Transcript Directed Numbers - Tak Sun Secondary School

Form 1 Mathematics Chapter 10
Lesson requirement
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Textbook 1B
Workbook 1B
Notebook
Before lessons start
Desks in good order!
No rubbish around!
No toilets!
Keep your folder at home
Prepare for Final Exam
Missing HW
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Detention
Ch 11 & Ch 12 OBQ Correction
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24 May (Fri)
Ch 11 & Ch 12 CBQ Correction and
signature
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24 May (Fri)
Acute angle
(larger than 0°
but smaller
than 90°)
Straight angle
(equal to 180°)
Right angle
(equal to 90°)
Obtuse angle
(larger than 90°
but smaller
than 180°)
Reflex angle
Round angle
(larger than 180°
but smaller than
360°)
(equal to 360°)
Classify the angles below.
A, D
acute angle: ________
C
right angle: ________
B, G
obtuse angle: _______
I
straight angle: ______
F, H
reflex angle: ________
E
round angle: ________
1. AB and CD lie in the same
plane and they never meet. We
say that they are a pair of
parallel lines, or ‘AB is
parallel to CD’. In symbols,
we write ‘AB // CD’.
(Parallel lines are usually
indicated by arrows.)
2. PQ and RS lie in the
same plane and intersect at
90°. We say that they are a
pair of perpendicular lines,
or ‘PQ is perpendicular to
RS’. In symbols, we write
‘PQ  RS’.
The sum of the interior angles of any triangle is 180°.
i.e. In the figure, a + b + c = 180°.
[Reference:  sum of ]
Example: Calculate the unknown angles in the following
triangles.
(a)
(b)
45°
(a) _______________
110°
(b) _______________
1. The two angles x and y have a
common vertex O, a common arm OB
and lie on opposite sides of the
common arm OB. We say that x and y
are a pair of adjacent angles (鄰角).
2. The sum of angles at a point is 360°.
e.g. In the figure, a + b + c + d = 360°.
[Reference: s at a pt.]
 Example
1:
Find x in the figure.
x + 210° + 90° = 360° (s at a pt)
x = 360° – 210° – 90°
= 60°
 Example
2:
Find AOB in the figure.
2x + 6x + 240° = 360° (s at a pt)
8x = 120°
x = 15°
∴
2x = 30°
i.e.
AOB = 30°
The sum of adjacent angles on a straight line is 180°.
e.g. In the figure, a + b + c = 180°.
[Reference:
adj. s on st. line]
 Example
1:
In the figure, POQ is a straight line. Find q.
q + 60° = 180° (adj. s on st. line)
q = 180° – 60°
= 120°
 Example
2:
In the figure, XOY is a straight line. Find .
30° + 90° +  = 180° (adj. s on st. line)
 = 180° – 30° – 90°
= 60°
 Example
3:
In the figure, a light ray SP strikes a mirror HK at point
P, and then reflects in the direction PR. It is known that
SPH = RPK. Suppose SPH = , SPR = .
(a) Express  in terms of .
(b) If  = 32°, find .
(a) RPK = SPH = 
(b) When  = 32°,
Since HPK is a straight line,
 = 180° – 2  32°
 +  +  = 180° (adj. s on st. line)
= 116°
∴
 = 180° – 2
 Example
4:
In the figure, AOB is a straight line.
(a) Find AOD.
(b) If AOE = 30°, determine
whether EOD is a straight line.
(a) 3a + 2a + a = 180° (adj. s on st. line)
6a = 180°
a = 30°
AOD = 3a + 2a
= 5a = 5  30°
= 150°
(b) EOD
= AOE + AOD
= 30° + 150°
= 180°
∴ EOD is a straight line.
When two straight lines intersect, the vertically
opposite angles formed are equal.
i.e. In the figure, a = b.
[Reference: vert. opp. s]
 Example
1:
Find x and y in the figure.
x = 45° (vert. opp. s)
y = 135° (vert. opp. s)
 Example
2:
In the figure, the straight lines
AE, BF and CG intersect at O,
and AE  CG. Find p.
BOA = 75° (vert. opp. s)
Consider all the adjacent angles on the upper side of CG.
COB + BOA + AOG = 180° (adj. s on st. line)
∴
p + 75° + 90° = 180°
p = 15°
 Example
3:
In the figure, the straight
lines PS and QT intersect
at R and TRS = PQR.
Find x and y.
In △PQR,
QPR + PQR + PRQ = 180°
( sum of )
∴ TRS = PQR = 50° (Given)
PRQ = TRS = 50° (vert. opp. s) y + 50° + 50° = 180°
y = 80°
x + 310° = 360° (s at a pt)
x = 50°
Pages 140 – 143 of Textbook 1B
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Questions 1 – 32
Pages 54 – 57 of Workbook 1B
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Question 1 - 13
Missing HW


Detention
Ch11 & Ch 12 OBQ Correction


24 May (Fri)
Ch 11 & Ch 12 CBQ Correction and
signature


24 May (Fri)
Ch 10 SHW(I)
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27 May (Mon)
Ch 10 OBQ
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29 May (Wed)
Enjoy the world of
Mathematics!
Ronald HUI