Transcript Slide 1

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 2 :
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Trigonometry
Statistics
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
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UNIT 2 :
Trigonometry
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TRIGONOMETRY : Question 1
Find the area of the following triangle to the nearest cm2.
B
40cm
A
25°
C
50cm
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TRIGONOMETRY : Question 1
Find the area of the following triangle to the nearest cm2.
B
40cm
A
25°
C
50cm
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1. For area of a
2. Always
triangle
questions
remember
to
with
an angle
round use:
your
present
answer
if the
1
Aquestions
 2 ab sinasks
C
you to.
TRIGONOMETRY : Question 1
Find the area of the following triangle to the nearest cm2.
B
40cm
A
25°
C
50cm
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Area of  = 423cm2
Question 1
1. For area of triangles questions
where an angle is present use:
Find the area of the
following triangle
Area of  = ½ bcsinA°
= 50 x 40 x sin25°  2
to the nearest cm2.
B
= 422.61…
40cm
A
2. Remember to round if asked to.
25°
C
50cm
= 423 to nearest unit
Area of  = 423cm2
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Markers Comments
Check formulae list for:
1. For area of triangles questions
where an angle is present use:
1
Area of triangle = 2 absinC
Area of  = ½ bcsinA°
= 50 x 40 x sin25°  2
(Note: 2 sides and the included
angle)
= 422.61…
Relate formula to labels being
used.
B
2. Remember to round if asked to.
40cm
c
= 423 to nearest unit
Area of  =
423cm2
A
25°
b
50cm
C
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TRIGONOMETRY : Question 1B
Find the area of the following triangle.
K
6.5m
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L
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150°
8m
M
TRIGONOMETRY : Question 1B
Find the area of the following triangle.
K
6.5m
L
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150°
8m
1. For area of a
2. Always
triangle
questions
remember
to
with
an angle
round use:
your
present
answer
if the
Aquestions
 12 ab sinasks
C
you to.
M
TRIGONOMETRY : Question 1B
Find the area of the following triangle.
K
6.5m
L
150°
8m
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Area of  = 13m2
M
Question 1B
1. For area of triangles questions
where an angle is present use:
Find the area of the
Area of  = ½ kmsinL°
= 8 x 6.5 x sin150°  2
following triangle.
K
= 13
6.5m
L
150
°
Area of  = 13m2
8m
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M
Markers Comments
Check formulae list for:
1. For area of triangles questions
where an angle is present use:
Area of  = ½ kmsinL°
= 8 x 6.5 x sin150°  2
absinC
(Note: 2 sides and the included
angle)
Relate formula to labels being
used.
= 13
Area of  =
Area of triangle =
13m2
K
6.5m m
L
150
°
k
8m
M
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TRIGONOMETRY : Question 2
Two helicopters leave an air base. The first flies on a bearing of 340°
at 160km/hr. The second flies due east at 200km/hr.
How far apart will they be after 21/2 hours?
Answer
N
to the nearest 10km.
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340°
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TRIGONOMETRY : Question 2
Two helicopters leave an air base. The first flies on a bearing of 340°
at 160km/hr. The second flies due east at 200km/hr.
How far apart will they be after 21/2 hours?
Answer
N
to the nearest 10km.
What would you like to do now?
Reveal answer only
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1. Identify
what you
2.
Calculate
as
3.
Make
a
sketch
to
340°
5.need
Substitute
known
4. Identify
which
trig
to findofand
many
the
clarify
matters.
values,
remembering
to
rule to use:
the missing
information
you
angles
as
use
brackets
as
Two
sides
+
two
angles
have topossible.
help you.
appropriate.
= sine rule
Go to Comments
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Three sides + one angle
= cosine rule
TRIGONOMETRY : Question 2
Two helicopters leave an air base. The first flies on a bearing of 340°
at 160km/hr. The second flies due east at 200km/hr.
How far apart will they be after 21/2 hours?
Answer
N
to the nearest 10km.
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340°
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Distance is 740km
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Question 2
1. Identify what needs to be found.
160km/hr.
2. Need distances travelled and angle
between flight paths.
N
340°
How far apart will
they be after
21/2 hours?
20°
90°
d1 = speed1 x time = 160 x 2.5 = 400km
d2 = speed2 x time = 200 x 2.5 = 500km
340° clockwise = 20° anti-clockwise
Full angle = 20° + 90° = 110°
3. Sketch triangle.
200km/hr
b
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110°
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a
400km
A
500km
c
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Question 2
4. Apply Cosine rule.
160km/hr.
