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

Topic Trigonometric Ratios of Complementary Angles

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**GENERAL OBJECTIVES OBJECTIVES**

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

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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

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**Text Book of Mathematics for Class 10th**

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

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

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**(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 =?

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**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

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**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

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**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

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**ROTATE THE TRIANGLE TOWARDS LEFT SIDE**

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**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

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**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

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**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

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**» 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

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**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˚ .

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**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˚

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**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

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CONTINUED……

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