Transcript Trigonometry

Trigonometry
Describe what a bearing is.
Describe what a bearing is.
• A bearing is a measurement of an angle
from North in a clockwise direction.
Do you know how to write a
vector?
Do you know how to write a
vector?
• A vector is written like this

x 
 
y 
An example of a vector.
x  3 
   
y  4
3
-4
An example of a vector.
x  3 
   
y  4
3
-4
Vector
Problems
A taut guy wire to the top of a
transmission mast is anchored in
the same horizontal plane as the
foot of the mast. The wire is 50 m
long and makes an angle of 62
degrees with the horizontal. How
far is the lower end of the wire
from the foot of the mast?
Draw a diagram
50 m
62
x
• A taut guy wire to the top
of a transmission mast is
anchored in the same
horizontal plane as the
foot of the mast. The wire
is 50 m long and makes an
angle of 62 degrees with
the horizontal. How far is
the lower end of the wire
from the foot of the mast?
Solve using cosine
x
cos62 
50
x  23.5m
50 m
62
x

An aeroplane is flying at 300
km/hr. How far (a) north (b) East
of its starting-point is the
aeroplane after one hour if the
direction of flight is (i) North;
(ii)) N 40 degrees E; (iii) N 60
degrees E.
(i)
• An aeroplane is flying
at 300 km/hr. How far
(a) north (b) East of its
starting-point is the
aeroplane after one
hour if the direction of
flight is (i) North; (ii))
N 40 degrees E; (iii) N
60 degrees E.
(a) 300 km
(b) 0 km
(a) (ii)
• An aeroplane is flying
at 300 km/hr. How far
(a) north (b) East of its
starting-point is the
aeroplane after one
hour if the direction of
flight is (i) North; (ii))
N 40 degrees E; (iii) N
60 degrees E.
N 40 300
(a) (ii)
• An aeroplane is flying
at 300 km/hr. How far 
(a) north (b) East of its
starting-point is the
aeroplane after one
hour if the direction of
flight is (i) North; (ii))
N 40 degrees E; (iii) N
60 degrees E.
N
300
N  300 cos40  229.8km
cos40 
N 40 300
N
300
N  300 cos60 150km
(a) (iii)
• An aeroplane is flying
at 300 km/hr. How far
(a) north (b) East of its
starting-point is the
aeroplane after one
hour if the direction of
flight is (i) North; (ii))
N 40 degrees E; (iii) N
60 degrees E.
cos60 

N 60 300
(b) (ii)
• An aeroplane is flying
at 300 km/hr. How far 
(a) north (b) East of its
starting-point is the
aeroplane after one
hour if the direction of
flight is (i) North; (ii))
N 40 degrees E; (iii) N
60 degrees E.
E
300
N  300 sin40 192.8km
sin40 
E
40
300
(b) (iii)
• An aeroplane is flying
at 300 km/hr. How far 
(a) north (b) East of its
starting-point is the
aeroplane after one
hour if the direction of
flight is (i) North; (ii))
N 40 degrees E; (iii) N
60 degrees E.
E
300
N  300 sin60  259.8km
sin60 
E
60
300
From a point 42 m above waterlevel at low tide the angle of
depression of a buoy in the water
was 57 degrees. At high tide the
angle of depression was 55
degrees. Find the horizontal
distance of the buoy from the
viewer and the rise of the tide.
Draw a diagram
• From a point 42 m above
water-level at low tide the
angle of depression of a
buoy in the water was 57
degrees. At high tide the
angle of depression was
55 degrees. Find the
horizontal distance of the
buoy from the viewer and
the rise of the tide.
57
42
x
Draw a diagram
• From a point 42 m above
water-level at low tide the
angle of depression of a
buoy in the water was 57
degrees. At high tide the
angle of depression was
55 degrees. Find the
horizontal distance of the
buoy from the viewer and
the rise of the tide.
42
57
33
x
x
tan33
42
x  27.3m
• From a point 42 m above
water-level at low tide the

angle of depression
of a
buoy in the water was 57
degrees. At high tide the
angle of depression was
55 degrees. Find the
horizontal distance of the
buoy from the viewer and
the rise of the tide.
42
57
33
x
x
27.3
x  39m  Rise of tide is 3 m
tan55
• From a point 42 m above
water-level at low tide the
angle
 of depression of a
buoy in the water was 57
degrees. At high tide the
angle of depression was
55 degrees. Find the
horizontal distance of the
buoy from the viewer and
the rise of the tide.
27.3
55
42
x
Fred wishes to estimate the
height of a building. He steps out
a distance of 60 m from the foot
of the building and finds the
angle of elevation of the top of
the building is 38 degrees. Find
the height of the building if his
eyes are at a height of 1.7 m.
Draw a diagram
h
38
60
1.7
• Fred wishes to estimate
the height of a building.
He steps out a distance of
60 m from the foot of the
building and finds the
angle of elevation of the
top of the building is 38
degrees. Find the height of
the building if his eyes are
at a height of 1.7 m.
Draw a diagram
h 1.7 60 tan38 48.6m
h

