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Video Presentation Link: https://www.youtube.com/watch?v=bRYWBbvOMpo
Reference: Grade 10 Mathematics LM
This learner's module discusses about the Six Trigonometric Ratios. It also teaches about the definition and characteristics of each of the Six Trigonometric Ratio.
MEASURES OF POSITION FOR UNGROUPED DATA : QUARTILES , DECILES , & PERCENTILESChuckry Maunes
MEASURES OF POSITION FOR UNGROUPED DATA : QUARTILES , DECILES , & PERCENTILES
Video Presentation Link: https://www.youtube.com/watch?v=bRYWBbvOMpo
Reference: Grade 10 Mathematics LM
The Law of Sines is a principle of trigonometry stating that the length of the sides of any triangle are proportional to the sines of the opposite angles.
"Plug into Power: The Key to Success."_CE101rey castro
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The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
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This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
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http://sandymillin.wordpress.com/iateflwebinar2024
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Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
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Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
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The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
2. Oblique Triangle
• There are several laws that can be use to solve
oblique triangle. These are the law of sines,
law of cosines and law of tangents.
• As in solving right triangles, you should know
three parts of an oblique triangle to find the
other three missing parts.
3. Oblique Triangle
Four Cases
1. ASA or SAA – Law of Sines
2. SSA – law of Sines ( ambiguous case)
3. SAS – Law of cosines
4. SSS- Law of Cosines
4. Oblique Triangle
1. Given: A, b, C Law of Sin
2. c, B, C Law of Sin
3. c, a, C Law of Sin
4. b,A,c Law of Cos
5. c, B, a Law of cos
6. a, b, c Law of cos
6. Law of Sines
• If A, B, and C, are the angles of any triangle,
and a,b, and c, are respectively, the measures
of the sides opposite these angles, then
𝑎
sin 𝐴
=
𝑏
sin 𝐵
=
𝑐
sin 𝐶
8. Law of Sines
1. Solve the triangle given.
Solution: A + B + C = 180
B = 180 – 51.2 – 48.6
B = 80.2o
From the law of sine
23.5
sin 51.2
=
𝑏
sin 80.2
b=
23.5(𝑠𝑖𝑛80.2)
𝑠𝑖𝑛51.2
b= 29.7
80.2o
29.7
9. Law of Sines
1. Solve the triangle given.
B= 80.2
b = 29.7
From the law of sine
23.5
sin 51.2
=
𝑐
sin 48.6
c=
23.5(𝑠𝑖𝑛48.6)
𝑠𝑖𝑛51.2
c= 22.6
22.6
80.2o
29.7
10. Solve the triangle ABC, given a=62.5, A=112o,
and C=42
B=180-112-42
B=26
62.5
sin 112
=
𝑐
sin 42
c=
62.5(𝑠𝑖𝑛42)
𝑠𝑖𝑛112
c= 45.1
11. Solve the triangle ABC, given a=62.5, A=112o, and C=42
B=180-112-42
B=26
62.5
sin 112
=
𝑏
sin 26
b=
62.5(𝑠𝑖𝑛26)
𝑠𝑖𝑛112
b= 29.5
Answers B=26o b= 45.1 c=29.5
12. If
𝑠𝑖𝑛𝐵
𝑏
=
sin 𝐶
𝑐
, then B =____
a) B = 𝑠𝑖𝑛−1 𝑏𝑠𝑖𝑛𝐶
𝑐
b) B = 𝑠𝑖𝑛−1 𝑐𝑠𝑖𝑛𝐶
𝑏
c)B = 𝑠𝑖𝑛−1 𝑏
𝑐𝑠𝑖𝑛 𝐶
Ans. a
13. If
𝑎
sin 𝐴
=
𝑏
sin 𝐵
, then a =____
a) a =
𝑏𝑠𝑖𝑛𝐵
sin 𝐴
b) a =
𝑠𝑖𝑛𝐵
bsin 𝐴
c) a =
𝑏𝑠𝑖𝑛𝐴
sin 𝐵
Ans. c
If
𝑎
sin 𝐴
=
𝑐
sin 𝐶
, then c =____
a)
𝑎𝑠𝑖𝑛𝐴
sin 𝐶
= 𝑐 b)
𝑎𝑠𝑖𝑛𝐶
sin 𝐴
= 𝑐 c)
𝑠𝑖𝑛𝐶
asin 𝐵
= 𝑐
Ans. B
14. Oblique Triangle
Give the appropriate law.
