WISKUNDE
GRAAD 11
NOG OEFENINGE
Oplos van driehoeke : antwoorde.
MATHEMATICS
GRADE 11
MORE EXERCISES
Solving triangles : answers.
Antwoord/Answer 1.1
In ΔABC : ∠A = 49°, ∠B = 72°, BC = 23 . . . gegee / given
Dus, 'n sy en die teenoorstaande hoek en 'n ander hoek word gegee sodat ons dus
die sin-reël kan toepas. /
Therefore, a side and the opposite angle and another angle are given so that we can apply the sine rule.
∠C = 180° − (49° + 72°) = 59°
AB
BC
AC
AB
23
AC
───── = ───── = ─────
en dus / and therefore
───── = ───── = ─────
sin C
sin A
sin B
sin 59°
sin 49°
sin 72°
BC × sin C
BC × sin B
BA = ──────────
AC = ──────────
sin A
sin A
23 × sin 59°
23 × sin 72°
= ────────────
= ────────────
sin 49°
sin 49°
= 26,12
= 28,98
C = 59°, AB = 26,12, AC = 28,98
Antwoord/Answer 1.2
In ΔDEF : ∠D = 67°, ∠E = 38°, EF = 56 . . . gegee / given
Dus, 'n sy en die teenoorstaande hoek en 'n ander hoek word gegee sodat ons dus
die sin-reël kan toepas. /
Therefore, a side and the opposite angle and another angle are given so that we can apply the sine rule.
∠F = 180° − (67° + 38°) = 75°
DE
EF
DF
DE
56
DF
───── = ───── = ─────
en dus / and therefore
───── = ───── = ─────
sin F
sin D
sin E
sin 75°
sin 67°
sin 38°
EF × sin F
EF × sin E
DE = ──────────
DF = ──────────
sin D
sin D
23 × sin 75°
23 × sin 38°
= ────────────
= ────────────
sin 67°
sin 67°
= 24,13
= 15,38
F = 75°, DE = 24,13, DF = 15,38
Antwoord/Answer 1.3
In ΔKLM : ∠L = 29,5°, ∠M = 66,3°, KM = 13 . . . gegee / given
Dus, 'n sy en die teenoorstaande hoek en 'n ander hoek word gegee sodat ons dus
die sin-reël kan toepas. /
Therefore, a side and the opposite angle and another angle are given so that we can apply the sine rule.
∠K = 180° − (29,5° + 66,3°) = 84,2°
KL
LM
KM
KL
LM
13
───── = ───── = ─────
en dus / and therefore
───── = ───── = ─────
sin M
sin K
sin L
sin 66,3°
sin 84,2°
sin 29,5°
KM × sin M
KM × sin K
DE = ──────────
DF = ──────────
sin L
sin L
13 × sin 66,3°
13 × sin 84,2°
= ────────────
= ────────────
sin 29,5°
sin 29,5°
= 24,17
= 26,26
K = 84,2°, KL = 24,17, LM = 26,26
Antwoord/Answer 1.4
In ΔPQR : ∠P = 67,2°, ∠R = 38,8°, PQ = 45 . . . gegee / given
Dus, 'n sy en die teenoorstaande hoek en 'n ander hoek word gegee sodat ons dus
die sin-reël kan toepas. /
Therefore, a side and the opposite angle and another angle are given so that we can apply the sine rule.
∠Q = 180° − (67,2° + 38,8°) = 74°
PQ
QR
PR
45
QR
PR
───── = ───── = ─────
en dus / and therefore
───── = ───── = ─────
sin R
sin P
sin Q
sin 38,8°
sin 84,2°
sin 67,2°
PQ × sin P
PQ × sin Q
QR = ──────────
PR = ──────────
sin R
sin R
45 × sin 67,2°
45 × sin 74°
= ────────────
= ────────────
sin 38,8°
sin 38,8°
= 66,20
= 69,03
∠Q = 74°, QR = 66,20, PR = 69,03
Antwoord/Answer 1.5
In ΔABC : ∠A = 59°, BC = 32, AC = 23 . . . gegee / given
Dus, 'n sy en die teenoorstaande hoek en 'n ander sy word gegee sodat ons dus
die sin-reël kan toepas. /
Therefore, a side and the opposite angle and another side are given so that we can apply the sine rule.
sin A
sin B
sin C
sin 59°
sin B
sin C
───── = ───── = ─────
en dus / and therefore
───── = ───── = ─────
BC
AC
AB
32
23
AB
AC × sin A
sin B = ──────────
BC
23 × sin 59°
= ──────────
32
= 0,616088.....