a2 = b2 + c2 – (2bccosA°)
N
340°
How far apart will
they be after
21/2 hours?
20°
90°
200km/hr
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5. Substitute known values and
remember to use brackets.
= 4002 + 5002 – (2 x 400 x 500 x cos110°)
= 546808.05..
a = 546808.05..
= 739.46....
6. Remember to round answer if
asked to.
= 740
Distance is 740km
Markers Comments
1. Identify what needs to be found.
2. Need distances travelled and
angle between flight paths.
Note:
Bearings are measured
clockwise from N.
d1 = speed1 x time = 160 x 2.5 = 400km
d2 = speed2 x time = 200 x 2.5 = 500km
340° clockwise = 20° anti-clockwise
Full angle = 20° + 90° = 110°
3. Sketch triangle.
400km
110°
500km
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Markers Comments
4. Apply Cosine rule.
a2 = b2 + c2 – (2bc cosA°)
Check formulae list for the
cosine rule:
a2 = b2 + c2 – 2bc cosA
5. Substitute known values and
remember to use brackets.
(2 sides and the included
angle)
= 4002 + 5002 – (2 x 400 x 500 x cos110°)
Relate to variables used
= 546808.05..
a2 = b2 + c2 – 2bc cosA
a = 546808.05..
= 739.46....
6. Remember to round answer if
asked to.
= 740
Distance is 740km
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TRIGONOMETRY : Question 2B
Two ships sail from a port. The first travels for two hours on a
bearing of 195° at a speed of 18mph. The second travels south-east
for three hours at a speed of 15mph.
N
How far apart will they be?
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195°
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SE
TRIGONOMETRY : Question 2B
Two ships sail from a port. The first travels for two hours on a
bearing of 195° at a speed of 18mph. The second travels south-east
for three hours at a speed of 15mph.
N
How far apart will they be?
What would you like to do now?
Reveal answer only
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2. Calculate
1. Identify
what you as
3. Make a sketch
toof the
many
need
to
find
and
4. Identify
which
trig as
clarify
matters.
missing
angles
therule
information
you
to use:
possible.
have
to
help
you.
Two sides
+ two angles
5. Substitute
known
= sine remembering
rule
values,
to
use brackets as
195° + one angle
Three sides
SE
appropriate.
= cosine rule
TRIGONOMETRY : Question 2B
Two ships sail from a port. The first travels for two hours on a
bearing of 195° at a speed of 18mph. The second travels south-east
for three hours at a speed of 15mph.
N
How far apart will they be?
Distance is 41.2 miles
What would you like to do now?
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195°
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SE
Question 2B
How far apart will
they be?
1. Identify what needs to be found.
N
2. Need distances travelled and angle
between paths.
1350
3hr@15mph
2hr@18mph
195°
d1 = speed1 x time = 18 x 2 = 36miles
600
SE
d2 = speed2 x time = 15 x 3 = 45miles
NB: SE = 135°
Angle = 195° - 135° = 60°
3. Sketch triangle.
A
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60°
45miles
c
36miles
b
a
Question 2B
How far apart will
they be?
4. Apply Cosine rule.
N
a2 = b2 + c2 – (2bc cosA°)
5. Substitute known values and
remember to use brackets.
1350
3hr@15mph
2hr@18mph
= 362 + 452 – (2 x 36 x 45 x cos60°)
= 1701
195°
600
SE
a = 1701
= 41.243....
= 41.2
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Distance is 41.2 miles
Comments
1. Identify what needs to be found.
2. Need distances travelled and
angle between flight paths.
Note:
Bearings are measured
clockwise from N.
d1 = speed1 x time = 18 x 2 = 36miles
d2 = speed2 x time = 15 x 3 = 45miles
NB: SE = 135°
Angle = 195° - 135° = 60°
3. Sketch triangle.
A
60°
45miles
c
36miles
b
a
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Comments
4. Apply Cosine rule.
a2 = b2 + c2 – (2bc cosA°)
5. Substitute known values and
remember to use brackets.
= 362 + 452 – (2 x 36 x 45 x cos60°)
= 1701
Check formulae list for the
cosine rule:
a2 = b2 + c2 – 2bc cosA
(2 sides and the included
angle)
Relate to variables used
a2 = b2 + c2 – 2bc cosA
a = 1701
= 41.243....
= 41.2
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Distance is 41.2 miles
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TRIGONOMETRY : Question 3
In the kite shown below PQ = 10cm, QR = 15cm &
diagonal PR = 22cm.
(a) Find the size of angle QPR.
(b) Hence find the area of the kite to the nearest square unit.
P
10cm
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Q
S
22cm
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15cm
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R
TRIGONOMETRY : Question 3
In the kite shown below PQ = 10cm, QR = 15cm &
diagonal PR = 22cm.
(a) Find the size of angle QPR.
(b) Hence find the area of the kite to the nearest square unit.
P
1. Identify which trig
3. For10cm
area of a
rule to use:
What would you like to do now?
triangle
with an
4.
Always
Two sides + two angles
angle
present to
use:
remember
2.
Substitute
known
Q
= sine rule
Reveal answer only
round
your 22cmto
values,
remembering
1 brackets
answer
if the
use
A

