Transcript Lesson plan Class 10th Time 35min. Subject Mathematics

Lesson plan
Class 10th
Subject
Time 35min.
Mathematics
Topic
Trigonometric Ratios of
Complementary Angles
OBJECTIVES
• GENERAL OBJECTIVES
• To inculcate the knowledge of
tratios of complementary angles
• To increase the logical thinking and
reasoning ability
SPECIFIC OBJECTIVES :
• To enable the students to establish the
relationship between the t-ratios of the angles
and t-ratios of their complementary angles
• To enable the students to convert the t- ratios
of angles (0˚ -45˚) to the t-ratios of (45˚ – 90˚)
& vice-versa.
• Enabling the students to solve the problems
related to the t-ratios of complementary
angles
MATERIAL REQUIRED
LCD Projector , Chalk , Duster , Writing
Board, Pointer
RESOURCES
Text Book of Mathematics for Class
10th
PK TESTING
• To asses the previous knowledge of students
following questions may be asked
(1) What do you mean by complementary angles?
(1) Look at the figure ;now answer
a) <A+<B+<C =……..
b)
<B = …….
c) <A+<C
=……..
(3 ) Are <A & <C complementary ?
(4) Define following t-ratios :
(a) sin A =? (b) sin C =?
(c) cos A =?
COMPLEMENTARY ANGLES
Since <A & <C are
complementary
<A+ <C= 90˚
Thus,
<C = 90˚- <A
<A = 90˚ - <C
LAUNCHING OF TOPIC
• Activity 1:
Divide the students in 5 groups &
provide each group a rt. angled triangle.
Ask students to write all
the t- ratios of <A
T-ratios of < A
• Look at this
C
sin A = BC/AC
cos A = AB/AC
tan A = BC/AB
sec A = AC/AB
cosec A = AC/BC
cot A = AB/BC
ROTATE THE TRIANGLE TOWARDS LEFT
SIDE
T-ratios of <C
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sin C =
cos C =
tan C =
sec C =
cosec C =
cot C =
AB/AC
BC/AC
AB/BC
AC/BC
AC/AB
BC/AB
A
C
B
COMPARING T-RATIOS OF <A& <C
T-RATIOS OF <C
T-RATIOS OF <A
• sin C
= AB/AC
sin A = BC/AC
• cos C
= BC/AC
cos A = AB/AC
• tan C
= AB/BC
tan A = BC/AB
•
= AC/BC
sec A = AC/AB
sec C
• cosec C = AC/AB
•
cot C = BC/AB
Sec A = AC/AB
cosec A = AC/BC
cot A = AB/BC
After comparison we get
Using C =90˚-A
sinC = AB/AC = cosA
cosC = BC/AC = sinA
tanC = AB/BC = cotA
secC = AC/BC = cosecA
cosecC = AC/AB = secA
cotC = BC/AB = tanA
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sin(90˚-A) = cosA
cos(90˚-A) = sinA
tan(90˚-A) = cotA
sec(90˚-A) = cosecA
cosec(90˚-A) = secA
cot(90˚-A) = tanA
THUS , WE HAVE
» sin(90˚-A) =
» cos(90˚-A) =
» tan(90˚-A) =
» sec(90˚-A) =
» cosec(90˚-A) =
» cot(90˚-A) =
cosA
sinA
cotA
cosecA
secA
tanA
RECAPTULATION
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1. What is sin (90˚-A) ?
2. What is sec (90˚-C) ?
3. Convert sin30˚ in terms of cos .
4. Convert tan 60˚ in terms of cot .
5 Express cos 70˚ in t-ratio of an angle
between 0˚ - 45˚ .
HOME ASSIGNMENT
• 1. Evaluate : sin18˚
cos 72˚
2.Show that : tan 30˚tan 42˚tan60˚tan48˚ = 1
3.If tanA = cotB ,
then prove that A+B = 90˚
LESSON PREPARED BY
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Narvir Singh Chandel
Ramesh Chand
Manoj Sharma
Onkar singh
Roop Lal
Vinod Kumar
Joginder Singh
TGT(NM)
TGT(NM)
TGT(NM)
TGT(NM)
TGT(NM)
TGT(NM)
TGT(NM)
GHS Panoh
GHS Pantehra
GSSS Jejwin
GSSS Bardin
GSSS Malyawar
GSSS Gwalmuthani
GSSS Chalehly
CONTINUED……