∠B = 38,0°
∠C = 180° − (59° + 38°) = 83°
BC × sin C
AB = ──────────
sin A
32 × sin 83°
= ────────────
sin 59°
= 37,05
∠B = 38°, ∠C = 83°, AB = 37,05
Antwoord/Answer 1.6
In ΔDEF : ∠D = 74°,
DE = 18, EF = 56 . . . gegee / given
Dus, 'n sy en die teenoorstaande hoek en 'n ander sy word gegee sodat ons dus
die sin-reël kan toepas. /
Therefore, a side and the opposite angle and another side are given so that we can apply the sine rule.
sin D
sin E
sin F
sin 74°
sin E
sin F
───── = ───── = ─────
en dus / and therefore
───── = ───── = ─────
EF
DF
DE
56
DF
18
DE × sin D
sin F = ──────────
EF
18 × sin 74°
= ──────────
56
= 0,308976.....
∠F = 18,0°
∠E = 180° − (74° + 18°) = 88°
EF × sin E
DF = ──────────
sin D
56 × sin 88°
= ────────────
sin 74°
= 58,22
∠E = 88°, ∠F = 18°, DF = 58,22
Antwoord/Answer 1.7
In ΔKLM : ∠L = 138°,
KL = 58, KM = 123 . . . gegee / given
Dus, 'n sy en die teenoorstaande hoek en 'n ander sy word gegee sodat ons dus
die sin-reël kan toepas. /
Therefore, a side and the opposite angle and another side are given so that we can apply the sine rule.
sin K
sin L
sin M
sin K
sin 138°
sin M
───── = ───── = ─────
en dus / and therefore
───── = ───── = ─────
LM
KM
KL
LM
123
58
KL × sin L
sin M = ──────────
KM
58 × sin 138°
= ──────────
123
= 0,315525.....
∠F = 18,4°
∠K = 180° − (138° + 18,4°) = 23,6°
KM × sin K
LM = ──────────
sin L
123 × sin 23,6°
= ────────────
sin 138°
= 73,59
∠K = 23,6°, ∠M = 18,4°, DF = 73,59
Antwoord/Answer 1.8
In ΔPQR : ∠Q = 140°,
QR = 48, PR = 74 . . . gegee / given
Dus, 'n sy en die teenoorstaande hoek en 'n ander sy word gegee sodat ons dus
die sin-reël kan toepas. /
Therefore, a side and the opposite angle and another side are given so that we can apply the sine rule.
sin P
sin Q
sin R
sin P
sin 140°
sin R
───── = ───── = ─────
en dus / and therefore
───── = ───── = ─────
QR
PR
PQ
48
74
PQ
QR × sin Q
sin P = ──────────
PR
48 × sin 140°
= ──────────
74
= 0,4169.....
∠P = 24,6°
∠R = 180° − (140° + 24,6°) = 15,4°
PR × sin R
PQ = ──────────
sin Q
74 × sin 15,4°
= ────────────
sin 140°
= 30,57
∠P = 24,6°, ∠R = 15,4°, PQ = 30,57
Antwoord/Answer 2.1
Antwoord/Answer 2.2
q
r
r
q
───── = ─────
───── = ─────
sin y
sin x
sin x
sin y
r sin y
q sin x
q = ──────
r = ───────
sin x
sin y
Antwoord/Answer 2.3
Antwoord/Answer 2.4
Onbekende hoek / Unknown angle = (180° − (x + y))
sin (180° − (x + y)) = sin (x + y)
p
q
p
r
──────── = ─────
──────── = ─────
sin (x + y)
sin y
sin (x + y)
sin x
q sin (x + y)
r sin (x + y)
p = ─────────
p = ─────────
sin y
sin x
Antwoord/Answer 2.5
Antwoord/Answer 2.6
q
p
r
p
───── = ────────
───── = ────────
sin y
sin (x + y)
sin x
sin (x + y)
p sin y
p sin x
q = ─────────
r = ────────
sin (x + y)
sin (x + y)
Antwoord/Answer 3.1
Antwoord/Answer 3.2
AC2 = AB2 + BC2 − 2 . AB . BC . cos B
DF2 = DE2 + EF2 − 2 . DE . EF . cos E
= 82 + 72 − 2 . 8 . 7 . cos 53°
= 112 + 52 − 2 . 11 . 5 . cos 144°
= 45,5967...