sin
C as
Three sides + one
2 abangle
Go to full solution
questions
appropriate.
= cosine
rule asks
15cm
you to.
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R
S
TRIGONOMETRY : Question 3
In the kite shown below PQ = 10cm, QR = 15cm &
diagonal PR = 22cm.
(a) Find the size of angle QPR. angleQPR = 35.3°
(b) Hence find the area of the kite to the nearest square unit.
P
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= 127cm2
10cm
Q
S
22cm
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15cm
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R
Question 3
P
1. To find angle when you have 3 sides
use 2nd version of cosine rule:
10cm
Q
S
22cm
cosP
r2 + q2 - p2
= 2rq
= 102 + 222 - 152
2 x 10 x 22
15cm
= (102 + 222 - 152)  (2 x 10 x 22)
R
(a) Find the size of angle QPR
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= 0.8159…
2. Remember to use inverse function
to find angle.
angleQPR = cos-1(0.8159..) = 35.3°
Question 3
P
1. Kite = 2 identical triangles. For area
of triangles where an angle is present
use:
10cm
Q
S
22cm
(b) Area QPR = ½ qrsinP°
= 10 x 22 x sin35.3°  2
15cm
= 63.56..cm2
2. Remember to double this and
round answer.
R
(b) Hence find the area of the kite
to the nearest square unit
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Area of kite = 2 x 63.56..
= 127.12..cm2
= 127cm2
Markers Comments
1. To find angle when you have 3
sides use 2nd version of cosine rule:
cosP
r2
q2
+
= p22rq
Check the formulae list for the
second form of the cosine rule:
2 + c 2 – a2
b
cosA =
2bc
( 3 sides)
= 102 + 222 - 152
2 x 10 x 22
= (102 + 222 - 152)  (2 x 10 x 22)
= 0.8159…
2. Remember to use inverse function
to find angle.
angleQPR = cos-1(0.8159..) = 35.3°
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Markers Comments
Relate to variables used:
1. To find angle when you have 3
sides use 2nd version of cosine rule:
cosP
cosP =
r 2 + q2 = p22rq
=
= 102 + 222 - 152
2 x 10 x 22
= (102 + 222 - 152)  (2 x 10 x 22)
q2 + r2 – p2
2qr
222 + 102 – 152
2x22x10
P
10cm
r
Q
q 22cm
= 0.8159…
2. Remember to use inverse function
to find angle.
angleQPR = cos-1(0.8159..) = 35.3°
15cm p
R
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Markers Comments
Note:
1. To find angle when you have 3
sides use 2nd version of cosine rule:
cosP
r 2 + q2 = p22rq
= 102 + 222 - 152
2 x 10 x 22
When keying in to calculator
work out the top line and the
bottom line before dividing or
use brackets.
= (102 + 222 - 152)  (2 x 10 x 22)
= 0.8159…
2. Remember to use inverse function
to find angle.
angleQPR = cos-1(0.8159..) = 35.3°
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TRIGONOMETRY : Question 3B
The sides of a rhombus are each 15cm while the main
diagonal is 25cm
Find the size of angle EFH and hence find the area of the
rhombus to the nearest square unit.
Get hint
E
15cm
Reveal answer only
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15cm
25cm
H
F
Go to Comments
15cm
15cm
Go to Trigonometry Menu
G
EXIT
TRIGONOMETRY : Question 3B
The sides of a rhombus are each 15cm while the main
diagonal is 25cm
Find the size of angle EFH and hence find the area of the
rhombus to the nearest square unit.
What would you like to do now?
E
Reveal answer only
15cm
1. Identify which trig 15cm
Go to full solution
H
Go to Comments
Go to Trigonometry Menu
EXIT
3. For area of a
rule to use:
triangle
with an
4. Always
Two sides + 25cm
two
angles
angle
present to
use:
2. remember
Substitute
known
= sine
rule
round
your
values,
remembering
to
1
if the
use
brackets
Aanswer
 one
abangle
sin
C as
Three
sides
+
2
15cm
15cm
questions
appropriate.
= cosine
rule asks
you to.
G
F
TRIGONOMETRY : Question 3B
The sides of a rhombus are each 15cm while the main
diagonal is 25cm
angleEFH = 33.6°
