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Trigonometry class10

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Sanjay Sahu

1) The document discusses trigonometric ratios and their values for specific angles like 0°, 45°, 90°, and 300°. It defines trigonometric functions like sine, cosine, tangent, cotangent, secant, and cosecant. 2) Examples of right triangles are used to derive the trigonometric ratios of angles like sin(45°)=1/√2 and tan(300°)=√3. 3) The key properties of trigonometric ratios at limit angles like sin(0°)=0 and undefined values like cot(90°) are also explained.

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Submitted by Sohanlal Rituraj Sandeep Sanjay Kabir Sanjay Sahu Satyendra Shubham Gupta Shubham Prajapati
 The Trigonometry is derived from a Greek word, trigonon and metron. The word trigonon means a triangle and the word metron means measure. Therefore, trigonometry means the science of measuring triangles.
 (I) Sin θ = Perpendicular Hypotenuse  (ii) Cosine θ = Base Hypotenuse  (iii) Tangent θ = Perpendicular  Base  (iv) Cosecant θ = Hypotenuse  Perpendicular  (v) Secant θ = Hypotenuse  Base  (vi) Cotangent θ = Base  Perpendicular
Let angle XAY = θ be an acute angle and P be a point on its terminal side AY. Draw perpendicular PM from P on AX . IN ΔAMP, we have Sin θ = PM, cos θ = AM AP AP and tan θ = PM AM
It is evident from ΔAMP that as θ becomes smaller and smaller, line segment PM also becomes smaller and smaller; finally when θ becomes O⁰; the point P will coincide with M. Consequently , we have PM = O and AP=AM Therefore, sin O⁰ = PM = O =O AP AP cos O⁰=AM =AP = 1 AP AP
 Therefore,  sin 90⁰ = PM = PM = 1 AP PM and cos 90⁰ = AM = O = O AP AP Thus, we have sin 90⁰ = 1 and cos 90⁰ =O Remark:- It is evident from the above discussion that tan 90⁰ = PM =O is not defined . Similarly, sec 90⁰, cosec 0⁰, cot o⁰ are not defined .
and , tan O⁰ = PM = O =O AP AP Thus, we have sin O⁰ = O, cos O⁰= 1 and tan O⁰ = O From ΔAMP, it is evident that as θ increase, line segment AM becomes smaller and smaller and finally when θ becomes 90⁰ the point M will coincide with A. Consequently, we have AM =O, AP =PM
 Considered an equilateral triangle ABC with each side of length 2a. Since each angle of an equilateral triangle is of 60 ⁰. Therefore, each angle of ΔABC is of 60⁰. Let AD be perpendicular from A on BC. Since the triangle is equilateral the triangle is equilateral. Therefore, AD is the bisector of angle A and D is the mid-point of BC.
Therefore, BD =DC=a and angle BAD =30⁰ Thus, in ΔABD , angle D is a right angle, hypotenuse AB =2a and BD = a. So, by Pythagoras theorem, we Have AB² = AD² + BD² (2a)²= AD² = a² AD²= 4a²- a² AD =√3a Trigonometric Ratios of 30⁰: In right Δ ADB, we have
 Base =AD = √3a , perpendicular = BD = a, Hypotenuse = AB = 2a and angle DAB= 30⁰ sin 30⁰ =BD = a = 1 AB 2a 2 Therefore, cos 30⁰ = AD = √3a = √3 AB 2a 2 tan 30⁰= BD = a = 1 AD √3a √3 cosec 30⁰ = 1 =2 sin 30⁰ sec 30⁰ = 1 = 2 and cot30⁰ = √3 cos 30⁰ √3
 Trigonometry ratios of 60⁰:  In right triangle ADB, we have Base = BD =a, Perpendicular =AD= √3a , Hypotenuse =AB = 2a and angle ABD = 60⁰ Therefore, Sin 60⁰= AD =√3a = √3, AB 2a 2 Cos 60⁰ = BD = a = 1 AB 2a 2 Tan 60⁰ = AD = √3a = √3 BD a Cosec, sec and cot are opposite of sin, cos, tan respectively.
 Consider a right angle triangle ABC with right at B such that angle A= 45 °. Then, Angle A + Angle B + angle C =180° 45 ° + 90° + Angle C= 180 ° angle C = 45 ° Therefore, angle A = angle C Therefore, AB = BC
 Let AB = BC =a. Then , by Pythagoras theorem , we have AC² = AB² = BC² AC² =a² +a² AC² = 2a² AC =√2a Thus, in we have triangle ABC, we have Angle = 45°, Base = AB =a, Perpendicular = BC =a, Hypotenuse = AC =√2a
 Therefore sin 45= BC = a = 1 AC √2 √2 cos 45° = AB = a = 1 AC √2a √2 tan 45° = BC =a =1 , AB a cosec 45° = 1 = √2 sin 45° sec 45° = 1 = √2 and cos 45° cot 45° = 1 = 1 tan 45°
Trigonometry class10
 An equation involving trigonometric ratios of an angle θ (say)is said to be a Trigonometric identity if it is satisfied for all values of θ for which the given trigonometric ratios are defined.  For example, cos² θ – 1cos θ = cos θ ( cos θ – 1) 2 2 is a trigonometric identity, whereas cos θ ( cos θ – 1 ) = 0 2 is an equation.
Trigonometry class10
Trigonometry class10
Trigonometry class10

