Transcript No Slide Title

Non-right-angled triangles and trigonometry
a) find the trigonometric ratios of obtuse angles either from redefinition of the
trigonometric relationships using the circle (usually the unit circle) or by use of a calculator
b) find the possible acute and/or obtuse angles, given a trigonometric ratio


c) establish and use the following relationships for obtuse angles (0°  A  90°):
sin (180° – A) = sin A
cos (180° – A) = – cos A
tan (180° – A) = – tan A

d) draw graphs of the sine and cosine curves for 0°  A  180°
e) solve angle and length problems using the sine rule in acute- and obtuse-angled triangles
SineRule :
a
b
c


Sin A Sin B Sin C

f) solve angle and length problems using the cosine rule in acuteand obtuse angled triangles
a 2  b2  c 2
2
2
2
Cosine rule: c = a + b – 2ab cos C, and Cos C 
2ab
g) use the area formula to find the area of a triangle
Area =½ab sin C
h) select and use appropriate trigonometric ratios and formulae to solve problems involving
trigonometry that require the use of more than one triangle (two dimensions), where the
diagram is provided.
(All these things are on your self-assessment sheet for this topic)
Finding the length of sides and size of angles in
non right-angled triangles
So far in your trig work, you have only used acute angles,
because the triangles you have been working with have been
only right-angled.
What do you do in these triangles?
Why can’t you use normal Sin, Cos
or Tan Trigonometry?
110°
35°
x m
12 m
Find x.
25°
28 m

15 m
Find .
There is no Hypotenuse because they aren’t right-angled triangles!
So how do you find the side marked x or the angle marked ?
A new rule is needed!

Let’s investigate these types of triangles!
First of all, we need
to be able to name
the sides of the
triangle.
We could use two letters,
e.g. AB, but it would be
quicker if it was only one
letter!
A
And, this
side is c
This side
is b
C
B
So, this
side is a
The convention used is that the side opposite an
angle(which is ALWAYS named with an upper case letter)
is named by a lower case of the same letter.
C
b
a
h
In this ABC, to develop the new
rule, first draw a perpendicular
from C to AB. Name the
intersecting point D and the line h.
Do it now on your diagram
Now, ABC is broken up into
2 triangles, BCD & ACD,
(and they are both right angled).
A
D
In ACD
Sin A =
h
b
And, by manipulating,
h = b  Sin A
c
B
And in BCD
Sin B =
h
a
And, by manipulating,
h = a  Sin B
We now have 2
statements which are
both equal to h.
So, we can make them
equal to each other….
b Sin A = a Sin B
And, by manipulating,
a
Sin A
=
b
Sin B
C
So, we have derived a rule which uses
2 sides and their opposite angles.
a
Sin A
=
b
b
Sin B
If we changed the perpendicular so
that it came from A and went to CB,
we would come up with this rule:
C
b
b
D
Sin B
=
D
Sin C
a
c
B
B
c
Both parts of the two
rules can be combined to
develop what is known as
the Sine rule:
c
Sin A
A
h
A
a
h
a
=
b
Sin B
=
c
Sin C
a
Sin A
b
=
Sin B
=
B
c
Sin C
When the rule is in this form, we
shall first use it to find a side:
But we will only ever need 4 of
the six parts of the rule in any
one use of the formula.
c
Sin A
=
b
Sin B
=
c
Sin C
Having placed the letters, substitute into the
relevant parts of the formula correctly.
x
Sin 40°
=
9
Sin 80°
And, by manipulating,
x =
9  Sin 40°
Sin 80°
(
a
9 cm
First,because the vertices of the triangle
have not been named, name each corner.
It doesn’t matter where you put each of
the vertex letters, but it DOES matter
where you put the side letters!
a
40°
A
Find x to 2
decimal
places.
80°
x cm
C
b
As we don’t know a, we won’t be
able to use it OR angle A.
x 
5.87 cm
(to 2 d.p.)
This is a reasonable answer as it opposite
the 40°, and the 9cm is opposite the 80°
Don’t presume that just because 40° is half
80° that the answer must be half 9cm!
)
a
Sin A
=
b
Sin B
=
c
Sin C
Find  to the nearest minute.
It will make our work much easier to
find an angle if we turn the Sine
rule upside down, so that the angles
are on the top. After all when we
found the sides they were on top!
Sin A
a
=
Sin B
b
=
Sin C
C
b
A
10°
°
6 cm
a
18 cm
B
c
c
This time, we don’t know b, so we
Once again we will only ever need 4 of the six
won’t be able to use it OR angle B.
parts of the rule in any one use of the formula.
It doesn’t matter which pair of the
Having placed the letters, substitute into the information is written first, as long as the
relevant parts of the formula correctly.
information in each fraction is opposite
each other in the triangle. Here C and c
Sin 10°
Sin °
have been written first.
=
18
6
 
And, by manipulating,
Sin   18  Sin 10°
6
Sin   0.52094….
(
31 24
Using the INV key on your calculator
This is a reasonable answer as it opposite
the 19cm, and the 6cm is opposite the 10°
Don’t presume that just because 6cm is one third
18cm that the answer must be 3 times 10°!
)
Your turn, Exercise 9F, page 344 Year 10 Text