Find the size of angle EFH and hence find the area of the
rhombus to the nearest square unit.
= 415cm2
E
What would you like to do now?
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15cm
15cm
25cm
H
F
Go to Comments
15cm
15cm
Go to Trigonometry Menu
G
EXIT
E
Question 3B
15
15
25
H
1. To find angle when you have 3 sides
use 2nd version of cosine rule:
F
e 2 + h2 - f 2
(a) cosF =
2eh
= 152 + 252 - 152
2 x 25 x 15
15
15
G
(a) Find the size of angle EFH
= (152 + 252 - 152)  (2 x 25 x 15)
= 0.8333…
2. Remember to use inverse function
to find angle.
angleEFH = cos-1(0.8333..) = 33.6°
Continue Solution
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E
Question 3B
15
15
25
H
33.60
15
15
G
(b) Hence find the area
to the nearest square unit
1. Rhombus = 2 identical triangles. For
area of triangles where an angle is
present use:
(b) Area EFH = ½ ehsinF°
F
= 25 x 15 x sin33.6°  2
= 207.52....cm2
2. Remember to double this and
round answer.
Area of kite = 2 x 207.52......
= 415.04..cm2
= 415cm2
Continue Solution
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Markers Comments
1. To find angle when you have 3
sides use 2nd version of cosine rule:
(a) cosF =
e2
h2
+
2eh
f2
Check the formulae list for the
second form of the cosine rule:
2 + c 2 – a2
b
cosA =
2bc
( 3 sides)
= 152 + 252 - 152
2 x 25 x 15
= (152 + 252 - 152)  (2 x 25 x 15)
= 0.8333…
2. Remember to use inverse function
to find angle.
angleEFH = cos-1(0.8333..) = 33.6°
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Markers Comments
Relate to variables used:
1. To find angle when you have 3
sides use 2nd version of cosine rule:
cosF =
e2 + h 2 - f 2
(a) cosF =
2eh
= 0.8333…
2. Remember to use inverse function H
to find angle.
222 + 102 – 152
2x22x10
=
= 152 + 252 - 152
2 x 25 x 15
= (152 + 252 - 152)  (2 x 25 x 15)
e 2 + h2 – f 2
2eh
E
15cm
h
f
15cm
e
25cm
F
angleEFH = cos-1(0.8333..) = 33.6°
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Markers Comments
Check formulae list for:
1. Rhombus = 2 identical triangles.
For area of triangles where an angle
is present use:
(b) Area EFH = ½ ehsinF°
= 25 x 15 x sin33.6°  2
= 207.52....cm2
2. Remember to double this and
round answer.
1
Area of triangle = 2 absinC
(Note: 2 sides and the included
angle)
Relate formula to labels being
used.
E
15cm
h
f
Area of kite = 2 x 207.52......
= 415.04..cm2
= 415cm2
15cm
e
H
25cm
F
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TRIGONOMETRY : Question 4
In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm
and UV = 10cm.
Find the perimeter to one decimal place.
T
Get hint
105°
5.9cm
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35°
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U
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V
10cm
TRIGONOMETRY : Question 4
In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm
and UV = 10cm.
Find the perimeter to one decimal place.
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35°
U
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T
3. Identify which trig
rule105°
to use:
5. Always
Two sides + two angles
5.9cm
1. For perimeter
remember
to
=2.sine
rule
Calculate
need
all three
round
your
unknown
sides.ifSo
answer
themust find
Three sides
+ one angle
angles.
TU.
questions
asks
4. Substitute
known
= cosine rule
you to.
values,
remembering to
use
brackets as
10cm
appropriate.
V
TRIGONOMETRY : Question 4
In triangle TUV, angle U = 35°, angle T = 105°, TV = 5.9cm
and UV = 10cm.
= 22.6cm
Find the perimeter to one decimal place.
T
105°
5.9cm
What would you like to do now?
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35°
U
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V
10cm
T
Question 4
1. Perimeter requires all three sides.
105°
So we need to find TU .
5.9c
m
U
35
°
