text: # LAW OF SINES AND COSINES EXAMPLES
  
  ## About "Law of Sines and Cosines Examples"
  
  Law of Sines and Cosines Examples :
  
  Here we are going to see some example problems based on law of sines and cosines.
  
  ## Law of Sines and Cosines Examples - Examples
  
  Question 1 :
  
  In a triangle ABC, if cos C = sin A / 2 sin B, show that the triangle is isosceles.
  
  Solution :
  
  cos C = sin A / 2 sin B
  
  cos C (2 sin B)  =  sin A
  
  2 sin B cos C  =  sin A
  
  sin (B + C) + sin (B - C)  =  sin A
  
  A + B + C  =  180
  
  B + C  =  180 - A
  
  sin (180 - A) + sin (B - C)  =  sin A
  
  sin A + sin (B - C)  =  sin A
  
  sin (B - C)  =  0
  
  B - C  =  0
  
  B  =  C
  
  Hence the given triangle is isosceles triangle.
  
  Question 2 :
  
  In a triangle ABC, prove that sin B/sinC  =  (c − a cosB)/(b − a cosC)
  
  Solution :
  
  sin B/sinC  =  (c − a cosB)/(b − a cosC)
  
  Laws of sine :
  
  a/sin A  =  b/sin B  =  c/sin C  =  2R
  
  a  =  2R sin A, b = 2R sin B and c = 2R sin C
  
  R.H.S
  
  =  (c − a cosB)/(b − a cosC)
  
  =  (2R sin C - 2R sin A cos B) / (2R sin B - 2R sin A cos C)
  
  =  (sin C - sin A cos B) / (sin B - sin A cos C)
  
  =  (sin C - (1/2) [sin (A+B) + sin (A-B)] / (sin B - (1/2)[sin(A+C) sin (A-C)]
  
  =  (sin C - (1/2) [sin C + sin (A-B)] / (sin B - (1/2)[sinB sin (A-C)]
  
  =  (1/2) [sin C - sin (A-B)]/(1/2)[sin B - sin (A-C)]
  
  =  [sin (A+B) - sin (A-B)]/[sin (A+C) - sin (A-C)]
  
  =  (2 cos A sin B)/(2 cos A sin C)
  
  =  sin B / sin C
  
  Hence proved.
  
  Question 3 :
  
  In a triangle ABC, prove that a cosA + b cosB + c cosC = 2a sinB sinC.
  
  Solution :
  
  L.H.S :
  
  a cosA + b cosB + c cosC
  
  a/sin A  =  b/sin B  =  c/sin C  =  2R
  
  a = 2R sin A, b = 2R sin B  and c = 2R sin C
  
  =  2R sin A cos A + 2R sin B cos B + 2R sin C cos C
  
  =  R[sin 2A + sin 2B + sin 2C]
  
  =  R[2 sin (A + B) cos (A - B) + sin 2C]
  
  =  R[2 sin (180 - C) cos (A - B) + sin 2C]
  
  =  R[2 sin C cos (A - B) + 2 sin C cos C]
  
  =  2R sin C [cos (A - B) + cos C]
  
  =  2R sin C [cos (A-B) + cos (180-(A+B))]
  
  =  2R sin C [cos (A-B) - cos (A+B)]
  
  =  2R sin C [2 sin A sin B]
  
  Instead of 2R sin A, we may apply "a".
  
  =  2a sin A sin B sin C
  
  Hence proved.
  
  After having gone through the stuff given above, we hope that the students would have understood, "Law of Sines and Cosines Examples"
  
  Apart from the stuff given in "Law of Sines and Cosines Examples", if you need any other stuff in math, please use our google custom search here.
  
  Kindly mail your feedback to v4formath@gmail.com
  
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explanation: The document provides examples and solutions related to the Law of Sines and Cosines in trigonometry. It contains three questions, each with a detailed solution showing the steps to prove or solve the given problem.
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distilabel_metadata.content: # LAW OF SINES AND COSINES EXAMPLES
  
  ## About "Law of Sines and Cosines Examples"
  
  Law of Sines and Cosines Examples :
  
  Here we are going to see some example problems based on law of sines and cosines.
  
