angles in intersecting and parallel lines
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Form 1 Mathematics Chapter 10. Angles in Intersecting and Parallel Lines. Reminder. Lesson requirement 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. Reminder. Missing HW - PowerPoint PPT PresentationTRANSCRIPT
Form 1 Mathematics Chapter 10
Lesson requirement 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 Detention
Ch 10 SHW(I) 28 May (Tue)
Ch 10 SHW(II) 31 May (Fri)
Ch 10 SHW(III) 31 May (Fri)
Ch 10 OBQ 31 May (Fri)
Ch 10 CBQ 4 June (Tue)
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)
(a) _______________ (b) _______________45° 110°
(b)
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 2:
i.e. AOB = 30°
Find AOB in the figure.
2x + 6x + 240° = 360° (s at a pt)
8x = 120°
x = 15°
∴ 2x = 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 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)
(b) EOD
= AOE + AOD
= 30° + 150°
= 180°
∴ EOD is a straight line.
6a = 180°
a = 30°
AOD = 3a + 2a
= 5a = 5 30°
= 150°
When two straight lines intersect, the vertically
opposite angles formed are equal.
i.e. In the figure, a = b.
[Reference: vert. opp. s]
Example 3:
In the figure, the straight
lines PS and QT intersect
at R and TRS = PQR.
Find x and y.
x + 310° = 360° (s at a pt)
x = 50°
∴ TRS = PQR
PRQ = TRS
= 50° (Given)
= 50° (vert. opp. s)
In △PQR,
QPR + PQR + PRQ = 180°
( sum of )
y + 50° + 50° = 180°
y = 80°
Pages 140 – 143 of Textbook 1B Questions 1 – 32
Pages 54 – 57 of Workbook 1B Question 1 - 13
According to the figure,
1. the straight line EF is called the transversal (截線 ) of AB and CD.
2. a and e, b and f, c and g, d and h are pairsof corresponding angles (同位角 ).
3. c and e, d and f are pairs of alternate angles (內錯角 ).
4. c and f, d and e are pairs of interior angles on the same side of the transversal (同旁內角 ).
The corresponding angles formed by parallel lines and a
transversal are equal.
i.e. In the figure, if AB // CD, then a = b.
[Reference: corr. s, AB // CD]
The alternate angles formed by parallel lines and a
transversal are equal.
i.e. In the figure, if AB // CD, then a = b.
[Reference: alt. s, AB // CD]
The sum of the interior angles of parallel lines on the
same side of the transversal is 180°.
i.e. In the figure, if AB // CD, then a + b = 180°.
[Reference: int. s, AB // CD]
Example 1: Find x in the figure.
x = 50° (corr. s, AB // CD)
Example 2: Find x in the figure.
x = 135° (alt. s, AB // CD)
OR x + 45° = 180° (int. s, AB // CD) x = 135°
Example 3: Find x in the figure.
∴ x = 135°
x + 45° = 180° (int. s, PQ // RS)
Example 4:Find a and b in the figure.
a + 120° = 180° (int. s, AD // BE)
a = 60°
EBC = DAC (corr. s, AD // BE)
b + 50° = 120°
b = 70°
Example 5:Find x and y in the figure.
x + 30° + 50° = 180° ( sum of )
x = 100°
y = x (alt. s, AB // CE)
∴ y = 100°
Example 6:
pq
r
In the figure, AB, QR and CD are parallel
lines, while PQ and RS are another pair of
parallel lines. If RSA = 66°, find QPD.
Using the notation in the figure,r + 66° = 180° (int. s, AD // QR) r = 114°∵ q = r (alt. s, PQ // RS)
∴ q = 114° ∵ p + q = 180° (int. s, QR // CD)
p + 114° = 180° p = 66°
∴ QPD = 66°
Example 7:Find the unknown angle x in the figure.
Draw the straight line AT such that AT // PQ.Since PQ // NS, we have AT // NS.Using the notation in the figure,
y + 145° = 180° (int. s, PQ // AT)
∴ y = 35° 67° + x + y = 180° (int. s, NS // AT)
67° + x + 35° = 180°
x = 78°
Ay
Pages 154 – 155 of Textbook 1B Questions 4 – 25
Pages 59 – 61 of Workbook 1B Question 1 - 8
Enjoy the world of Mathematics!
Ronald HUI