= 234,9918...
───────
───────
AC = √45,5967...
DF = √234,9918...
= 6,75
= 15,33
Antwoord/Answer 3.3
Antwoord/Answer 3.4
KM2 = KL2 + LM2 − 2 . KL . LM . cos L
PQ2 = QR2 + PR2 − 2 . QR . PR . cos R
= 4,22 + 3,72 − 2 . 4,2 . 3,7 . cos 130°
= 6322 + 5432 − 2 . 632 . 543 . cos 56°
= 51,3078...
= 310 469,8323...
───────
───────
AC = √51,3078...
DF = √310 469,8323...
= 7,16
= 557,20
Antwoord/Answer 3.5
Antwoord/Answer 3.6
YZ2 = XY2 + XZ2 − 2 . XY . XZ . cos X
AB2 = AC2 + BC2 − 2 . AC . BC . cos C
= 98,32 + 76,92 − 2 . 98,3 . 76,9 . cos 98,2°
= 9,62 + 4,72 − 2 . 9,6 . 4,7 . cos 144°
= 17 732,84124...
= 187,2556...
───────
───────
AC = √17 732,84124...
DF = √187,2556...
= 133,16
= 13,68
Antwoord/Answer 3.7
Antwoord/Answer 3.8
EF2 = DE2 + DF2 − 2 . DE . DF . cos E
KM2 = KL2 + LM2 − 2 . KL . LM . cos L
Skryf cos D i.t.v. die sye. / Write cos D in terms of the sides.
DE2 + DF2 − EF2
KL2 + LM2 − KL2
cos D = ────────────────
cos L = ────────────────
2 . DE . DF
2 . KL . LM
102 + 142 − 52
92 + 112 − 52
= ─────────────
= ─────────────
2 . 10 . 14
2 . 9 . 11
= 0,9678...
= 0,8939...
∠D = 14,57°
∠L = 26,63°
Antwoord/Answer 3.9
Antwoord/Answer 3.10
QR2 = PQ2 + PR2 − 2 . PQ . PR . cos P
YZ2 = XY2 + XZ2 − 2 . XY . XZ . cos X
PQ2 + PR2 − QR2
XY2 + XZ2 − YZ2
cos P = ────────────────
cos X = ────────────────
2 . PQ . PR
2 . XY . XZ
62 + 32 − 42
92 + 52 − 112
= ─────────────
= ─────────────
2 . 6 . 3
2 . 9 . 5
= 0,8055...
= −0,1666...
∠P = 36,34°
cos X is negatef en dus is X 'n stomphoek.
cos X is negative and therefore X is
an obtuse angle.
skerphoek / acute angle = 80,4059...°
∠X = 180° − 80,4059...°
= 99,59°
Antwoord/Answer 3.11
Antwoord/Answer 3.12
DF2 = DE2 + EF2 − 2 . DE . EF . cos E
q2 = p2 + r2 − 2 . p . r . cos X
DE2 + EF2 − DF2
p2 + r2 − q2
cos E = ───────────────
cos X = ───────────
2 . DE . EF
2 . p . r
102 + 52 − 142
4,32 + 42 − 62
= ─────────────
= ─────────────
2 . 10 . 5
2 . 4,3 . 4
= −0,71
= −0,043895...