  ## Law of Sines and Cosines Examples - Examples
  
  Question 1 :
  
  In a triangle ABC, if cos C = sin A / 2 sin B, show that the triangle is isosceles.
  
  Solution :
  
  cos C = sin A / 2 sin B
  
  cos C (2 sin B)  =  sin A
  
  2 sin B cos C  =  sin A
  
  sin (B + C) + sin (B - C)  =  sin A
  
  A + B + C  =  180
  
  B + C  =  180 - A
  
  sin (180 - A) + sin (B - C)  =  sin A
  
  sin A + sin (B - C)  =  sin A
  
  sin (B - C)  =  0
  
  B - C  =  0
  
  B  =  C
  
  Hence the given triangle is isosceles triangle.
  
  Question 2 :
  
  In a triangle ABC, prove that sin B/sinC  =  (c − a cosB)/(b − a cosC)
  
  Solution :
  
  sin B/sinC  =  (c − a cosB)/(b − a cosC)
  
  Laws of sine :
  
  a/sin A  =  b/sin B  =  c/sin C  =  2R
  
  a  =  2R sin A, b = 2R sin B and c = 2R sin C
  
  R.H.S
  
  =  (c − a cosB)/(b − a cosC)
  
  =  (2R sin C - 2R sin A cos B) / (2R sin B - 2R sin A cos C)
  
  =  (sin C - sin A cos B) / (sin B - sin A cos C)
  
  =  (sin C - (1/2) [sin (A+B) + sin (A-B)] / (sin B - (1/2)[sin(A+C) sin (A-C)]
  
  =  (sin C - (1/2) [sin C + sin (A-B)] / (sin B - (1/2)[sinB sin (A-C)]
  
  =  (1/2) [sin C - sin (A-B)]/(1/2)[sin B - sin (A-C)]
  
  =  [sin (A+B) - sin (A-B)]/[sin (A+C) - sin (A-C)]
  
  =  (2 cos A sin B)/(2 cos A sin C)
  
  =  sin B / sin C
  
  Hence proved.
  
  Question 3 :
  
  In a triangle ABC, prove that a cosA + b cosB + c cosC = 2a sinB sinC.
  
  Solution :
  
  L.H.S :
  
  a cosA + b cosB + c cosC
  
  a/sin A  =  b/sin B  =  c/sin C  =  2R
  
  a = 2R sin A, b = 2R sin B  and c = 2R sin C
  
  =  2R sin A cos A + 2R sin B cos B + 2R sin C cos C
  
  =  R[sin 2A + sin 2B + sin 2C]
  
  =  R[2 sin (A + B) cos (A - B) + sin 2C]
  
  =  R[2 sin (180 - C) cos (A - B) + sin 2C]
  
  =  R[2 sin C cos (A - B) + 2 sin C cos C]
  
  =  2R sin C [cos (A - B) + cos C]
  
  =  2R sin C [cos (A-B) + cos (180-(A+B))]
  
  =  2R sin C [cos (A-B) - cos (A+B)]
  
  =  2R sin C [2 sin A sin B]
  
  Instead of 2R sin A, we may apply "a".
  
  =  2a sin A sin B sin C
  
  Hence proved.
  
  After having gone through the stuff given above, we hope that the students would have understood, "Law of Sines and Cosines Examples"
  
  Apart from the stuff given in "Law of Sines and Cosines Examples", if you need any other stuff in math, please use our google custom search here.
  
  Kindly mail your feedback to v4formath@gmail.com
  
  ## Recent Articles
  
  1. ### SAT Math Videos
  
  May 22, 24 06:32 AM
  
  SAT Math Videos (Part 1 - No Calculator)
  
  2. ### Simplifying Algebraic Expressions with Fractional Coefficients
  
  May 17, 24 08:12 AM
  
  Simplifying Algebraic Expressions with Fractional Coefficients
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document provides examples and solutions related to the Law of Sines and Cosines in trigonometry. It contains three questions, each with a detailed solution showing the steps to prove or solve the given problem.
  
  Label: multiple-question-and-answer