2. Whether you use Sine or Cosine
rule need angle V.
400
Angle V = 180° - 35° - 105° = 40°
V
10cm
Find the perimeter to one
decimal place.
3. If we use Sine rule:
v
=
sinV°
t
sinT°
4. Substitute known values:
Continue Solution
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v
sin40°
=
10
sin105°
T
Question 4
4. Substitute known values:
105°
v
= 10
sin40°
sin105°
5.9c
m
5. Cross multiply:
U
35
°
400
V
v x sin105°
= 10 x sin40°
10cm
v = 10 x sin40°  sin105°
Find the perimeter to
= 6.654…
one decimal place.
= 6.7cm
6. Answer the question:
Continue Solution
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Perim of  = (6.7 + 10 + 5.9)cm
= 22.6cm
Markers Comments
1. Perimeter requires all three sides.
So we need to find TU .
2. Whether you use Sine or Cosine
rule need angle V.
Since we can pair off two angles
with the opposite sides
Sine Rule
Refer to the Formulae List :
a
b
c
=
=
Sine A
Sine B
Sine C
T
Angle V = 180° - 35° - 105° = 40°
3. If we use Sine rule:
v =
sinV°
105°
t
sinT°
4. Substitute known values:
v
= 10
sin40°
sin105°
5.9cm
v
U
35°
t
10cm
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V
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Markers Comments
4. Substitute known values:
v
= 10
sin40°
sin105°
Go straight to values :
10
v
=
Sine 105˚
Sine 40˚
5. Cross multiply:
v x sin105°
v = 10 x sin40°  sin105°
= 6.654…
= 6.7cm
6. Answer the question:
Perim of  = (6.7 + 10 + 5.9)cm
= 22.6cm
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TRIGONOMETRY : Question 5
Lighthouse C is on a bearing of 050° from lighthouse A
and northwest of lighthouse B. Lighthouse B is 12km
due east of lighthouse A. A ship(S) is sailing directly from
lighthouse B to lighthouse A. How close does it come to
lighthouse C?
Hints
C
N
Answer only
Full solution
Comments
50°
Trig Menu
A
EXIT
12km
B
S
TRIGONOMETRY : Question 5
Lighthouse C is on a bearing of 050° from lighthouse A
and northwest of lighthouse B. Lighthouse B is 12km
due east of lighthouse A. A ship(S) is sailing directly from
lighthouse B to lighthouse A. How close does it come to
lighthouse C?
C
What would you like to do now?
2. Identify which trig
1.4.Calculate
missing
N
rule to
use:
Closest
Answer only
Full solution
Comments
50°
Trig Menu
A
EXIT
angles.
Two sides
distance
= + two angles
(NB.
North= West
= 3150)
sine rule
perpendicular
distance.
3. Substitute known
Three
+ one angle
Create
asides
rightvalues,
remembering
to
= cosine
ruleas
angleduse
triangle
brackets
and use SOH
CAH
appropriate.
TOA.
12km
S
B
TRIGONOMETRY : Question 5
Lighthouse C is on a bearing of 050° from lighthouse A
and northwest of lighthouse B. Lighthouse B is 12km
due east of lighthouse A. A ship(S) is sailing directly from
lighthouse B to lighthouse A. How close does it come to
lighthouse C?
C
Closest distance is 5.48km
What would you like to N
do now?
Full solution
Comments
50°
Trig Menu
A
EXIT
12km
B
S
Question 5
C
1. Calculate missing angles.
NW = 3150
AngleA = 90° - 50° = 40°
N
& angleB = 315° - 270° = 45°
950
So angle C = 180° - 40° - 45° = 95°
A
5
0
°
2. Draw a sketch
400
450
C
b
12
km
B
A
95°
40°
45°
12km
S
3. If we use Sine rule:
Continue Solution
Comments
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b =
sinB°
c
sinC°
B
Question 5
4. Substitute known values:
C
N
b
= 12
sin45°
sin95°
950
5. Cross multiply:
A
5
0
°
400
b x sin95°
450
= 12 x sin45°
b = 12 x sin45°  sin95°
12
km
B
S
Continue Solution
Comments
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= 8.5176.…
= 8.52km
Question 5
6. Closest point is perpendicular so
sketch right angled triangle:
C
N
C
950
8.52km
= 8.52km
A
A
5
0
°
400
a
40°
S
450
7. Now using SOHCAH TOA:
12
km
B
sin40°
=
a
8.52
S
a = 8.52 x sin40°
Continue Solution
Comments
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= 5.476….