skerphoek / acute angle = 44,76508....°
skerphoek / acute angle = 87,4841....°
∠E = 180° − 44,76508...°
∠Q = 180° − 87,4841...°
= 135,23°
= 92,52°
Antwoord/Answer 4.1
Antwoord/Answer 4.2
Opp. / Area ΔABC = ½ . AC . BC . sin C
Opp. / Area ΔDEF = ½ . DE . DF . sin D
= ½ . 5 . 8 . sin 35°
= ½ . 18 . 24 . sin 76°
= 11,472 cm2
= 209,584 cm2
Antwoord/Answer 4.3
Antwoord/Answer 4.4
Opp. / Area ΔFGH = ½ . h . f . sin G
Opp. / Area ΔKLM = ½ . m . l . sin K
= ½ . 14,3 . 18,6 . sin 141°
= ½ . 3,1 . 5,2 . sin 123°
= 83,693 cm2
= 6,760 cm2
Antwoord/Answer 4.5
Antwoord/Answer 4.6
Opp. / Area ΔPQR = ½ . p . r . sin Q
Opp. / Area ΔXYZ = ½ . x . y . sin Z
= ½ . 8 . 5 . sin 72°
= ½ . 21,3 . 15,8 . sin 48°
= 19,021 cm2
= 125,049 cm2
Antwoord/Answer 4.7
Antwoord/Answer 4.8
Gebruik die sinus-reël om 'n tweede sy,
Use the sine rule to calculate a second side,
bv. AC, b te bereken.
e.g. DE, f.
Bereken die derde hoek, ∠A
Calculate the third angle, ∠E
∠A = 180° − (∠B + ∠C)
∠E = 180° − (∠D + ∠F)
= 180° − (65° + 43°)
= 180° − (38° + 75°)
= 72°
= 67°
b
e
a
d
────── = ──────
────── = ──────
sin B
sin E
sin B
sin D
a sin B
d sin E
b = ───────
e = ───────
sin A
sin D
13 sin 65°
16 sin 67°
= ───────
= ───────
sin 72°
sin 38°
Moenie b bereken nie - doen dit in die
Do not calculate e - do that in the
finale berekening.
final calculation.
Bereken nou die oppervlakte.
Now calculate the area.
Opp. / Area = ½ . b . a . sin C
Opp. / Area = ½ . e . d . sin F
1 . 13 sin 65° . 13 . sin 43°
1 . 16 sin 67° . 16 . sin 75°
= ──────────────────
= ──────────────────
2 . sin 72°
2 . sin 38°
= 54,917 cm2
= 184,858 cm2
Antwoord/Answer 4.9
Antwoord/Answer 4.10
Gebruik die sinus-reël om 'n tweede sy,
Use the sine rule to calculate a second side,
bv. h te bereken.
e.g. k.
Bereken die derde hoek, ∠F
Calculate the third angle, ∠M
∠F = 180° − (∠G + ∠H)
∠M = 180° − (∠K + ∠L)
= 180° − (32° + 125°)
= 180° − (23° + 138°)
= 23°
= 19°
h
f
k
m
────── = ──────
────── = ──────
sin H
sin F
sin K
sin M
f sin H
m sin K
h = ───────
k = ───────
sin F
sin M
9,5 sin 125°
28,3 sin 23°
= ───────
= ───────
sin 23°
sin 19°
Moenie h bereken nie - doen dit in die
Do not calculate k - do that in the
finale berekening.
final calculation.
Bereken nou die oppervlakte.
Now calculate the area.
Opp. / Area = ½ . h . f . sin G
Opp. / Area = ½ . k . m . sin K
1 . 9,5 sin 125° . 9,5 . sin 32°
1 . 28,3 sin 23° . 28,3 . sin 138°
= ─────────────────────
= ──────────────────
2 . sin 23°
2 . sin 19°
= 50,132 cm2
= 321,581 cm2
Antwoord/Answer 4.11
Antwoord/Answer 4.12
Gebruik die oppervlakte-reël om die
Use the area rule to calculate the requested
gevraagde sy te bereken.
side.