1. SAS –
Law of Cos
2. SSS –
Law of cos
3. SSA –
law of Sines ( ambiguous case)
4. ASA -
Law of sin
5. SAA –
law of sin
15. Law of Cosines
• For any Triangle ABC with sides a,b, and c,
Use to solve the missing sides
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
16. Law of Cosines
• For any Triangle ABC with sides a,b, and c,
Use to solve the missing sides
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
17. Law of Cosines
• For any Triangle ABC with sides a,b, and c,
Use to solve the missing angles
cos A=
b2 + c2 − a2
2𝑏𝑐
cos B=
a2 + c2 − b2
2𝑎𝑐
cos C=
a2 + b2 − c2
2𝑎𝑏
18. Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
a2 = b2 + c2 – 2bc cos A
a2 = 12 + 32 – 2(1)(3) cos 80
a2 = 1 + 9 – 6cos 80
a2 = 10 – 1.04
a2 = 8.96
a= 8.96
a =2.99
a=2.99
19. Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
sin 𝐴
𝑎
=
sin 𝐵
𝑏
sin 80
2.99
=
sin 𝐵
1
(1 )sin 80
2.99
= sin 𝐵
0.3294 = sin B
𝑠𝑖𝑛−1
0.3294 = 𝐵
19.2o = B
a=2.99
19.2o
20. Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
Since A+B+C = 180o
Then 80o +19.2o +C=180o
99.2o +C =180o
C =180o -99.2o
C = 80.8o
a=2.99
19.2o
80.8o
21. Law of
• Solve the triangle with a = 5, b = 8, and c=9
a2 = b2 + c2 – 2bc cos A
52 = 82 + 92 – 2(8)(9) cos A
25=64+81-144cosA
25=145-144cosA
144cosA=145-25
cosA=120/144
cosA = 0.8333
A= 𝑐𝑜𝑠−1
0.8333
A = 33.6o
33.6o
22. Law of Cos
• Solve the triangle with a = 5, b = 8, and c=9
sin 𝐴
𝑎
=
sin 𝐵
𝑏
sin 33.6
5
=
sin 𝐵
8
(8 )sin 33.6
5
= sin 𝐵
0.8854 = sin B
𝑠𝑖𝑛−1
0.8854 = 𝐵
62.3o = B
33.6o62.3o
23. Law of Cos
• Solve the triangle with a = 5, b = 8, and c=9
Since A+B+C = 180o
33.6o +62.3o +C=180o
95.9o +C =180o
C =180o -95.9o
C = 84.1o
33.6o62.3o
84.1o
24. • A Triangular lot has dimensions 20.6m, 31.4m,
and 38.3m. Find the angles at the corners of the
property.
20.62=31.42+38.32 -2(31.4)(38.3)cosA
424.36=2452.85-2405.24cosA
cosA=2028.49/2405.24
cosA=0.8434
A= 32.5o
25. • A Triangular lot has dimensions 20.6m, 31.4m,
and 38.3m. Find the angles at the corners of
the property.
sin 32.5/20.6 = sin B/31.4
0.8190 =sin B
B = 54. 98o or 55
C =180 – 32.5 – 54.98 = 92.52o
29. The diagram shows part of a logo design.
There is one known angle of 142°.
Calculate the sizes of the other two angles.
30. • Mrs Jones goes on a round trip from Town A
to Town B to Town C and back to Town A, as
shown in the following diagram. All roads are
straight. To the nearest mile. How long is the
round trip?
31. • What is the length of side c?
c2=5.32+3.62-2(5.3)(3.6)cos59
c=4.63
34. • Ayton is 25 miles due north of Beeton. Ceeton lies to the east
side the road joining Ayton to Beeton, and is 47 miles from
Ayton and 63 miles from Beeton. (All roads are straight.)
Calculate the three-figure bearing of Ceeton from Ayton.
Note A three-figure bearing is always measured in a clockwise
sense from the direction North.