Transcript File

Chapter 9
Non-right Angled Triangle Trigonometry
Opening problem
A triangular sail is to be cut from a section of
cloth. Two of the sides must have lengths 4 m
and 6 m as illustrated. The total area for the
sail must be 11.6 m2, the maximum allowed for
the boat to race in its class.
a. Can you find the size of the angle q between
the two sides of given length?
b. Can you find the length of the third side of
the sail?
6
q
4
Included angle
Knowing two sides and the included angle
between them
8 cm
49o
10 cm
No angles
Knowing all three sides
9 cm
8 cm
10 cm
Using the included angle
A
c
B
b
h
49o
C
10 cm
a
A
c
b
h
180o - C
B
a
C
D
sin C 
h
sin( 180   C )
b
 h  b sin C
 h  b sin( 180   C )
 h  b sin C
because
A
1
2
ah 
1
2
ab sin C
sin C  sin( 180   C )
Exercise 9A #1
Exercise 9A #1
A
1
ab sin C
2
A
1
(10 )( 9 ) sin 40 
2
A  45 sin 40 
A  28 . 9 cm
2
Exercise 9A #1
A
1
ab sin C
2
A
1
( 25 )( 31 ) sin 82 
2
A  387 . 5 sin 82 
A  383 . 73 km
2
Exercise 9A #1
A
1
ab sin C
2
A
1
(10 . 2 )( 6 . 4 ) sin
2
A  32 . 64 sin
3
2
3
A  28 . 3 cm
2
2
Exercise 9A #3
6. A parallelogram has two adjacent sides of
length 4 cm and 6 cm respectively. If the
included angle measures 52o , find the area
of the parallelogram.
6 cm
4 cm
h
52o
Exercise 9A #3
6. A parallelogram has two adjacent sides of
length 4 cm and 6 cm respectively. If the
included angle measures 52o , find the area
of the parallelogram.
6 cm
h  4 sin 52 
4 cm
h
52o
A  bh
A  ( 6 )( 4 sin 52  )
A  18 . 91 cm
2
A.
Area
1
2
sector
 Area
(1 . 5 )(12 ) 
2
1
triangle
(12 )(12 ) sin( 1 . 5 )
2
108  71 . 82  36 . 18
B.
Area
1
 Area
triangle
sector
( 30 )(12 ) sin . 66 
2
62 . 8 cm
1
2
2
(. 66 )(12 )
2
C.
Area
135 
360 
sector
 Area
 (7 ) 
2
1
triangle
( 7 )( 7 ) sin( 135  )
2
57 . 73  17 . 32  40 . 4 mm
2
Cosine rule
The cosine rule involves the sides and
angles of a triangle.
A
a  b  c  2 bc cos A
2
2
2
or
b  a  c  2 ac cos B
2
2
2
b
c
or
c  a  b  2 ab cos C
2
2
2
B
a
C
The cosine rule can be used to solve
problems involving triangles given:
• two sides and an included angle
• three sides.
Ambiguous case
If we are given two sides and a non-included angle, then when
we try to find the third side we obtain a quadratic equation.
This is an ambiguous case where there may be two plausible
solutions. We may not be able to solve for the length uniquely if
there are two positive, plausible solutions to the quadratic
equation.
b
a
b
B
a
B
b c a
2
cos A 
2
2
2 bc
c a b
2
cos B 
2
2
2 ca
a b c
2
cos C 
2
2 ab
2
Exercise 9B #1a
A
105o
21 cm
C
15 cm
B
Exercise 9B #4
11. a. Find the smallest angle of a triangle
with sides 11 cm, 13 cm and 17 cm.
11 cm
13 cm
q
17 cm
b. Find the largest angle of a triangle with
sides 4 cm, 7 cm and 9 cm.
q
4 cm
9 cm
7 cm
Exercise 9B#5
a. Find cos q but not q.
2 cm
5 cm
q
4 cm
b. Find the value of x
x cm
1 cm
Exercise 9B #6
The Sine Rule
The sine rule is a set of equations which connects the lengths
of the sides of any triangle with the sines of the angles of the
triangle. The triangle does not have to be right angled for the
sine rule to be used.
sin A

a
sin B

b
sin C
c
b
A
or
C
a
sin A

b
sin B

c
c
a
sin C
B
Exercise 9C.1#2
15. In triangle ABC find:
a. a if A = 63o , B = 49o and b = 18 cm
a

sin A
sin B
a
sin 63 
a
b

18
sin 49 
18 sin 63 
sin 49 
 21 . 25
b. b if A = 82 o , C = 25 o and c = 34 cm
A + B + C = 180 o
82 o + B + 25 o = 180
B = 73 o
c

sin C
sin B
34
sin 25 
b
b

b
sin 73 
34 sin 73 
sin 25 
 76 . 94
c. c if B = 21 o , C = 48 o and a = 6.4 cm.
A + B + C = 180 o
A + 21o + 48 o = 180
A = 111 o
c

sin C
sin A
c
sin 48 
c
a

6 .4
sin 111 
6 . 4 sin 48 
sin 111 
 5 . 09
Ambiguous case
The problem of finding angles using the sine
rule is more complicated because there
may be two possible answers. For example,
if
sin q 
3
2
then
q  60  or q  120 
We call this situation an ambiguous case.
This occurs when we have two sides and a
non-included angle.
investigation: the ambiguous case.
Exercise 9C.2#1
17. Triangle ABC has angle B = 40o, b = 8 cm, and
c = 11 cm. Find the two possible values for angle
C.
11
11
B
8
8
8
B
40o
40o
C
ACUTE
8
sin 40 
sin C 
11

sin C
11 sin 40 
8
C  sin
1
 11 sin 40  


8


C  62 . 1 
C
ym
95o
xm
30o
118o
22 m
Use the cosine rule when given:
• three sides
• two sides and an included angle.
Use the sine rule when given:
• one side and two angles
• two sides and a non-included angle, but
beware of an ambiguous case which can occur
when the smaller of the two given sides is
opposite the given angle.
20. Rodrigo wishes to determine the height
of a flagpole. He takes a sighting to the top
of the flagpole from point P. He then moves
further away from the flagpole by 20
meters to point Q and takes a second
sighting. The information is shown in the
diagram alongside. How high is the flagpole?
Q
28o
20 m
P
53o
21. A golfer played his tee shot a distance of
220 m to point A. He then played a 165 m
six iron to the green. If the distance from
tee to green is 340 m, determine the
number of degrees the golfer was off line
with his tee shot.