YEAR 10 Chapter 5 Expressions, equations and linear relationships Summary notes - Cambridge Maths NSW 5.1/5.2/5.3
YEAR 10 Chapter 4 Single variable and bivariate statistics Summary notes - Cambridge Maths NSW Stage 5.1/5.2/5.3
YEAR 10 Chapter 3 Probability Summary notes - Cambridge Maths NSW Stage 5.1/5.2/5.3
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Chapter 7 n o te s
TAtrigonometric ratios
->
The hypotenuse is the longests i d e in a right-angled triangle Iti s opposite
the right-angle.
-> Given a right-angled triangle containing an angle 8, the t h re e trigometric
ratios a re :
r a t i o :S i n length oft h e opposite side
·
Th e sine O -
length of the hypoten use nypo+enuse
↑
opposite
7
Th e ratio:C o S -length of the adjacent
side
C -
cosine O
·
adjacent
length of
the hypotenuse
·
The r a t i o :t a n t length of
the opposite side
tangent =
length of
the adjacents i d e
I
to find an u n k n ow n length on a right-angled triangle:
trigometric
·
choose a ratio l i n ks
that one k n ow n an g le an d a
kn ow n side length
with the u n k n ow n side length.
·
solve fo r the u n k n ow n side length.
Find the value of a in these right-angled triangles, c o r re c t
to 2 decimal places.
x
a) 11 < Wi b) mm
T 270 6038mm
xc m
= c o s t :adjacent
a) coso Choose the ratio
hypotenuse
=
cos2
Multiply both sides by 11. then use a c a l c u l a t o r.
... x 11
=
x
cos(2)0)
9.8 0
=
(to 2 decimal
places) Round
your answer as required.
=
b) +
an = The tangentr a t i o uses the opposite and the
adjacents i d e s .
=
tan690
Multiply both sides by 38.
... 38
=
+
x an690
- 9 8.9 9 (to 2 decimal
places)
, Chapter 7 n o te s
ratios
TA trigonometric
Find the value of
a in these right-angled triangles, rounding your an swe r to 2 d.p
xc m
a) L
b) E
(330 15340
21 m
xm
-1.3
cm
a) sinc 1
=
H
Sin 330 12
=
choose the sine ratio since the side
adjacent is not
3
m a r ke d .
x sin
x 330 I
21
21
x - Multiply both sides by to re m ove the
fraction, then divide
Sin 330
both sides by sin 330.
=38.56 ( +0 2 decimal
places)
b)
= tangentr a t i o.
ano
+
The hypotenuse is unmarked, so use the
53040' 71.3
+an
=
multiply both sides by x, then solve by dividing both
x
sides by tan 53040'
an53040 - 1.3
=
xx +
x -
71.3
tan 53040'
=52.44 ( to 2 decimal
places)
LB Finding Unknown Angles
->
The sin", cos" and tab" buttons on c a l c u l a to rs a re used to find angles when the
trigonometric ratio is known.
ex
·
if sine
=
then 8 300 A
0.5 =
I
wise=
·
If COS & =
0.5 then 8
=
000
O Pp0Si e
+
C-E-
·
If tan & 0.5
=
then 8 26034'
=
( to the nearest minute). R -x
0
adjacent
run
->
On the Car tesian plane, gradient(m) can calculated
be
using the fo r m u l a m=tant, where o is the angle between a
line a n dt h e positive direction of the x-axis W
rise opposite
gradient O
=
= =+an
m -
run adjacent
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