= 5.48
Closest distance is 5.48km
Markers Comments
1. Calculate missing angles.
AngleA = 90° - 50° = 40°
NW = 3150
Since we can pair off two angles
with the opposite sides
Sine Rule
& angleB = 315° - 270° = 45°
So angle C = 180° - 40° - 45° = 95°
2. Draw a sketch
Refer to the Formulae List :
a
b
c
Sine A = Sine B = Sine C
C
95°
40
45
° 12km °
A
B
3. If we use Sine rule:
b
sinB°
=
c
sinC°
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Markers Comments
6. Closest point is perpendicular so
sketch right angled triangle:
Note: the shortest distance from
a point to a line is the
perpendicular distance.
C
8.52km
A
C
a
40°
S
A
B
7. Now using SOHCAH TOA:
sin40°
=
a
8.52
a = 8.52 x sin40°
= 5.476….
= 5.48
Closest distance is 5.48km
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TRIGONOMETRY : Question 6
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill tower is 2.9°
while from the deck of ship B it is 3.8°.
The ships are 400m apart.
D
How far from the platform is each vessel ?
Hints
Answer only
2.9°
Full solution
3.8°
400m
O
Comments
A
Trig Menu
EXIT
B
TRIGONOMETRY : Question 6
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill tower is 2.9°
while from the deck of ship B it is 3.8°.
The ships are 400m apart.
D
How far from the platform is each vessel ?
What would you like to do now?
Answer only
2.9°
Full solution
400m
Comments
A
Trig Menu
EXIT
4. Then use SOH
2. Identify
trig
1.
Calculate
missing
CAH
TOA to which
find
rulefrom
to angles.
use
to
find
length
rig
to
3. Substitute
known
common
side to both
first
ship.
Second
values, remembering
to
triangles:
ship
is then 400
m
3.8°
use
brackets
as
Two further.
sides + two angles
appropriate.
= sine rule
O
Three sides + one angle
= cosine rule
B
TRIGONOMETRY : Question 6
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill tower is 2.9°
while from the deck of ship B it is 3.8°.
The ships are 400m apart.
D
How far from the platform is each vessel ?
What would you like to do now?
Try another like this
2.9°
Full solution
3.8°
400m
O
Comments
A
Trig Menu
EXIT
B
ShipA is 1685m away
& ShipB is 1285m away
D
Question 6
1. Calculate missing angles.
Angle B = 180° - 3.8° = 176.2°
Angle D = 180° - 2.9° - 176.2° = 0.9°
0.9°
2. Draw a sketch
2.9
°
400
m
D
3.8
°
0.9°
O
A
2.9° 176.2°
400m
A
B
Continue Solution
Comments
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B
3. BD is common link to both
triangles so find it using sine rule.
a
=
sinA°
d
sinD°
D
Question 6
4. Substitute known values:
a
= 400
sin2.9°
sin0.9°
0.9°
5. Cross multiply:
2.9
°
400
m
3.8
°
a x sin0.9° = 400 x sin2.9°
a = 400 x sin2.9°  sin0.9°
O
= 1288m
A
B
Continue Solution
Comments
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to nearest metre
D
Question 6
6. Sketch right angled triangle:
D
1288m
0.9°
3.8°
B
d
2.9
°
400
m
3.8
°
7. Now using SOHCAH TOA:
cos3.8°
1285 m
A
B
Continue Solution
O
O
=
d
1288
d = 1288 x cos3.8°
= 1285m to nearest metre
Try another like this
OA is 1285m+400m = 1685m
Comments
ShipA is 1685m away
& ShipB is 1285m away
Trigonometry Menu
Back to Home
Markers Comments
Since we can pair off two angles
with the opposite sides
1. Calculate missing angles.
Angle B = 180° - 3.8° = 176.2°
Angle D = 180° - 2.9° - 176.2° = 0.9°
Sine Rule
Refer to the Formulae List :
2. Draw a sketch
D
a
b
c
=
=
Sine A
Sine B
Sine C
0.9°
A
2.9° 176.2°
400m
B
3. BD is common link to both
triangles so find it using sine rule.
a =
sinA°
d
sinD°
Next Comment
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Markers Comments
6. Sketch right angled triangle:
D
Since angle BOD is right - angled
use SOHCAHTOA in triangle BOD
1288m
3.8°
B
d
O
7. Now using SOHCAH TOA:
cos3.8°
=
d
1288
d = 1288 x cos3.8°
= 1285m to nearest metre
OA is 1285m+400m = 1685m
ShipA is 1685m away
