1. 28 December 2012
Trigonometry
Learning Objective:
To be able to describe the sides of right-
angled triangle for use in trigonometry.
Trigonometry is concerned with the
connection between the sides and
angles in any right angled triangle.
Angle
2. The sides of a right -angled triangle are
given special names:
The hypotenuse, the opposite and the
adjacent.
The hypotenuse is the longest side and is
always opposite the right angle.
The opposite and adjacent sides refer to
another angle, other than the 90o.
A
A
3. There are three formulae involved in
trigonometry:
sin A=
cos A=
tan A =
S OH C AH T OA
4. Using trigonometry on the calculator
28 December 2012
Learning Objective:
To be able to use a scientific calculator to
find decimal values and angles in
trigonometry.
All individual angles have different sine, cosine
and tangent ratios (or decimal values).
Scientific calculators store information about
every angle.
We need to be able to access this
information in the correct manner.
5. Finding the ratios
The simplest form of question is finding the
decimal value of the ratio of a given angle.
Find:
1) sin 32 =
sin 32 =
2) cos 23 =
3) tan 78 =
4) tan 27 =
5) sin 68 =
6. Using ratios to find angles
We have just found that a scientific
calculator holds the ratio information
for sine (sin), cosine (cos) and
tangent (tan) for all angles.
It can also be used in reverse, finding
an angle from a ratio.
To do this we use the sin-1, cos-1 and
tan-1 function keys.
7. Example:
1. sin x = 0.1115 find angle x.
sin-1 0.1115 =
( shift sin )
x = sin-1 (0.1115)
x = 6.4o
2. cos x = 0.8988 find angle x
cos-1 0.8988 =
( shift cos )
x = cos-1 (0.8988)
x = 26o
9. Finding an angle from a triangle
To find a missing angle from a right-angled
triangle we need to know two of the sides of
the triangle.
We can then choose the appropriate ratio,
sin, cos or tan and use the calculator to
identify the angle from the decimal value of
the ratio.
1. Find angle C
a) Identify/label the
14 cm names of the sides.
b) Choose the ratio that
C contains BOTH of the
6 cm letters.
10. 1. We have been given
H the adjacent and
14 cm hypotenuse so we use
COSINE:
C adjacent
Cos A =
6 cm hypotenuse
A
Cos A = a
h
Cos C = 6
14
Cos C = 0.4286
C = cos-1 (0.4286)
C = 64.6o
11. 2. Find angle x
Given adj and opp
x need to use tan:
3 cm
A opposite
Tan A = adjacent
8 cm
O
Tan A = o
a
Tan x = 8
3
Tan x = 2.6667
x = tan-1 (2.6667)
x = 69.4o
12. 3.
Given opp and hyp
12 cm need to use sin:
10 cm
opposite
y Sin A = hypotenuse
sin A = o
h
sin x = 10
12
sin x = 0.8333
x = sin-1 (0.8333)
x = 56.4o
14. Finding a side from a triangle
To find a missing side from a right-angled
triangle we need to know one angle and one
other side.
Note: If
x
Cos45 = 13
To leave x on its own we need to
move the ÷ 13.
It becomes a “times” when it moves.
Cos45 x 13 = x
15. 1. We have been given
H the adj and hyp so we
7 cm use COSINE:
adjacent
Cos A = hypotenuse
30o
k
A
Cos A = a
h
Cos 30 = k
7
Cos 30 x 7 = k
6.1 cm = k
16. 2. We have been given
the opp and adj so we
50o
use TAN:
4 cm
A
Tan A =
r
O
Tan A = o
a
Tan 50 =
r
4
Tan 50 x 4 = r
4.8 cm = r
17. 3. We have been given
H the opp and hyp so we
k 12 cm use SINE:
O
Sin A =
25o
sin A = o
h
sin 25 = k
12
Sin 25 x 12 = k
5.1 cm = k
18. Finding a side from a triangle
There are occasions when the unknown
letter is on the bottom of the fraction after
substituting.
Cos45 = 13
u
Move the u term to the other side.
It becomes a “times” when it moves.
Cos45 x u = 13
To leave u on its own, move the cos 45
to other side, it becomes a divide.
u =
13
Cos 45
19. 28 December
Trigonometry
Learning Objective:
To be able to use trigonometry to find an
unknown side when the unknown letter is
on the bottom of the fraction.
When the unknown letter is on the bottom
of the fraction we can simply swap it with
the trig (sin A, cos A, or tan A) value.
Cos45 = 13
u
u =
13
Cos 45
20. 1. Cos A = a
h
H
x Cos 30 = 5
x
5
30 o x = cos 30
5 cm
A x = 5.8 cm
sin A = o
2. h
H
8 cm m sin 25 = 8
m
O
m= 8
25o sin25
m = 18.9 cm
22. 1. Cos A =
a
h
H
x Cos 30 = 5
x
5
30o x =
cos 30
5 cm
A x = 5.8 cm
2.
Tan A = o
a
50o
Tan 50 =r
4 cm 4
A
Tan 50 x 4 = r
r 4.8 cm = r
O
3. o
sin A =
h
12 cm sin y = 10
10 cm 12
y sin y = 0.8333
y = sin-1 (0.8333)
y = 56.4o
