS
GRADE 10
MORE EXERCISES
Definitions of the trigonometric functions : answers.
Antwoord / Answer 1.
teenoorstaande / opposite
aangrensende / adjacent
1.1
sin C = —————————————
1.2
cos C = —————————————
skuinssy / hypotenuse
skuinssy / hypotenuse
t / o
a
AB
BC
= ——— = ——
= ——— = ——
s / h
s / h
BC
AC
6
3
8
4
= —— = —
= —— = —
10
10
teenoorstaande / opposite
teenoorstaande / opposite
1.3
tan C = —————————————
1.4
sin A = —————————————
aangrensende / adjacent
skuinssy / hypotenuse
t / o
t / o
AB
BC
= ——— = ——
= ——— = ——
a
s / h
BC
AC
6
3
8
4
= —— = —
= —— = —
10
10
aangrensende / adjacent
teenoorstaande / opposite
1.5
cos A = —————————————
1.6
tan A = —————————————
skuinssy / hypotenuse
aangrensende / adjacent
a / a
t / o
AB
BC
= ——— = ——
= ——— = ——
s / h
a / a
AC
AB
6
3
8
4
= —— = —
= —— = —
10
8
1
skuinssy / hypotenuse
1
skuinssy / hypotenuse
1.7
cosec C = ——— = —————————————
1.8
sec C = ———— = ————————————
sin C
teenoorstaande / opposite
cos C
aangrensende / adjacent
a / a
t / o
AB
BC
= ——— = ——
= ——— = ——
s / h
a / a
AC
AB
6
3
8
4
= —— = —
= —— = —
10
8
1
aangrensende / adjacent
1
skuinssy / hypotenuse
1.9
cot C = ——— = —————————————
1.10
cosec A = ———— = —————————————
tan C
teenoorstaande / opposite
sin A
teenoorstaande / opposite
t / o
t / o
AB
AB
= ——— = ——
= ——— = ——
t / o
t / o
AB
AB
8
4
10
5
= —— = —
= —— = —
6
6
1
skuinssy / hypotenuse
1
teenoorstaande / opposite
1.11
sec A = ——— = —————————————
1.12
cot A = ———— = ————————————
cos A
teenoorstaande / opposite
tan A
aangrensende / adjacent
a / a
t / o
AB
BC
= ——— = ——
= ——— = ——
s / h
a / a
AC
AB
6
3
8
4
= —— = —
= —— = —
10
8
Antwoord / Answer 2.
t / o
DF
a / a
EF
t / o
DF
2.1
sin θ = ——— = ——
2.2
cos α = ——— = ———
2.3
tan θ = —— = ———
s / h
DE
s / h
DE
a / a
EF
8
15
17
= ———
= ———
= ———
s
DE
s
DE
a
DF
2.4
cosec α = —— = ———
2.5
sec θ = —— = ———
2.6
cot α = —— = ———
t
EF
a
EF
t
EF
17
17
8
= ——
= ——
= ——
t / o
EF
a / a
EF
t / o
EF
2.7
sin α = —— = ———
2.8
cos θ = —— = ———
2.9
tan α = —— = ———
s / h
DE
s / h
DE
a / a
DF
15
15
15
= ——
= ——
= ——
s / h
DE
s / h
DE
a / a
EF
2.10
cosec θ = ——— = ———
2.11
sec α = ——— = ———
2.12
cot θ = ——— = ———
t / o
DF
a / a
DF
t / o
DF
17
17
15
= ——
= ——
= ———
Antwoord / Answer 3.
t / o
PR
s / h
PQ
a / a
QR
3.1
sin Q = ——— = ———
3.2
sec P = ——— = ———
3.3
cot Q = ——— = ———
s / h
PQ
a / a
PR
t / o
PR
5
13
12
= ——
= ——
= ——
a / a
PR
s / h
PQ
t / o
QR
3.4
cos P = ——— = ———
3.5
cosec Q = ——— = ———
3.6
tan P = ——— = ———
s / h
PQ
t / o
PR
a / a
PR
5
13
12
= ——
= ——
= ——
t / o
PR
s / h
PQ
a / a
PR
3.7
tan Q = ——— = ———
3.8
sec Q = ——— = ———
3.9
cot P = ——— = ———
a / a
QR
a / a
QR
t / o
QR
5
13
5
= ——
= ——
= ——
a / a
QR
t / o
QR
s / h
PQ
3.10
cos Q = ——— = ———
3.11
sin P = ——— = ———
3.12
cosec P = ——— = ———
s / h
PQ
s / h
PQ
t / o
QR
12
12
13
= ——
= ——
= ——
Antwoord / Answer 4.
