Mathematics AA SL IB Diploma Program - Topic 3: Geometry and Trigonometry
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Course
Math AA SL
Institution
Book
Oxford IB Diploma Programme: IB Mathematics: Analysis and Approaches, Standard Level, Print and Enhanced Online Course Book Pack
Detailed notes for the third topic of the mathematics analysis and approaches SL course of the IB diploma program. Includes worked examples and IB-style questions.
Summary Oxford IB Diploma Programme: IB Mathematics: Analysis and Approaches, Standard Level, Print and Enhanced Online Course Book Pack - International Baccalaureate (IB)
Internal Assessment IB Diploma Mathematics A&A SL: Relationship between Different Cinematographical Factors and a Film's Opening Weekend Revenue
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topics 3 :
Geometry
and
trigonometry
, Wednesday ,
13
January
-
circles raeecsranrd sectors ,
⑨ For
a circle with radius r :
⑤
the circumference G- 2T r '
radius
⑤
the area A- = IT r2
⑨
An arc is a part of a circle which joins any 2
different points
beg
.
r
⑤
O subtended by
o
It be measured the the points at the center
can
using .
. O
⑤ O
Arc length
-
-
x stir
360
⑨
and the
Agysector
is the
region between 2 radii of a circle arc between them .
Perimeter of
3%
RF
a sector -
-
2rt X 21T r
r
⑨ O O
Area of a sector = X Thr
2
.
360
⑨
Worked example for : the given figure find to ,
3 significant figures :
0
⑤ the × 2K r
length of the arc : 360
48
×
360 2 IT (3) =
2.51cm
0
⑤ X 21T r
the perimeter of the sector : 2 r t 360
ago
213) t 34680 ×
217131=8.51 cm
3cm
O
⑨ Z
the area of the sector : 360 X Mr
48
2
360 X IT (3) =
3.77cm
Exercise 6A : SL H ease core 2019 .
1. Find the
length of :
A. the blue arc .
b. the red arc .
O
Arc length 36% x 21T r Arc length = 360 X 2Thr
60 142
360 X 21T (5) 360 X 21T (5)
=
5.24cm =
12.4cm
,2. Find the perimeter of :
o o o
A. f- 2Mt b. PoS=2rt360X2Tr c. Perimeter of a sector --2rt360X2Tr d. Perimeter of a sector --2rt360X2TLr
180 125 32
18.2
( 21T 213.41+360×21713.4) =2( 8) +360×21718 )
e-
2141+360×21714)
'
=
2 -
C- 21719.1 ) 217.5cm -33.5cm 10.2cm
'
'
-
G- 57.2mm
e. Perimeter
of a sector --2rt3%X2Tr
260=216.21+360×21716.2)
-
-
40.5M
3. Find the radius of the circle .
4. Find the perimeter of the sector .
41.38 er 36=360×2 Thr
36
127 =p
360×212
16.24mm er
5. Find the area of :
TLRZ O O
TLr2
A. A=TLr2 b. A= 2 C. A= 4 d. Area of a sector -_ 360 Xtra e. Area of a sector -_ 360 XTLRZ
2
Z TLC 8.212 TLC 3. 5) 67 215
A- IT 123 A= 2 A- =
4 =
360×1711012 =
360 XTC ( 85
A IT (6.512 9.6mi
'
58.5cm
' '
-_ A -_ 705.6mm A -
-
= =
120.1M
'
A' 132.7cm
-
o
f. Area of a sector -_ 360 XTLRZ
260
0=360 -
100 = 360×1719.212
0=2600 492cm
?
6. Find the radius : 7. Find the perimeter .
O O
*
0=670 Area of a sector 360 XTLRZ * 0=1360 Perimeter of a sector --2rt 360 XZTLR
67 136
*
16.2=360 XTLRZ *
? ?
Area -_
16.2cm A- 28.8cm =2rt 360 x2TLr
16.2 O 136
×t=rZ Area of sector 360 XTLRZ =2( 4.931+360×2544.93)
36670 a --
, b. Calculate Jason 's !
Determine the diameter of the semicircular ends
average speed in
'
a. . MS
distance
TP -1600M d=2r Speed
-
-
time Time --4m25sec
1600M
1 circle --
600M D= 95.5×2 f- 2655 -265 Sec
'
2TLr=6OO d- 791M f- 6.04ms "
600
✓ = 21T
re 95.5 Thursday ,14 January
thecieecle
⑤ Radian measure of angles :
⑨
The circle is divided into 360 sectors and the central each sector is 19
equal angle of
⑨
This unit of does not relate to the circle itself ; it
measure is
only dividing it into 360 equal parts .
⑨
There is another unit of measure for angles and it is derived from the relationship between the arc
length and the radius of the
circle This unit of
. measure is called the radian .
⑨
Radian when the : arc of a circle is the same length as the radius then the, angle subtended bythearcis 't radian .
⑤ The radian is the SI unit of
angle .
⑨
Itis the subtended circular the radius
measure of a central angle by a arc which has the same length as .
⑨ When the arc
length is equal to half the circumference of a circle 11809 the , corrispondihg angle ist radians .
⑨ the circle 13609 the radians
If arc
length is equal to the whole circumference of a , corrisponding angle is 21T .
⑤ Conversion between
degrees and radians :
⑤ 1800 IB notation :
IT radians =
⑨ 360°
*
Radians -_ rad
21T radians =
⑨
I radian = 790=57.30
*
Angle given in terms oft .
Example 1 : An
angle is given as 40 ? Example 2. An angle is given as
4Th radians .
¥ TL TL
radians ( 3 significant
=
*
Find its value in .
figures) Find its value in
degrees .
A
3180
°
X TL
7¥
4 IT
=
1rad
*
= 400 -1800 IT radians -- 1800 *
A = IT
401T 180
1800×34
3
Xrad =
400 *
X =
180 4Th radians -
*
A 1350
*
X = GET =135°
*
Xx 0.698rad
Exercise 1 Exercise 2 Exercise 3 Exercised
What is 1200in radians ? What equation represents the what is 0.15rad in degrees ? Convert 75 to radians .
R IT 180 180 R IT
= 0.15 TL
=
=
degrees ? 1rad 1rad
* *
1200 180 conversion of 1.2rad to -_ IT A 180 = IT 73 180
1201T 180 1.2 IT 0.15×180 731T
=
R= 180 1rad =
IT
*
A 180 0.15rad -_
A * A- =
IT Rrad=73° *
R= 180
1. 2×180
12=-23 'T
7.2rad -
A
*
A = IT * A- 28.590
*
12=1.270
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