Oppv. / Area = ½ . q . r . sin P
Oppv. / Area = ½ . x . y . sin Z
685 = ½ . q . 54 . sin 61°
9,5 = ½ . x . 6 . sin 27°
2 . 685
2 . 9,5
q = ─────────
x = ─────────
54 . sin 61°
6 . sin 27°
= 29
= 6,975
Antwoord/Answer 4.13
Antwoord/Answer 4.14
Gebruik die oppervlakte-reël om die
Use the area rule to calculate the requested
gevraagde sy te bereken.
side.
Oppv. / Area = ½ . a . c . sin B
Oppv. / Area = ½ . d . e . sin F
30,12 = ½ . 9 . c . sin 123,2°
168,53 = ½ . d . 28 . sin 28°
2 . 30,12
2 . 168,53,5
c = ─────────
d = ─────────
9 . sin 123,2°
28 . sin 28°
= 8,0
= 25,64
Antwoord/Answer 4.15
Antwoord/Answer 4.16
Gebruik die oppervlakte-reël om die
Use the area rule to calculate the requested
gevraagde hoek te bereken.
angle.
Oppv. / Area = ½ . q . r . sin P
Oppv. / Area = ½ . f . h . sin G
19 = ½ . 8 . 5 . sin P
83,7 = ½ . 14 . 19 . sin G
2 . 19
2 . 83,7
sin P = ──────
sin G = ───────
8 . 5
14 . 19
= 0,95
= 0,6293...
P = 71,8°
G = 39°
Antwoord/Answer 5.1
Antwoord/Answer 5.2
q
r
───── = ─────
g2 = f2 + h2 − 2 . f . h . cos x
sin x
sin y
r sin y
g = ──────
sin x
Antwoord/Answer 5.3
Antwoord/Answer 5.4
f
g
f2 = g2 + h2 − 2 . g . h . cos (180° − (x + y))
────────────── = ─────
sin (180° − (x + y))
sin x
= g2 + h2 − 2 . g . h . (− cos (x + y))
g sin (x + y)
= g2 + h2 + 2 . g . h . cos (x + y)
f = ──────────
sin x
Antwoord/Answer 5.5
Antwoord/Answer 5.6
A = ½ . f . h . sin x
A = ½ . f . h . sin (180° − (x + y))
= ½ . f . h . sin (x + y)
Antwoord/Answer 5.7
Antwoord/Answer 5.8
A = ½ . f . h . sin x
A = ½ . f . h . sin (180° − (x + y))
2 A
= ½ . f . h . sin (x + y)
f = ──────
h sin x
2 A
g = ────────
h sin (x + y)
Antwoord/Answer 5.9
Antwoord/Answer 5.10
sin x
sin y
────── = ─────
h2 = f2 + g2 − 2 . f . g . cos y
g
h
g sin y
f2 + g2 − h2
sin x = ──────
cos y = ─────────
h
2fg
Antwoord/Answer 6.
Gebruik die cos-reël maar omdat ons net iets
Use the cosine rule, but seeing that something
van ∠L weet, begin ons met KM2
of ∠L only is known, we start with KM2
KM2 = KL2 + LM2 − 2 . KL . LM . cos L
72 = 52 + LM2 − 2 . 5 . LM . ½
LM2 − 5 . LM + (25 − 49) = 0 . . . herrangskik / rearrange
LM2 − 5 . LM − 24 = 0
(LM − 8)(LM + 3) = 0
LM = 8 OF / OR LM = − 3
LM = 8 lengte is positief / length is positive
Antwoord/Answer 7.
a . sin β
Beskou / Consider : MR = ──────────────
cos α . cos (θ − β)
a en α is elemente van ΔPQR terwyl β in
a and α are elements of ΔPQR while β is in
ΔPMR is en θ deel is van ∠PRN
ΔPMR and θ is part of ∠PRN
Skryf MR in terme van PR, sin ∠MPR
Express MR in terms of PR, sin ∠MPR
en sin ∠M
and sin ∠M
Bepaal eers die grootte van ∠M
First determine the size of ∠M
∠PRM = 90° − θ
∠M = 180° − (∠MPR + PRM)
= 180° − (β + (90° − θ))
= 180° − β − 90° + θ
= 90° + θ − β
Bepaal PR. / Determine PR.
