Presentation on theme: "Lesson plan Class 10th Time 35min. Subject Mathematics"— Presentation transcript:
1 Lesson plan Class 10th Time 35min. Subject Mathematics TopicTrigonometric Ratios of Complementary Angles
2 GENERAL OBJECTIVES OBJECTIVES To inculcate the knowledge of t-ratios of complementary anglesTo 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 anglesTo 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 REQUIREDLCD Projector , Chalk , Duster , Writing Board, PointerRESOURCESText Book of Mathematics for Class 10th
5 PK TESTINGTo 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 answera) <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˚ - <CSince A & C are complementaryThus, <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 allthe t- ratios of <A
12 T-ratios of <C sin C = AB/AC A cos C = BC/AC tan C = AB/BC sec C = AC/BCcosec C = AC/AB C Bcot C = BC/AB
13 COMPARING T-RATIOS OF <A& <C T-RATIOS OF <CT-RATIOS OF <Asin C = AB/ACcos C = BC/ACtan C = AB/BCsec C = AC/BCcosec C = AC/ABcot C = BC/ABsin A = BC/AC cos A = AB/AC tan A = BC/AB sec A = AC/AB cosec A = AC/BC cot A = AB/BCSec A = AC/AB
14 After comparison we get Using C =90˚-AsinC = AB/AC = cosA » sin(90˚-A) = cosAcosC = BC/AC = sinA » cos(90˚-A) = sinAtanC = AB/BC = cotA » tan(90˚-A) = cotAsecC = AC/BC = cosecA » sec(90˚-A) = cosecAcosecC = AC/AB = secA » cosec(90˚-A) = secAcotC = BC/AB = tanA » cot(90˚-A) = tanAsinC = cosANow as we proved above C =90-ATherefore sinC = sin(90-A) = AB/AC = cosA
Your consent to our cookies if you continue to use this website.