& ShipB is 1285m away
Next Comment
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TRIGONOMETRY : Question 6B
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill
tower is 27° while from the deck of vessel B it is 35°.
The ships are 80m apart and the height of their decks
is 5m.
D
How high is the drill tower?
Hints
Answer only
27°
Full solution
35°
80m
O
Comments
A
Trig Menu
EXIT
B
TRIGONOMETRY : Question 6B
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill
tower is 27° while from the deck of vessel B it is 35°.
The ships are 80m apart and the height of their decks
is 5m.
D
How high is the drill tower?
What would you like to do now?
Answer only
27°
Full solution
80m
Comments
A
Trig Menu
EXIT
4. Then use SOH
2. Identify
trig
1.
Calculate
missing
CAH
TOA to which
find
ruleoftorig
use
to find
angles.
Height
from
3. Substitute
known
common
side to both
deck
level.
values, remembering
to
triangles:
Remember
decks
use
brackets
as
35°
Two
sides
+
two
angles
are
5m
above
sea
appropriate.
= sine rule
level.
O
Three sides + one angle
= cosine rule
B
TRIGONOMETRY : Question 6B
Two supply vessels are approaching an oil platform.
From the deck of ship A the angle of elevation of the drill
tower is 27° while from the deck of vessel B it is 35°.
The ships are 80m apart and the height of their decks
is 5m.
D
How high is the drill tower?
What would you like to do now?
27°
Full solution
35°
80m
O
Comments
A
Trig Menu
EXIT
B
Height of tower is 123m
D
Question 6B
1. Calculate missing angles.
Angle B = 180° - 35° = 145°
Angle D = 180° - 145° - 27° = 8°
8°
2. Draw a sketch
D
27
°
80m
8°
35
°
O
A
B
Continue Solution
Comments
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A
27°
80m
145°
B
3. BD is common link to both
triangles so find it using sine rule.
a
=
sinA°
d
sinD°
D
Question 6B
4. Substitute known values:
a
= 80
sin27°
sin8°
8°
5. Cross multiply:
a x sin8°
27
°
80m
35
°
a = 80 x sin27°  sin8°
O
A
B
Continue Solution
Comments
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= 80 x sin27°
= 261m
to nearest metre
D
Question 6B
6. Sketch right angled triangle:
D
261m
8°
b
27°
B
O
7. Now using SOHCAH TOA:
27
°
80m
35
°
sin27°
O
=
b
261
b = 261 x sin27°
A
B
Continue Solution
Comments
Trigonometry Menu
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= 118m to nearest metre
Height = OD + deck height
= 118 + 5 = 123m
Height of tower is 123m
Comments
1. Calculate missing angles.
Since we can pair off two angles
with the opposite sides
Angle B = 180° - 35° = 145°
Angle D = 180° - 145° - 27° = 8°
Sine Rule
Refer to the Formulae List :
2. Draw a sketch
D
a
b
c
=
=
Sine A
Sine B
Sine C
8°
A
27°
80m
145°
B
3. BD is common link to both
triangles so find it using sine rule.
a
=
sinA°
d
sinD°
Next Comment
Trigonometry Menu
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Markers Comments
6. Sketch right angled triangle:
D
261m
Since angle BOD is right - angled
use SOHCAHTOA in triangle BOD
b
27°
B
O
7. Now using SOHCAH TOA:
sin27°
=
b
261
b = 261 x sin27°
= 118m to nearest metre
Height = OD + deck height
Next Comment
= 118 + 5 = 123m
Trigonometry Menu
Height of tower is 123m
Back to Home
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
Statistics
Please choose a question to attempt from the following:
1
EXIT
2
3
4
Back to
Unit 2 Menu
5
STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
10
7
7
15
9
(a) Find the mean and standard deviation for this data.
(b) A similar experiment was conducted with a second
company. The results for this were….
mean = 12
and standard deviation = 1.88
How does the second company compare to the first?
Get hint
Reveal answer only
Go to full solution
EXIT
Go to Comments
Go to Statistics Menu
STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
(a)

