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- 1. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
- 2. The word ‘trigonometry’ is derived from the Greek words ‘tri’(meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the sides and angles of a triangle. Early astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved
- 3. A triangle in which one angle is equal to 90 is called right triangle. The side opposite to the right angle is known as hypotenuse. AC is the hypotenuse The other two sides are known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB © iTutor. 2000-2013. All Rights Reserved
- 4. In any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 = (Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved
- 5. Pythagoras Theorem Proof: Given: Δ ABC is a right angled triangle where B = 900 And AB = P, BC= b and AC = h. To Prove: h2 = p2 + b2 Construction : Draw a BD from B to AC , where AD = x and CB = h-x , Proof : In Δ ABC and Δ ABD, Δ ABC Δ ABD --------(AA) In Δ ABC and Δ BDC both are similar So by these similarity, p b h A B C
- 6. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h (h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h p And p b h A B C
- 7. Let us take a right triangle ABC Here, ∠ ACB ( ) is an acute angle. The position of the side AB with respect to angle . We call it the side opposite to angle . AC is the hypotenuse of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000-2013. All Rights Reserved
- 8. The trigonometric ratios of the angle C in right ABC as follows : Sine of C = = Cosine of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved
- 9. Tangent of C = = Cosecant of C= = Secant of C = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights Reserved
- 10. Cotangent of C Above Trigonometric Ratio arbitrates as sin C, cos C, tan C , cosec C , sec C, Cot C . If the measure of angle C is ‘ ’ then the ratios are : sin , cos , tan , cosec , sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved
- 11. Tan = Cosec = 1 / Sin Sec = 1 / Cos Cot = Cos / Sin = 1 / Tan A CB p b h © iTutor. 2000-2013. All Rights Reserved cos sin
- 12. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p b h © iTutor. 2000-2013. All Rights Reserved
- 13. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 450 © iTutor. 2000-2013. All Rights Reserved
- 14. Sin 450 = = = = 1/ 2 Cos 450 = = = = 1/ 2 Tan 450 = = = = 1 Cosec 450 = 1 / sin 450 = 1 / 1/ 2 = 2 Sec 450 = 1 / cos 450 = 1 / 1/ 2 = 2 Cot 450 = 1 / tan 450 = 1 / 1 = 1 Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights Reserved
- 15. Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT) Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved
- 16. As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - (a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved
- 17. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600 AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved
- 18. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved
- 19. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved
- 20. Relation of with Sin when 00 900 The greater the value of ‘ ’, the greater is the value of Sin . Smallest value of Sin = 0 Greatest value of Sin = 1 Relation of with Cos when 00 900 The greater the value of ‘ ’, the smaller is the value of Cos . Smallest value of Cos = 0 Greatest value of Cos = 1 © iTutor. 2000-2013. All Rights Reserved
- 21. Relation of with tan when 00 900 Tan increases as ‘ ’ increases But ,tan is not defined at ‘ ’ = 900 Smallest value of tan = 0 © iTutor. 2000-2013. All Rights Reserved
- 22. If 00 900 1. Sin(900- ) = Cos 2. Cos(900- ) = Sin If 00< 900 1. Tan(900- ) = Cot 2. Sec(900- ) = Cosec If 00 < 900 1. Cot(900- )= Tan 2. Cosec(900- ) = Sec A CB p b h © iTutor. 2000-2013. All Rights Reserved
- 23. Sin2 +Cos2 = 1 Sec2 -Tan2 = 1 Cosec2 - Cot2 = 1 © iTutor. 2000-2013. All Rights Reserved
- 24. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit

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