a / a
KL
s / h
KM
t / o
KL
4.1
cot β = ——— = ———
4.2
cosec Φ = ——— = ———
4.3
tan Φ = ——— = ———
t / o
LM
t / o
KL
a / a
LM
7
16
7
= ——
= ——
= ——
t / o
LM
a / a
LM
s / h
KM
4.4
sin β = ——— = ———
4.5
cos Φ = ——— = ———
4.6
cosec β = ——— = ———
s / h
KM
s / h
KM
t / o
LM
11
11
16
= ——
= ——
= ——
t / o
LM
s / h
KM
a / a
KL
4.7
tan β = ——— = ———
4.8
sec β = ——— = ———
4.9
cos β = ——— = ———
a / a
KL
a / a
KL
s / h
KM
11
16
7
= ——
= ——
= ——
s / h
KM
t / o
KL
a / a
LM
4.10
sec Φ = ——— = ———
4.11
sin Φ = ——— = ———
4.12
cot Φ = ——— = ———
a / a
LM
s / h
KM
t / o
KL
16
7
11
= ——
= ——
= ——
Antwoord / Answer 5.
Figuur / Figure 5.1
y
4
r
5
y
4
(a)
sin θ = —— = ——
(b)
sec θ = —— = ——
(c)
tan θ = —— = ——
r
5
x
3
x
r
5
x
3
x
3
(d)
cosec θ = —— = ———
(e)
cos θ = —— = ———
(f)
cot θ = —— = ———
y
4
r
5
y
Figuur / Figure 5.2
y
12
r
15
y
12
(a)
sin θ = —— = ———
(b)
sec θ = —— = ———
(c)
tan θ = —— = ———
r
15
x
− 9
x
−9
4
5
4
= ——
= − ——
= − ——
5
3
r
15
x
− 9
x
− 9
(d)
cosec θ = —— = ———
(e)
cos θ = —— = ———
(f)
cot θ = —— = ———
y
12
r
15
y
12
5
3
3
= ——
= − ——
= − ——
4
5
Figuur / Figure 5.3
r
17
x
− 8
x
− 8
(a)
cosec θ = —— = ———
(b)
cos θ = —— = ———
(c)
cot θ = —— = ———
y
− 15
r
17
y
− 15
17
8
8
= − ——
= − ——
= ——
15
17
y
− 15
y
− 15
r
17
(d)
tan θ = —— = ———
(e)
sin θ = —— = ———
(f)
sec θ = —— = ———
x
− 8
r
17
x
−8
15
15
17
= ——
= − ——
= − ——
8
17
Figuur / Figure 5.4
r
√5
x
2
x
2
(a)
cosec θ = —— = ———
(b)
cos θ = —— = ———
(c)
cot θ = —— = ———
y
− 1
r
√5
y
− 1
2
= − √5
= ——
= − 2
√5
y
− 1
y
− 1
r
√5
(d)
tan θ = —— = ———
(e)
sin θ = —— = ———
(f)
sec θ = —— = ——
x
2
r
√5
x
2
1
1
√5
= − ——
= − ——
= ——
2
√5
Antwoord / Answer 6.
QR
QS
6.1
In ΔPQR; sin P = ——— en
in ΔPSQ; sin P = ———
PR
PQ
PQ
RS
6.2
In ΔPQR; cos R = ——— en
in ΔQSR; cos R = ———
PR
QR
QR
QS
6.3
In ΔPQR; tan P = ——— en
in ΔPSQ; tan P = ———
PQ
PS
Antwoord / Answer 7.
y
12
1
7.1
OP2 = PQ2 + OQ2 . . . Pythagoras
7.2
sin α = ─── = ──── = ───
25
144
12
= 122 + 52
= 144 + 25
13
12
1
7.3
13 cos α = —— X —— = —— = 12
= 169 = 132
1
13
1
OP = r = 13
12
13
25
7.4
tan α + sec α = ——
+ —— = ——
5
5
5
= 5
┌
┐2
┌
┐2
┌
┐2
│
5
12
│
5
7.5
sin2 α + cos2 α
=
7.6
tan2 α + 1
=
├───┤ + ├───┤
├───┤ + 1
13
13
│
12
│
└
┘
└
┘
└
┘
25
144
25
144
= —— +
——
= —— +
——
169
169
169
169
169
169
= —— = 1
= —— = 1
169
169
Antwoord / Answer 8.