SD  (b)



EXIT
10
15
9
Draw a7table 7
n
data deviation for this data.
2 comparing
Find the mean
and
standard
mean 
x  x  to mean.
 Then square
A similar experiment
was conducted with a second
n
Thevalues.
n 1
company.
results for this were….

When comparing

data sets= always
mean = 12 and standard deviation
1.88
make comment on
How does the second companythe
compare
to(which
the first?
average
is
bigger etc.) and the
spread of the data.



Reveal answer only
Go to Comments
Go to full solution
Go to Statistics Menu
What would you like to do now?
STATISTICS : Question 1
A taxi company was phoned each night of the week and the
response time in minutes of their cars were noted. They were ….
35
10
7
7
15
9
(a) Find the mean and standard deviation for this data.
(b) A similar experiment was conducted with a second
company. The results for this were….
mean = 12
and standard deviation = 1.88
How does the second company compare to the first?
(a) Mean = 8
SD  3.87
(b)The second company has a longer average response time.
The smaller standard deviation means their arrival time is
more predictable.
Go to Comments
What would you like to do now?
EXIT
Go to full solution
Go to Statistics Menu
Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
Begin Solution
Continue Solution
Comments
Statistics Menu
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1. Calculate mean.
(a)
Mean = (3+5+10+7+7+15+9)7 = 8
So x = 8
and no.pieces of data = n = 7
Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
x 8
Begin Solution
Continue Solution
Comments
Statistics Menu
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2. Draw table comparing data to mean.
x

3
5
10
7
7
15
9
-5
-3
2
-1
-1
7
1
25
9
4
1
1
49
1
 (x - x )2 =
90
xx


xx

2
Question 1
A taxi company was phoned each
night of the week and the
response time in minutes of their
cars were noted.
They were ….
3
5
10
7
7
15
9
(a) Find the mean and standard
deviation for this data.
Begin Solution
Continue Solution
Comments
Statistics Menu
Back to Home
3. Use formula to calculate standard
deviation.


xx


SD 
 n 1


 90 
SD   
 6 
SD  3.87

2





Just found!!
Question 1
(b) A similar experiment was
conducted with a second
company. The results for this
were….
mean = 12 and
standard deviation =