Trigonometry

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Trigonometry

  1. 1. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
  2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 23. Sin2 +Cos2 = 1 Sec2 -Tan2 = 1 Cosec2 - Cot2 = 1 © iTutor. 2000-2013. All Rights Reserved
  24. 24. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit

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