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

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Presentation on theme: "Lesson plan Class 10 th Time 35min. Subject Mathematics Topic Trigonometric Ratios of Complementary Angles."— Presentation transcript:

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

2 O BJECTIVES GENERAL 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 MATERIAL REQUIRED LCD Projector, Chalk, Duster, Writing Board, Pointer RESOURCES Text Book of Mathematics for Class 10 th

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

6 (1) What do you mean by complementary angles? (1)Look at the figure ;now answer a)

7 COMPLEMENTARY ANGLES Since

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

9

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

13 COMPARING T-RATIOS OF

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

15 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……CONTINUED……


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