60
38
1.7
A step ladder with both sets of
legs 3.5 m long and hinged at the
top is tied with a rope to prevent
the feet of the ladder being more
than 1.7 m apart. What is the
angle between the two parts when
the feet are fully apart?
Draw a diagram
3.5
3.5
1.7
• A step ladder with
both sets of legs 3.5 m
long and hinged at the
top is tied with a rope
to prevent the feet of
the ladder being more
than 1.7 m apart. What
is the angle between
the two parts when the
feet are fully apart?
Draw a diagram
3.5
3.5
1.7
x
3.5
0.85
0.85
tanx 
 x 13.6
3.5
2x  27.3

3.5
3.5
1.7
x
3.5
0.85
From a point A on a straight and
level road the angle of elevation
of the top of a tower at the end of
the road is 30 degrees. After
walking along the road to B the
angle of elevation of the top of
the tower is 50 degrees. How
long is AB if the tower is 45 m
high?
Draw a diagram
45
50
30
x
• From a point A on a
straight and level road the
angle of elevation of the
top of a tower at the end
of the road is 30 degrees.
After walking along the
road to B the angle of
elevation of the top of the
tower is 50 degrees. How
long is AB if the tower is
45 m high?
Draw a diagram
45
45
 y
 y  77.9
y
tan30
45
45
tan50 
 z
 z  37.8
z
tan50
tan30 
45
z
50

y
30
x
x = 40.1 m
45
45
 y
 y  77.9
y
tan30
45
45
tan50 
 z
 z  37.8
z
tan50
tan30 
45
z
50

y
30
x
B
A
C
D
E
Angle BAC is 32 degrees and AB
and CB are 30 and 20 m resp.
Find angle BCD
B
30
20
A
h
32
C
D
E
Angle BAC is 32 degrees and AB
and CB are 30 and 20 m resp.
Find angle BCD
B
30
20
A
h
32
C
D
h
tan 32 
 h  30 tan 32  18.746...
30
E
B
30
20
A
h
32
C
D
h
tan32 
 h  30tan32  18.746...
30
18.746...
sin BCD 
 BCD  69.6
20
E
From the top of a building 120 m
above the ground the angles of
depression of the top and bottom
of another building are 40 and 70
degrees respectively. Find the
distance apart of the buildings
and the height of the lower one.
Draw a diagram
70 40
120
h
x
• From the top of a building
120 m above the ground
the angles of depression of
the top and bottom of
another building are 40
and 70 degrees
respectively. Find the
distance apart of the
buildings and the height of
the lower one.
Find x
70
20
120
x
tan20 
120
x  43.7
x
Find y
y
50
120
40
43.7
tan50 
y
43.7
h
43.7

y  36.7
 h  120 36.7  83.3m
Two small pulleys are placed 8 cm
apart in a horizontal line and an
inextensible string of length 16 cm
is placed over the pulleys. Equal
masses hang symmetrically at each
end of the string and the middle
point is pulled down vertically until
it is in line with the masses. How far
does each mass rise?
Draw a diagram
8 cm
x
y
• Two small pulleys are placed 8
cm apart in a horizontal line and
an inextensible string of length
16 cm is placed over the
pulleys. Equal masses hang
symmetrically at each end of
the string and the middle point
is pulled down vertically until it
is in line with the masses. How
far does each mass rise?
Originally the masses are
hanging down 4 cm.
8 cm
x
y
• Two small pulleys are placed 8
cm apart in a horizontal line and
an inextensible string of length
16 cm is placed over the
pulleys. Equal masses hang
symmetrically at each end of
the string and the middle point
is pulled down vertically until it
is in line with the masses. How
far does each mass rise?
Originally the masses are
hanging down 4 cm.
2x  2 y  16
 x y 8
8 cm
x
y

Originally the masses are
hanging down 4 cm.
2x  2 y  16
 x y 8
4
x
y
y  x 4
2


2
2
4
x
y
Solve the equations by
substitution
x y 8
 y  8  x 
y  x  16
2
2
8  x   x
2
2
 16
x y 8
 y  8  x 
4
y  x  16
y
8  x   x
2
2
2
x
2
 16
6416x  x 2  x 2  16
48  16x  x  3
Height that
it rises is 1 cm

In a right-angled triangle, one of the sides
including the right angle is 7 cm longer than
the other. If the perimeter is 40 cm, find the
lengths of the three sides.
x+7
y
x
• In a right-angled
triangle, one of the
sides including the
right angle is 7 cm
longer than the other.
If the perimeter is 40
cm, find the lengths of
the three sides.
2x  y  7  40
 y  33 2x

x+7
y
x
• In a right-angled
triangle, one of the
sides including the
right angle is 7 cm
longer than the other.
If the perimeter is 40
cm, find the lengths of
the three sides.
2x  y  7  40
 y  33 2x
• Use Pythagoras’

x+7
y
x

y  x  x  7
2
2
2
2x  y  7  40
 y  33 2x
y  x  x  7
2

x+7
y
33 2x  x  x  7
2
x

2
2
2
1089132x  4x 2 
x  x  14x  49
2
2
2
1089 132 x  4x  x  x  14x  49
2
2
2
2x 146x  1040 0
x  65, 8
x 8
2
x+7
y
x

Lengths of sides are
8cm, 15cm and 17 cm