1. Lesson plan Class 10th Time 35min. Subject Mathematics
Topic
Trigonometric Ratios of Complementary Angles

2. GENERAL OBJECTIVES OBJECTIVES
To inculcate the knowledge of t-ratios of complementary angles
To increase the logical thinking and reasoning ability

3. 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

4. Text Book of Mathematics for Class 10th
MATERIAL REQUIRED
LCD Projector , Chalk , Duster , Writing Board, Pointer
RESOURCES
Text Book of Mathematics for Class 10th

5. PK TESTING
To asses the previous knowledge of students following questions may be asked

6. (1) What do you mean by complementary angles?
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 :
sin A =? (b) sin C =?
(c) cos A =?

7. COMPLEMENTARY ANGLES Since <A & <C are complementary
Thus,
<C = 90˚- <A
<A = 90˚ - <C
Since A & C are complementary
Thus, <A+ < C= 90˚
< A = 90˚ - < C

8. 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

10. 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

11. ROTATE THE TRIANGLE TOWARDS LEFT SIDE

12. T-ratios of <C sin C = AB/AC A cos C = BC/AC tan C = AB/BC
sec C = AC/BC
cosec C = AC/AB C B
cot C = BC/AB

13. COMPARING T-RATIOS OF <A& <C
T-RATIOS OF <C
T-RATIOS OF <A
sin C = AB/AC
cos C = BC/AC
tan C = AB/BC
sec C = AC/BC
cosec C = AC/AB
cot C = BC/AB
sin A = BC/AC cos A = AB/AC tan A = BC/AB sec A = AC/AB cosec A = AC/BC cot A = AB/BC
Sec A = AC/AB

14. After comparison we get
Using C =90˚-A
sinC = AB/AC = cosA » sin(90˚-A) = cosA
cosC = BC/AC = sinA » cos(90˚-A) = sinA
tanC = AB/BC = cotA » tan(90˚-A) = cotA
secC = AC/BC = cosecA » sec(90˚-A) = cosecA
cosecC = AC/AB = secA » cosec(90˚-A) = secA
cotC = BC/AB = tanA » cot(90˚-A) = tanA
sinC = cosA
Now as we proved above C =90-A
Therefore sinC = sin(90-A) = AB/AC = cosA

15. » sin(90˚-A) = cosA » cos(90˚-A) = sinA » tan(90˚-A) = cotA
THUS , WE HAVE
» sin(90˚-A) = cosA
» cos(90˚-A) = sinA
» tan(90˚-A) = cotA
» sec(90˚-A) = cosecA
» cosec(90˚-A) = secA
» cot(90˚-A) = tanA

16. RECAPTULATION 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˚ .

17. 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˚

18. LESSON PREPARED BY Narvir Singh Chandel TGT(NM) GHS Panoh
Ramesh Chand TGT(NM) GHS Pantehra
Manoj Sharma TGT(NM) GSSS Jejwin
Onkar singh TGT(NM) GSSS Bardin
Roop Lal TGT(NM) GSSS Malyawar
Vinod Kumar TGT(NM) GSSS Gwalmuthani
Joginder Singh TGT(NM) GSSS Chalehly

19. CONTINUED……