QR
QR
In ΔPQR : cos α = ────
sodat / so that PR = ─────
PR
cos α
MR
PR
In ΔPMR : ────── = ─────
sin MPR
sin M
PR . sin MPR
PR . sin β
MR = ────────── = ─────────────
sin β
sin (90° + (θ − β))
QR
sin β
= ────── × ─────────
cos α
cos (θ − β)
a . sin β
= ──────────────
cos α . cos (θ − β)
Antwoord/Answer 8.
8.1
Bereken AD in ΔAED. Bereken eers die groottte van elke hoek in ΔAED. /
Calculate AD in ΔAED. First calculate the size of each angle in ΔAED.
∠AED = 90° + β, ∠ADE = 90° − α,
∠EAD = 180° − (∠AED + ∠ADE)
= 180° − ((90° + β) + (90° − α))
= (α − β)
AD
DE
──────── = ───────
sin ∠AED
sin ∠DAE
DE . sin (90° + β)
AD = ──────────────
sin (α − β)
a . cos β
= ─────────
sin (α − β)
8.2
Bereken AC in ΔACD. / Calculate AD in ΔACD.
AC
AD
──────── = ───────
sin ∠ADC
sin ∠ACD
AD . sin α
AC = ────────────
sin (180° − θ)
a . cos β
sin α
= ──────── × ──────
sin (α − β)
sin θ
a . cos β . sin α
= ─────────────
sin (α − β) . sin θ
Antwoord/Answer 9.
sin R
sin P
──────── = ───────
PQ
RQ
PQ . sin P
sin R = ────────
RQ
2x . sin (90° − θ)
sin α = ─────────────
x
─────
sin θ
2x . cos θ
sin θ
= ──────── × ─────
1
x
= 2 . cos θ . sin θ
Antwoord/Answer 10.
DB
In ΔDBC : ──── = tan DBC
BC
h
──── = tan α
BC
h
BC = ────
tan α
AB
In ΔABC : ──── = tan ABC
BC
AB
k + h
BC = ──── = ────
tan β
tan β
k + h
h
──── = ────
tan β
tan α
k . tan α + h . tan α = h . tan β
k . tan α = h . tan β − h . tan α
= h(tan β − tan α)
Antwoord/Answer 11.
∠K = 180° − (∠L + ∠M)
= 180° − (α + β)
LM
KL
──── = ────
sin K
sin M
KL . sin (180° − (α + β))
LM = ─────────────────
sin β
x
sin (α + β)
= ───── × ─────────
sin θ
sin β
x . sin (α + β)
= ──────────
sin θ . sin β
Antwoord/Answer 12.
DB
12.1
In ΔDBC : ───── = tan ∠DCB
BC
DB = BC . tan ∠DCB
= x . tan α
12.2
In ΔABD : AD2 = AB2 + DB2 − 2 . AB . DB . cos ∠ABD
= x2 + (x . tan α)2 − 2 . x . x . tan α . cos β
= x2 + x2 . tan2 α − 2 . x2 . tan α . cos β
= x2(1 + tan2 α − 2 . tan α . cos β)
Antwoord/Answer 13.
13.1
In ΔLMN : LM2 = LN2 + NM2 − 2 . LN . NM . cos ∠N
= 102 + 62 − 2 . 10 . 6 . cos 120°
= 136 + 60
= 196
LM = 14 cm
sin ∠K
sin ∠KML
13.2
In ΔKLM : ────── = ──────
LM
KL
LM . sin ∠KML
sin ∠K = ───────────
KL
14 . sin 38,1°
= ───────────
10
= 0,8638....
∠K = 59,75°
13.3
In ΔKLM : ∠KLM = 180° − (∠K + ∠KML)
= 180° − (59,75° + 38,1°)
= 82,15°
Oppervlakte / Area KLNM = Oppervlakte / Area ΔKLM + Oppervlakte / Area ΔLNM
= ½ . KL . LM . sin ∠KLM + ½ . LN . NM . sin ∠N cm2
= ½ . 10 . 14 . sin 82,15° + ½ . 10 . 6 . sin 120° cm2
= 95,325 cm2