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Trigonometry class10

  • 1. Submitted by Sohanlal Rituraj Sandeep Sanjay Kabir Sanjay Sahu Satyendra Shubham Gupta Shubham Prajapati
  • 2.  The Trigonometry is derived from a Greek word, trigonon and metron. The word trigonon means a triangle and the word metron means measure. Therefore, trigonometry means the science of measuring triangles.
  • 3.  (I) Sin θ = Perpendicular Hypotenuse  (ii) Cosine θ = Base Hypotenuse  (iii) Tangent θ = Perpendicular  Base  (iv) Cosecant θ = Hypotenuse  Perpendicular  (v) Secant θ = Hypotenuse  Base  (vi) Cotangent θ = Base  Perpendicular
  • 4. Let angle XAY = θ be an acute angle and P be a point on its terminal side AY. Draw perpendicular PM from P on AX . IN ΔAMP, we have Sin θ = PM, cos θ = AM AP AP and tan θ = PM AM
  • 5. It is evident from ΔAMP that as θ becomes smaller and smaller, line segment PM also becomes smaller and smaller; finally when θ becomes O⁰; the point P will coincide with M. Consequently , we have PM = O and AP=AM Therefore, sin O⁰ = PM = O =O AP AP cos O⁰=AM =AP = 1 AP AP
  • 6.  Therefore,  sin 90⁰ = PM = PM = 1 AP PM and cos 90⁰ = AM = O = O AP AP Thus, we have sin 90⁰ = 1 and cos 90⁰ =O Remark:- It is evident from the above discussion that tan 90⁰ = PM =O is not defined . Similarly, sec 90⁰, cosec 0⁰, cot o⁰ are not defined .
  • 7. and , tan O⁰ = PM = O =O AP AP Thus, we have sin O⁰ = O, cos O⁰= 1 and tan O⁰ = O From ΔAMP, it is evident that as θ increase, line segment AM becomes smaller and smaller and finally when θ becomes 90⁰ the point M will coincide with A. Consequently, we have AM =O, AP =PM
  • 8.  Considered an equilateral triangle ABC with each side of length 2a. Since each angle of an equilateral triangle is of 60 ⁰. Therefore, each angle of ΔABC is of 60⁰. Let AD be perpendicular from A on BC. Since the triangle is equilateral the triangle is equilateral. Therefore, AD is the bisector of angle A and D is the mid-point of BC.
  • 9. Therefore, BD =DC=a and angle BAD =30⁰ Thus, in ΔABD , angle D is a right angle, hypotenuse AB =2a and BD = a. So, by Pythagoras theorem, we Have AB² = AD² + BD² (2a)²= AD² = a² AD²= 4a²- a² AD =√3a Trigonometric Ratios of 30⁰: In right Δ ADB, we have
  • 10.  Base =AD = √3a , perpendicular = BD = a, Hypotenuse = AB = 2a and angle DAB= 30⁰ sin 30⁰ =BD = a = 1 AB 2a 2 Therefore, cos 30⁰ = AD = √3a = √3 AB 2a 2 tan 30⁰= BD = a = 1 AD √3a √3 cosec 30⁰ = 1 =2 sin 30⁰ sec 30⁰ = 1 = 2 and cot30⁰ = √3 cos 30⁰ √3
  • 11.  Trigonometry ratios of 60⁰:  In right triangle ADB, we have Base = BD =a, Perpendicular =AD= √3a , Hypotenuse =AB = 2a and angle ABD = 60⁰ Therefore, Sin 60⁰= AD =√3a = √3, AB 2a 2 Cos 60⁰ = BD = a = 1 AB 2a 2 Tan 60⁰ = AD = √3a = √3 BD a Cosec, sec and cot are opposite of sin, cos, tan respectively.
  • 12.  Consider a right angle triangle ABC with right at B such that angle A= 45 °. Then, Angle A + Angle B + angle C =180° 45 ° + 90° + Angle C= 180 ° angle C = 45 ° Therefore, angle A = angle C Therefore, AB = BC
  • 13.  Let AB = BC =a. Then , by Pythagoras theorem , we have AC² = AB² = BC² AC² =a² +a² AC² = 2a² AC =√2a Thus, in we have triangle ABC, we have Angle = 45°, Base = AB =a, Perpendicular = BC =a, Hypotenuse = AC =√2a
  • 14.  Therefore sin 45= BC = a = 1 AC √2 √2 cos 45° = AB = a = 1 AC √2a √2 tan 45° = BC =a =1 , AB a cosec 45° = 1 = √2 sin 45° sec 45° = 1 = √2 and cos 45° cot 45° = 1 = 1 tan 45°
  • 16.  An equation involving trigonometric ratios of an angle θ (say)is said to be a Trigonometric identity if it is satisfied for all values of θ for which the given trigonometric ratios are defined.  For example, cos² θ – 1cos θ = cos θ ( cos θ – 1) 2 2 is a trigonometric identity, whereas cos θ ( cos θ – 1 ) = 0 2 is an equation.