− 12
12
8.1
OQ2 = x2 + y2 . . . Pythagoras
8.2
cos θ = ─── = − ────
13
13
= (−12)2 + (−5)2 = 144 + 25
−5
13
5
13
= 169
8.3
tan θ + sec θ = ─── + ─── = ─── − ────
−12
−12
12
12
OQ = r = √169
8
= 13
= − ───
12
┌
┐2
┌
┐2
13
−12
144
8.4
cosec2 θ − 1 =
├───┤ − 1
8.5
cot2 θ =
├───┤ = ────
−5
−5
25
└
┘
└
┘
169
25
= ─── − ───
25
25
144
= ───
25
Antwoord / Answer 9.
−√13
6
9.1
OK2 = MO2 + KM2 . . . Pythagoras
9.2
cos Φ + sin Φ = ──── + ────
7
7
= (−√13)2 + (6)2 = 13 + 36
6 − √13
= ───────
7
OK = r = √49
= 7
┌
┐2
┌
┐2
6
6
9.3
1 + tan2 = 1 +
├────┤
9.4
cot2 Φ + 1 =
├────┤ + 1
−√13
−√13
└
┘
└
┘
13
36
13
36
= ─── + ───
= ─── + ───
13
13
13
13
49
49
= ───
= ───
13
13
┌
┐2
7
49
9.5
cosec2 Φ =
├────┤ = ───
−√13
36
└
┘
Antwoord / Answer 10.
10.1
5 − 12 tan θ = 0 en / and sin θ < 0
−5
tan θ = —— en / and sin θ < 0
−12
∴ θ is in die derde kwadrant
∴ θ is in the third qudarant
of θ ∈ [180° ; 270°]
or θ ∈ [180° ; 270°]
en x en y is dus albei
and x and y are both
negatief, sodat
negative, so that
x = −12 en y = −5.
x = −12 and y = −5.
OQ2 = (−12)2 + (−5)2 . . . Pythagoras
= 169
r = OQ = √169 = 13
13
− 5
5
13
5
−12
10.2
13 sin θ = —— X ——
10.3
5 cosec Θ + 5 cot θ = ——— × ——— + ——— × ———
1
13
1
−5
1
−5
= −5
= −13 + 12 = −1
−5
13
8
10.4
tan Θ + sec θ = ——— + ——— = ———
−12
−12
−12
2
= − ——
3
Antwoord / Answer 11.
11.1
17 cos α = 8 en α ∈[180° ; 360°]
8
cos α = —— en α ∈[180° ; 360°]
17
∴ θ is in die vierde kwadrant
∴ θ is in the fourth qudarant
of α ∈ [270° ; 360°]
of α ∈ [270° ; 360°]
en x is positief en y is
and x is positive and y is
negatief, sodat x = 8
negative, so that x = 8
OQ2 = (8)2 + y2 . . . Pythagoras
172 = (8)2 + y2 . . . Pythagoras
y2 = 289 − 64 = 225
y = − √225 = − 15 . . . y < 0
17
11.2
cosec α = ——
8
┌
┐2
┌
┐2
− 15
8
11.3
sin2 α + cos2 α
=
├────┤
+
├───┤
17
17
└
┘
└
┘
┌
┐2
− 15
225
64
289
11.4
1 + tan2 α = 1 +
├────┤
= —— + ——— = ———
8
289
289
289
└
┘
64
225
289
= —— + —— = ———
= 1
64
64
64
−15
17
17
−15
15
8
11.5
tan α + sec α = —— + ——
11.6
17 sin α + 15 cot α = — X ——
+ — X ——
8
8
1
17
1
−15
1
= ——
= −15 − 8 = −23
4
Antwoord / Answer 12.
12.1
In ΔPQS ∠P + ∠PQS = 90° sodat / so that ∠PQS = (90° − α)
In ΔPQR ∠P + ∠R = 90° sodat / so that ∠R = (90° − α)
In ΔQSR ∠RQS + ∠R = 90° sodat / so that ∠RQS = α
Daarom / Therefore ∠RQS = α
QS
QR
SR
12.2
In ΔPQS tan α = —— en in / and in
ΔPQR tan α = —— en in / and in
ΔQSR tan α = ——
PS
PQ
QS
┌
┐2
┌
┐2
QS
PS
12.3
sin2 α + cos2 α =
├────┤
+
├────┤
PQ
PQ
└
┘
└
┘
QS2 + PS2
= ——————
PQ2
PQ2
= ——— . . . Pythagoras
PQ2
= 1